Chapter 15 Acids and Bases Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Chapter 15 Acids and Bases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

Stomach Acid & Heartburn The cells that line your stomach produce hydrochloric acid to kill unwanted bacteria to help break down food to activate enzymes that break down food If the stomach acid backs up into your esophagus, it irritates those tissues, resulting in heartburn acid reflux Tro, Chemistry: A Molecular Approach, 2/e

Curing Heartburn Mild cases of heartburn can be cured by neutralizing the acid in the esophagus swallowing saliva, which contains bicarbonate ion taking antacids that contain hydroxide ions and/or carbonate ions Tro, Chemistry: A Molecular Approach, 2/e

GERD Chronic heartburn is a problem for some people GERD (gastroesophageal reflux disease) is chronic leaking of stomach acid into the esophagus In people with GERD, the muscles separating the stomach from the esophagus do not close tightly, allowing stomach acid to leak into the esophagus Physicians diagnose GERD by attaching a pH sensor to the esophagus to measure the acidity levels of the fluids over time Tro, Chemistry: A Molecular Approach, 2/e

Properties of Acids Sour taste React with “active” metals i.e., Al, Zn, Fe, but not Cu, Ag, or Au 2 Al + 6 HCl ® 2 AlCl3 + 3 H2 corrosive React with carbonates, producing CO2 marble, baking soda, chalk, limestone CaCO3 + 2 HCl ® CaCl2 + CO2 + H2O Change color of vegetable dyes blue litmus turns red React with bases to form ionic salts Tro, Chemistry: A Molecular Approach, 2/e

Common Acids Tro, Chemistry: A Molecular Approach, 2/e

Structures of Acids Binary acids have acid hydrogens attached to a nonmetal atom HCl, HF Tro, Chemistry: A Molecular Approach, 2/e

Structure of Acids Oxy acids have acid hydrogens attached to an oxygen atom H2SO4, HNO3 Tro, Chemistry: A Molecular Approach, 2/e

Structure of Acids Carboxylic acids have COOH group HC2H3O2, H3C6H5O7 Only the first H in the formula is acidic the H is on the COOH Tro, Chemistry: A Molecular Approach, 2/e

Properties of Bases Also known as alkalis Taste bitter alkaloids = plant product that is alkaline often poisonous Solutions feel slippery Change color of vegetable dyes different color than acid red litmus turns blue React with acids to form ionic salts neutralization Tro, Chemistry: A Molecular Approach, 2/e

Common Bases Tro, Chemistry: A Molecular Approach, 2/e

Structure of Bases Most ionic bases contain OH− ions NaOH, Ca(OH)2 Some contain CO32− ions CaCO3 NaHCO3 Molecular bases contain structures that react with H+ mostly amine groups Tro, Chemistry: A Molecular Approach, 2/e

Indicators Chemicals that change color depending on the solution’s acidity or basicity Many vegetable dyes are indicators anthocyanins Litmus from Spanish moss red in acid, blue in base Phenolphthalein found in laxatives red in base, colorless in acid Tro, Chemistry: A Molecular Approach, 2/e

Arrhenius Theory Bases dissociate in water to produce OH− ions and cations ionic substances dissociate in water NaOH(aq) → Na+(aq) + OH−(aq) Acids ionize in water to produce H+ ions and anions because molecular acids are not made of ions, they cannot dissociate they must be pulled apart, or ionized, by the water HCl(aq) → H+(aq) + Cl−(aq) in formula, ionizable H written in front HC2H3O2(aq) → H+(aq) + C2H3O2−(aq) Tro, Chemistry: A Molecular Approach, 2/e

Arrhenius Theory HCl ionizes in water, producing H+ and Cl– ions NaOH dissociates in water, producing Na+ and OH– ions Tro, Chemistry: A Molecular Approach, 2/e

Hydronium Ion The H+ ions produced by the acid are so reactive they cannot exist in water H+ ions are protons!! Instead, they react with water molecules to produce complex ions, mainly hydronium ion, H3O+ H+ + H2O  H3O+ there are also minor amounts of H+ with multiple water molecules, H(H2O)n+ Tro, Chemistry: A Molecular Approach, 2/e

Arrhenius Acid–Base Reactions The H+ from the acid combines with the OH− from the base to make a molecule of H2O it is often helpful to think of H2O as H-OH The cation from the base combines with the anion from the acid to make a salt acid + base → salt + water HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Tro, Chemistry: A Molecular Approach, 2/e

Problems with Arrhenius Theory Does not explain why molecular substances, such as NH3, dissolve in water to form basic solutions – even though they do not contain OH– ions Does not explain how some ionic compounds, such as Na2CO3 or Na2O, dissolve in water to form basic solutions – even though they do not contain OH– ions Does not explain why molecular substances, such as CO2, dissolve in water to form acidic solutions – even though they do not contain H+ ions Does not explain acid–base reactions that take place outside aqueous solution Tro, Chemistry: A Molecular Approach, 2/e

Another Definition: Brønsted-Lowry Acid-Base Theory Brønsted and Lowry redefined acids and bases based on what happens in a reaction. Any reaction that involves H+ being transferred from one molecule to another is an acid–base reaction regardless of whether it occurs in aqueous solution, or if there is OH− present All reactions that fit the Arrhenius definition also fit the Brønsted-Lowry definition, but many more do as well Tro, Chemistry: A Molecular Approach, 2/e

Brønsted-Lowry Theory In a Brønsted-Lowry acid–base reaction, an H+ is transferred The acid is an H donor The base is an H acceptor base structure must contain an atom with an unshared pair of electrons In a Brønsted-Lowry acid-base reaction, the acid molecule gives an H+ to the base molecule H–A + :B  :A– + H–B+ Tro, Chemistry: A Molecular Approach, 2/e

Brønsted-Lowry Acids HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq) Brønsted-Lowry acids are H+ donors any material that has H can potentially be a Brønsted-Lowry acid because of the molecular structure, often one H in the molecule is easier to transfer than others When HCl dissolves in water, the HCl is the acid because HCl transfers an H+ to H2O, forming H3O+ ions water acts as base, accepting H+ HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq) acid base Tro, Chemistry: A Molecular Approach, 2/e

Brønsted-Lowry Bases NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq) base acid Brønsted-Lowry bases are H+ acceptors any material that has atoms with lone pairs can potentially be a Brønsted-Lowry base because of the molecular structure, often one atom in the molecule is more willing to accept H+ transfer than others When NH3 dissolves in water, the NH3(aq) is the base because NH3 accepts an H+ from H2O, forming OH–(aq) water acts as acid, donating H+ NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq) base acid Tro, Chemistry: A Molecular Approach, 2/e

A Warning! Because chemists know common bonding patterns, we often do not draw lone pair electrons on our structures. You need to be able to recognize when an atom in a molecule has lone pair electrons and when it doesn’t! Tro, Chemistry: A Molecular Approach, 2/e

Practice – Draw structures of the following that include lone pairs of electrons HClO HCO3− Tro, Chemistry: A Molecular Approach, 2/e

Amphoteric Substances Amphoteric substances can act as either an acid or a base because they have both a transferable H and an atom with lone pair electrons Water acts as base, accepting H+ from HCl HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq) Water acts as acid, donating H+ to NH3 NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq) Tro, Chemistry: A Molecular Approach, 2/e

Brønsted-Lowry Acid-Base Reactions One of the advantages of Brønsted-Lowry theory is that it allows reactions to be reversible H–A + :B  :A– + H–B+ The original base has an extra H+ after the reaction, so it will act as an acid in the reverse process And the original acid has a lone pair of electrons after the reaction – so it will act as a base in the reverse process :A– + H–B+  H–A + :B Tro, Chemistry: A Molecular Approach, 2/e

Conjugate Pairs In a Brønsted-Lowry acid–base reaction, the original base becomes an acid in the reverse reaction, and the original acid becomes a base in the reverse process Each reactant and the product it becomes is called a conjugate pair The original base becomes its conjugate acid; and the original acid becomes its conjugate base Tro, Chemistry: A Molecular Approach, 2/e

Brønsted-Lowry Acid–Base Reactions H–A + :B  :A– + H–B+ acid base conjugate conjugate base acid HCHO2 + H2O  CHO2– + H3O+ acid base conjugate conjugate base acid H2O + NH3  HO– + NH4+ acid base conjugate conjugate base acid Tro, Chemistry: A Molecular Approach, 2/e

Conjugate Pairs In the reaction H2O + NH3  HO– + NH4+ H2O and HO– constitute an acid/conjugate base pair NH3 and NH4+ constitute a base/conjugate acid pair Tro, Chemistry: A Molecular Approach, 2/e

Practice – Write the formula for the conjugate acid of the following H2O H3O+ NH3 NH4+ CO32− HCO3− H2PO41− H3PO4 H2O NH3 CO32− H2PO41− Tro, Chemistry: A Molecular Approach, 2/e

Practice – Write the formula for the conjugate base of the following H2O HO− NH3 NH2− CO32− because CO32− does not have an H, it cannot be an acid H2PO41− HPO42− H2O NH3 CO32− H2PO41− Tro, Chemistry: A Molecular Approach, 2/e

Example 15.1a: Identify the Brønsted-Lowry acids and bases, and their conjugates, in the reaction H2SO4 + H2O  HSO4– + H3O+ When the H2SO4 becomes HSO4, it loses an H+ so H2SO4 must be the acid and HSO4 its conjugate base When the H2O becomes H3O+, it accepts an H+ so H2O must be the base and H3O+ its conjugate acid H2SO4 + H2O  HSO4– + H3O+ acid base conjugate conjugate base acid Tro, Chemistry: A Molecular Approach, 2/e

Example 15.1b: Identify the Brønsted-Lowry acids and bases and their conjugates in the reaction HCO3– + H2O  H2CO3 + HO– When the HCO3 becomes H2CO3, it accepts an H+ so HCO3 must be the base and H2CO3 its conjugate acid When the H2O becomes OH, it donats an H+ so H2O must be the acid and OH its conjugate base HCO3– + H2O  H2CO3 + HO– base acid conjugate conjugate acid base Tro, Chemistry: A Molecular Approach, 2/e

Practice – Identify the Brønsted-Lowry acid, base, conjugate acid, and conjugate base in the following reaction HSO4−(aq) + HCO3−(aq)  SO42−(aq) + H2CO3(aq) Conjugate Base Conjugate Acid Acid Base Tro, Chemistry: A Molecular Approach, 2/e

HBr + H2O  Br− + H3O+ HBr HSO4− HSO4− + H2O  SO42− + H3O+ Practice—Write the equations for the following reacting with water and acting as a monoprotic acid & label the conjugate acid and base HBr + H2O  Br− + H3O+ Acid Base Conj. Conj. base acid HBr HSO4− HSO4− + H2O  SO42− + H3O+ Acid Base Conj. Conj. base acid Tro, Chemistry: A Molecular Approach, 2/e

Practice—Write the equations for the following reacting with water and acting as a monoprotic-accepting base and label the conjugate acid and base I− CO32− I− + H2O  HI + OH− Base Acid Conj. Conj. acid base CO32− + H2O  HCO3− + OH− Base Acid Conj. Conj. acid base Tro, Chemistry: A Molecular Approach, 2/e

Comparing Arrhenius Theory and Brønsted-Lowry Theory HCl(aq)  H+(aq) + Cl−(aq) HF(aq)  H+(aq) + F−(aq) NaOH(aq)  Na+(aq) + OH−(aq) NH4OH(aq)  NH4+(aq) + OH−(aq) Brønsted–Lowry theory HCl(aq) + H2O(l)  Cl−(aq) + H3O+(aq) HF(aq) + H2O(l)  F−(aq) + H3O+(aq) NaOH(aq)  Na+(aq) + OH−(aq) NH3(aq) + H2O(l)  NH4+(aq) + OH−(aq) Tro, Chemistry: A Molecular Approach, 2/e

Arrow Conventions  in these notes Chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions A single arrow indicates all the reactant molecules are converted to product molecules at the end A double arrow indicates the reaction stops when only some of the reactant molecules have been converted into products  in these notes Tro, Chemistry: A Molecular Approach, 2/e

Strong or Weak A strong acid is a strong electrolyte practically all the acid molecules ionize, → A strong base is a strong electrolyte practically all the base molecules form OH– ions, either through dissociation or reaction with water, → A weak acid is a weak electrolyte only a small percentage of the molecules ionize,  A weak base is a weak electrolyte only a small percentage of the base molecules form OH– ions, either through dissociation or reaction with water,  Tro, Chemistry: A Molecular Approach, 2/e

Strong Acids The stronger the acid, the more willing it is to donate H we use water as the standard base to donate H to Strong acids donate practically all their H’s 100% ionized in water strong electrolyte [H3O+] = [strong acid] [X] means the molarity of X HCl ® H+ + Cl− HCl + H2O ® H3O+ + Cl− 0.10 M HCl = 0.10 M H3O+ Tro, Chemistry: A Molecular Approach, 2/e

Weak Acids Weak acids donate a small fraction of their H’s most of the weak acid molecules do not donate H to water much less than 1% ionized in water [H3O+] << [weak acid] HF Û H+ + F− HF + H2O Û H3O+ + F− 0.10 M HF ≠ 0.10 M H3O+ Tro, Chemistry: A Molecular Approach, 2/e

Strong & Weak Acids Tro, Chemistry: A Molecular Approach, 2/e

Tro, Chemistry: A Molecular Approach, 2/e

Strengths of Acids & Bases Commonly, acid or base strength is measured by determining the equilibrium constant of a substance’s reaction with water HAcid + H2O  Acid− + H3O+ Base: + H2O  HBase+ + OH− The farther the equilibrium position lies toward the products, the stronger the acid or base The position of equilibrium depends on the strength of attraction between the base form and the H+ stronger attraction means stronger base or weaker acid Tro, Chemistry: A Molecular Approach, 2/e

General Trends in Acidity The stronger an acid is at donating H, the weaker the conjugate base is at accepting H Higher oxidation number = stronger oxyacid H2SO4 > H2SO3; HNO3 > HNO2 Cation stronger acid than neutral molecule; neutral stronger acid than anion H3O+ > H2O > OH−; NH4+ > NH3 > NH2− trend in base strength opposite Tro, Chemistry: A Molecular Approach, 2/e

Acid Ionization Constant, Ka Acid strength measured by the size of the equilibrium constant when reacts with H2O HAcid + H2O  Acid− + H3O+ The equilibrium constant for this reaction is called the acid ionization constant, Ka larger Ka = stronger acid Tro, Chemistry: A Molecular Approach, 2/e

Tro, Chemistry: A Molecular Approach, 2/e

Autoionization of Water Water is actually an extremely weak electrolyte therefore there must be a few ions present About 2 out of every 1 billion water molecules form ions through a process called autoionization H2O Û H+ + OH– H2O + H2O Û H3O+ + OH– All aqueous solutions contain both H3O+ and OH– the concentration of H3O+ and OH– are equal in water [H3O+] = [OH–] = 10−7M @ 25 °C Tro, Chemistry: A Molecular Approach, 2/e

Ion Product of Water The product of the H3O+ and OH– concentrations is always the same number The number is called the Ion Product of Water and has the symbol Kw aka the Dissociation Constant of Water [H3O+] x [OH–] = Kw = 1.00 x 10−14 @ 25 °C if you measure one of the concentrations, you can calculate the other As [H3O+] increases the [OH–] must decrease so the product stays constant inversely proportional Tro, Chemistry: A Molecular Approach, 2/e

Acidic and Basic Solutions All aqueous solutions contain both H3O+ and OH– ions Neutral solutions have equal [H3O+] and [OH–] [H3O+] = [OH–] = 1.00 x 10−7 Acidic solutions have a larger [H3O+] than [OH–] [H3O+] > 1.00 x 10−7; [OH–] < 1.00 x 10−7 Basic solutions have a larger [OH–] than [H3O+] [H3O+] < 1.00 x 10−7; [OH–] > 1.00 x 10−7 Tro, Chemistry: A Molecular Approach, 2/e

Practice – Complete the table [H+] vs. [OH−] Tro, Chemistry: A Molecular Approach, 2/e

Practice – Complete the table [H+] vs. [OH−] Acid Base [H+] 100 10−1 10−3 10−5 10−7 10−9 10−11 10−13 10−14 OH− H+ [OH−]10−14 10−13 10−11 10−9 10−7 10−5 10−3 10−1 100 Even though it may look like it, neither H+ nor OH− will ever be 0 The sizes of the H+ and OH− are not to scale because the divisions are powers of 10 rather than units Tro, Chemistry: A Molecular Approach, 2/e

Example 15. 2b: Calculate the [OH] at 25 °C when the [H3O+] = 1 Example 15.2b: Calculate the [OH] at 25 °C when the [H3O+] = 1.5 x 10−9 M, and determine if the solution is acidic, basic, or neutral Given: Find: [H3O+] = 1.5 x 10−9 M [OH] Conceptual Plan: Relationships: [H3O+] [OH] Solution: Check: the units are correct; the fact that the [H3O+] < [OH] means the solution is basic Tro, Chemistry: A Molecular Approach, 2/e

Practice – Determine the [H3O+] when the [OH−] = 2.5 x 10−9 M Tro, Chemistry: A Molecular Approach, 2/e

Practice – Determine the [H3O+] when the [OH−] = 2.5 x 10−9 M Given: Find: [OH] = 2.5 x 10−9 M [H3O+] Conceptual Plan: Relationships: [OH] [H3O+] Solution: Check: the units are correct; the fact that the [H3O+] > [OH] means the solution is acidic Tro, Chemistry: A Molecular Approach, 2/e

Measuring Acidity: pH The acidity or basicity of a solution is often expressed as pH pH = −log[H3O+] exponent on 10 with a positive sign pHwater = −log[10−7] = 7 need to know the [H3O+] concentration to find pH pH < 7 is acidic; pH > 7 is basic, pH = 7 is neutral [H3O+] = 10−pH Tro, Chemistry: A Molecular Approach, 2/e

Sig. Figs. & Logs When you take the log of a number written in scientific notation, the digit(s) before the decimal point come from the exponent on 10, and the digits after the decimal point come from the decimal part of the number log(2.0 x 106) = log(106) + log(2.0) = 6 + 0.30303… = 6.30303... Because the part of the scientific notation number that determines the significant figures is the decimal part, the sig figs are the digits after the decimal point in the log log(2.0 x 106) = 6.30 Tro, Chemistry: A Molecular Approach, 2/e

What Does the pH Number Imply? The lower the pH, the more acidic the solution; the higher the pH, the more basic the solution 1 pH unit corresponds to a factor of 10 difference in acidity Normal range of pH is 0 to 14 pH 0 is [H3O+] = 1 M, pH 14 is [OH–] = 1 M pH can be negative (very acidic) or larger than 14 (very alkaline) Tro, Chemistry: A Molecular Approach, 2/e

Example 15. 3b: Calculate the pH at 25 °C when the [OH] = 1 Example 15.3b: Calculate the pH at 25 °C when the [OH] = 1.3 x 10−2 M, and determine if the solution is acidic, basic, or neutral Given: Find: [OH] = 1.3 x 10−2 M pH Conceptual Plan: Relationships: [H3O+] [OH] pH Solution: Check: pH is unitless; the fact that the pH > 7 means the solution is basic Tro, Chemistry: A Molecular Approach, 2/e

Practice – Determine the pH @ 25 ºC of a solution that has [OH−] = 2 Practice – Determine the pH @ 25 ºC of a solution that has [OH−] = 2.5 x 10−9 M Tro, Chemistry: A Molecular Approach, 2/e

Practice – Determine the pH @ 25 ºC of a solution that has [OH−] = 2 Practice – Determine the pH @ 25 ºC of a solution that has [OH−] = 2.5 x 10−9 M Given: Find: [OH] = 2.5 x 10−9 M pH Conceptual Plan: Relationships: [H3O+] [OH] pH Solution: Check: pH is unitless; the fact that the pH < 7 means the solution is acidic Tro, Chemistry: A Molecular Approach, 2/e

Practice – Determine the [OH−] of a solution with a pH of 5.40 Tro, Chemistry: A Molecular Approach, 2/e

Practice – Determine the [OH−] of a solution with a pH of 5.40 Given: Find: pH = 5.40 [OH−], M Conceptual Plan: Relationships: [H3O+] pH [OH] Solution: Check: because the pH < 7, [OH−] should be less than 1 x 10−7; and it is Tro, Chemistry: A Molecular Approach, 2/e

pOH Another way of expressing the acidity/basicity of a solution is pOH pOH = −log[OH], [OH] = 10−pOH pOHwater = −log[10−7] = 7 need to know the [OH] concentration to find pOH pOH < 7 is basic; pOH > 7 is acidic, pOH = 7 is neutral pH + pOH = 14.0 Tro, Chemistry: A Molecular Approach, 2/e

pH and pOH Practice – Complete the table Tro, Chemistry: A Molecular Approach, 2/e

pH and pOH Practice – Complete the table Acid Base pH 0 1 3 5 7 9 11 13 14 [H+] 100 10−1 10−3 10−5 10−7 10−9 10−11 10−13 10−14 OH− H+ [OH−]10−14 10−13 10−11 10−9 10−7 10−5 10−3 10−1 100 pOH 14 13 11 9 7 5 3 1 0 Tro, Chemistry: A Molecular Approach, 2/e

Relationship between pH and pOH pH + pOH = 14.00 at 25 °C you can use pOH to find pH of a solution Tro, Chemistry: A Molecular Approach, 2/e

Example: Calculate the pH at 25 °C when the [OH] = 1 Example: Calculate the pH at 25 °C when the [OH] = 1.3 x 10−2 M, and determine if the solution is acidic, basic, or neutral Given: Find: [OH] = 1.3 x 10−2 M pH Conceptual Plan: Relationships: pOH [OH] pH Solution: Check: pH is unitless; the fact that the pH > 7 means the solution is basic Tro, Chemistry: A Molecular Approach, 2/e

Practice – Determine the pOH @ 25 ºC of a solution that has [H3O+] = 2 Practice – Determine the pOH @ 25 ºC of a solution that has [H3O+] = 2.5 x 10−9 M Tro, Chemistry: A Molecular Approach, 2/e

Practice – Determine the pOH @ 25 ºC of a solution that has [H3O+] = 2 Practice – Determine the pOH @ 25 ºC of a solution that has [H3O+] = 2.5 x 10−9 M Given: Find: [H3O+] = 2.5 x 10−9 M pOH Conceptual Plan: Relationships: pH [H3O+] pOH Solution: Check: pH is unitless; the fact that the pH < 7 means the solution is acidic Tro, Chemistry: A Molecular Approach, 2/e

pK A way of expressing the strength of an acid or base is pK pKa = −log(Ka), Ka = 10−pKa pKb = −log(Kb), Kb = 10−pKb The stronger the acid, the smaller the pKa larger Ka = smaller pKa because it is the –log The stronger the base, the smaller the pKb larger Kb = smaller pKb Tro, Chemistry: A Molecular Approach, 2/e

[H3O+] and [OH−] in a Strong Acid or Strong Base Solution There are two sources of H3O+ in an aqueous solution of a strong acid – the acid and the water There are two sources of OH− in an aqueous solution of a strong acid – the base and the water For a strong acid or base, the contribution of the water to the total [H3O+] or [OH−] is negligible the [H3O+]acid shifts the Kw equilibrium so far that [H3O+]water is too small to be significant except in very dilute solutions, generally < 1 x 10−4 M Tro, Chemistry: A Molecular Approach, 2/e

Finding pH of a Strong Acid or Strong Base Solution For a monoprotic strong acid [H3O+] = [HAcid] for polyprotic acids, the other ionizations can generally be ignored for H2SO4, the second ionization cannot be ignored 0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00 For a strong ionic base, [OH−] = (number OH−)x[Base] for molecular bases with multiple lone pairs available, only one lone pair accepts an H, the other reactions can generally be ignored 0.10 M Ca(OH)2 has [OH−] = 0.20 M and pH = 13.30 Tro, Chemistry: A Molecular Approach, 2/e

Finding the pH of a Weak Acid There are also two sources of H3O+ in an aqueous solution of a weak acid – the acid and the water However, finding the [H3O+] is complicated by the fact that the acid only undergoes partial ionization Calculating the [H3O+] requires solving an equilibrium problem for the reaction that defines the acidity of the acid HAcid + H2O  Acid + H3O+ Tro, Chemistry: A Molecular Approach, 2/e

Example 15.6: Find the pH of 0.200 M HNO2(aq) solution @ 25 °C HNO2 + H2O  NO2 + H3O+ write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H3O+] from water is ≈ 0 [HNO2] [NO2−] [H3O+] initial 0.200 ≈ 0 change equilibrium [HNO2] [NO2−] [H3O+] initial change equilibrium because no products initially, Qc = 0, and the reaction is proceeding forward Tro, Chemistry: A Molecular Approach, 2/e

Example 15.6: Find the pH of 0.200 M HNO2(aq) solution @ 25 °C represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HNO2] [NO2−] [H3O+] initial 0.200 change equilibrium x +x +x x x 0.200 x HNO2 + H2O(l)  NO2−(aq) + H3O+(aq) Tro, Chemistry: A Molecular Approach, 2/e

Example 15.6: Find the pH of 0.200 M HNO2(aq) solution @ 25 °C Ka for HNO2 = 4.6 x 10−4 determine the value of Ka from Table 15.5 because Ka is very small, approximate the [HNO2]eq = [HNO2]init and solve for x [HNO2] [NO2−] [H3O+] initial 0.200 ≈ 0 change −x +x equilibrium x 0.200 x Tro, Chemistry: A Molecular Approach, 2/e

Example 15.6: Find the pH of 0.200 M HNO2(aq) solution @ 25 °C Ka for HNO2 = 4.6 x 10−4 check if the approximation is valid by seeing if x < 5% of [HNO2]init [HNO2] [NO2−] [H3O+] initial 0.200 ≈ 0 change −x +x equilibrium x x = 9.6 x 10−3 the approximation is valid Tro, Chemistry: A Molecular Approach, 2/e

Example 15.6: Find the pH of 0.200 M HNO2(aq) solution @ 25 °C Ka for HNO2 = 4.6 x 10−4 substitute x into the equilibrium concentration definitions and solve [HNO2] [NO2−] [H3O+] initial 0.200 ≈ 0 change −x +x equilibrium 0.190 0.0096 [HNO2] [NO2−] [H3O+] initial 0.200 ≈ 0 change −x +x equilibrium 0.200−x x x = 9.6 x 10−3 Tro, Chemistry: A Molecular Approach, 2/e

Example 15.6: Find the pH of 0.200 M HNO2(aq) solution @ 25 °C Ka for HNO2 = 4.6 x 10−4 substitute [H3O+] into the formula for pH and solve [HNO2] [NO2−] [H3O+] initial 0.200 ≈ 0 change −x +x equilibrium 0.190 0.0096 Tro, Chemistry: A Molecular Approach, 2/e

Example 15.6: Find the pH of 0.200 M HNO2(aq) solution @ 25 °C Ka for HNO2 = 4.6 x 10−4 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HNO2] [NO2−] [H3O+] initial 0.200 ≈ 0 change −x +x equilibrium 0.190 0.0096 though not exact, the answer is reasonably close Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0 Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? (Ka = 1.4 x 10−5 @ 25 °C) Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0 Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H3O+] from water is ≈ 0 HC6H4NO2 + H2O  C6H4NO2 + H3O+ [HA] [A−] [H3O+] initial 0.012 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0 Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression HC6H4NO2 + H2O  C6H4NO2 + H3O+ [HA] [A−] [H3O+] initial 0.012 change equilibrium x +x +x x x 0.012 x Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0 Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 10−5 @ 25 °C HC6H4NO2 + H2O  C6H4NO2 + H3O+ determine the value of Ka because Ka is very small, approximate the [HA]eq = [HA]init and solve for x [HA] [A−] [H3O+] initial 0.012 ≈ 0 change −x +x equilibrium x 0.012 x Tro, Chemistry: A Molecular Approach, 2/e

the approximation is valid Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 10−5 @ 25 °C Ka for HC6H4NO2 = 1.4 x 10−5 check if the approximation is valid by seeing if x < 5% of [HC6H4NO2]init [HA] [A−] [H3O+] initial 0.012 ≈ 0 change −x +x equilibrium x x = 4.1 x 10−4 the approximation is valid Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0 Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 10−5 @ 25 °C substitute x into the equilibrium concentration definitions and solve [HA] [A−] [H3O+] initial 0.012 ≈ 0 change −x +x equilibrium 0.012−x x x = 4.1 x 10−4 Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0 Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 10−5 @ 25 °C substitute [H3O+] into the formula for pH and solve [HA] [A−] [H3O+] initial 0.012 ≈ 0 change −x +x equilibrium 0.00041 Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0 Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 10−5 @ 25 °C check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HA] [A−] [H3O+] initial 0.012 ≈ 0 change −x +x equilibrium 0.00041 the values match Tro, Chemistry: A Molecular Approach, 2/e

Example 15.7: Find the pH of 0.100 M HClO2(aq) solution @ 25 °C write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H3O+] from water is ≈ 0 HClO2 + H2O  ClO2 + H3O+ [HClO2] [ClO2−] [H3O+] initial 0.100 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

Example 15.7: Find the pH of 0.100 M HClO2(aq) solution @ 25 °C represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HClO2] [ClO2−] [H3O+] initial 0.100 ≈ 0 change −x +x equilibrium 0.100−x x Tro, Chemistry: A Molecular Approach, 2/e

Example 15.7: Find the pH of 0.100 M HClO2(aq) solution @ 25 °C Ka for HClO2 = 1.1 x 10−2 determine the value of Ka from Table 15.5 because Ka is very small, approximate the [HClO2]eq = [HClO2]init and solve for x [HClO2] [ClO2−] [H3O+] initial 0.100 ≈ 0 change −x +x equilibrium 0.100−x x Tro, Chemistry: A Molecular Approach, 2/e

Example 15.7: Find the pH of 0.100 M HClO2(aq) solution @ 25 °C Ka for HClO2 = 1.1 x 10−2 check if the approximation is valid by seeing if x < 5% of [HNO2]init [HClO2] [ClO2−] [H3O+] initial 0.100 ≈ 0 change −x +x equilibrium 0.100−x x x = 3.3 x 10−2 the approximation is invalid Tro, Chemistry: A Molecular Approach, 2/e

Example 15.7: Find the pH of 0.100 M HClO2(aq) solution @ 25 °C Ka for HClO2 = 1.1 x 10−2 if the approximation is invalid, solve for x using the quadratic formula Tro, Chemistry: A Molecular Approach, 2/e

Example 15.7: Find the pH of 0.100 M HClO2(aq) solution @ 25 °C Ka for HClO2 = 1.1 x 10−2 substitute x into the equilibrium concentration definitions and solve [HClO2] [ClO2−] [H3O+] initial 0.100 ≈ 0 change −x +x equilibrium 0.100−x x [HClO2] [ClO2−] [H3O+] initial 0.100 ≈ 0 change −x +x equilibrium 0.072 0.028 x = 0.028 Tro, Chemistry: A Molecular Approach, 2/e

Example 15.7: Find the pH of 0.100 M HClO2(aq) solution @ 25 °C Ka for HClO2 = 1.1 x 10−2 substitute [H3O+] into the formula for pH and solve [HClO2] [ClO2−] [H3O+] initial 0.100 ≈ 0 change −x +x equilibrium 0.072 0.028 Tro, Chemistry: A Molecular Approach, 2/e

Example 15.7: Find the pH of 0.100 M HClO2(aq) solution @ 25°C Ka for HClO2 = 1.1 x 10−2 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HClO2] [ClO2-] [H3O+] initial 0.100 ≈ 0 change −x +x equilibrium 0.072 0.028 the answer matches Tro, Chemistry: A Molecular Approach, 2/e

Example 15. 8: What is the Ka of a weak acid if a 0 Example 15.8: What is the Ka of a weak acid if a 0.100 M solution has a pH of 4.25? use the pH to find the equilibrium [H3O+] write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations and [H3O+]equil HA + H2O  A + H3O+ [HA] [A−] [H3O+] initial change equilibrium [HA] [A−] [H3O+] initial 0.100 ≈ 0 change equilibrium 5.6E-05 Tro, Chemistry: A Molecular Approach, 2/e

Example 15. 8: What is the Ka of a weak acid if a 0 Example 15.8: What is the Ka of a weak acid if a 0.100 M solution has a pH of 4.25? HA + H2O  A + H3O+ fill in the rest of the table using the [H3O+] as a guide if the difference is insignificant, [HA]equil = [HA]initial substitute into the Ka expression and compute Ka [HA] [A−] [H3O+] initial 0.100 change equilibrium −5.6E-05 +5.6E-05 +5.6E-05 0.100  5.6E-05 0.100 5.6E-05 5.6E-05 Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the Ka of nicotinic acid, HC6H4NO2, if a 0 Practice – What is the Ka of nicotinic acid, HC6H4NO2, if a 0.012 M solution of nicotinic acid has a pH of 3.40? Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the Ka of nicotinic acid, HC6H4NO2, if a 0 Practice – What is the Ka of nicotinic acid, HC6H4NO2, if a 0.012 M solution of nicotinic acid has a pH of 3.40? use the pH to find the equilibrium [H3O+] write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations and [H3O+]equil HA + H2O  A + H3O+ [HA] [A−] [H3O+] initial change equilibrium [HA] [A−] [H3O+] initial 0.012 ≈ 0 change equilibrium 4.0E-04 Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the Ka of nicotinic acid, HC6H4NO2, if a 0 Practice – What is the Ka of nicotinic acid, HC6H4NO2, if a 0.012 M solution of nicotinic acid has a pH of 3.40? fill in the rest of the table using the [H3O+] as a guide if the difference is insignificant, [HA]equil = [HA]initial substitute into the Ka expression and compute Ka HA + H2O  A + H3O+ [HA] [A−] [H3O+] initial 0.012 change equilibrium −4.0E-04 +4.0E-04 +4.0E-04 0.012  4.0E-04 4.0E-04 4.0E-04 0.012 Tro, Chemistry: A Molecular Approach, 2/e

Polyprotic Acids Acid molecules often have more than one ionizable H – these are called polyprotic acids the ionizable H’s may have different acid strengths or be equal 1 H = monoprotic, 2 H = diprotic, 3 H = triprotic HCl = monoprotic, H2SO4 = diprotic, H3PO4 = triprotic Polyprotic acids ionize in steps each ionizable H is removed sequentially Removing of the first H automatically makes removal of the second H harder H2SO4 is a stronger acid than HSO4 Tro, Chemistry: A Molecular Approach, 2/e

Percent Ionization Because [ionized acid]equil = [H3O+]equil Another way to measure the strength of an acid is to determine the percentage of acid molecules that ionize when dissolved in water – this is called the percent ionization the higher the percent ionization, the stronger the acid Because [ionized acid]equil = [H3O+]equil Tro, Chemistry: A Molecular Approach, 2/e

Example 15.9: What is the percent ionization of a 2.5 M HNO2 solution? write the reaction for the acid with water construct an ICE table for the reaction enter the Initial Concentrations define the change in concentration in terms of x sum the columns to define the equilibrium concentrations HNO2 + H2O  NO2 + H3O+ [HNO2] [NO2−] [H3O+] initial 2.5 ≈ 0 change equilibrium [HNO2] [NO2−] [H3O+] initial 2.5 ≈0 change equilibrium x +x +x 2.5  x x x Tro, Chemistry: A Molecular Approach, 2/e

Example 15.9: What is the percent ionization of a 2.5 M HNO2 solution? Ka for HNO2 = 4.6 x 10−4 determine the value of Ka from Table 15.5 because Ka is very small, approximate the [HNO2]eq = [HNO2]init and solve for x [HNO2] [NO2−] [H3O+] initial 2.5 ≈ 0 change −x +x equilibrium 2.5−x ≈2.5 x Tro, Chemistry: A Molecular Approach, 2/e

Example 15.9: What is the percent ionization of a 2.5 M HNO2 solution? HNO2 + H2O  NO2 + H3O+ substitute x into the equilibrium concentration definitions and solve [HNO2] [NO2−] [H3O+] initial 2.5 ≈ 0 change −x +x equilibrium 0.034 2.5  x x x x = 3.4 x 10−2 Tro, Chemistry: A Molecular Approach, 2/e

Example 15.9: What is the percent ionization of a 2.5 M HNO2 solution? HNO2 + H2O  NO2 + H3O+ apply the definition and compute the percent ionization [HNO2] [NO2−] [H3O+] initial 2.5 ≈ 0 change −x +x equilibrium 0.034 because the percent ionization is < 5%, the “x is small” approximation is valid Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the percent ionization of a 0 Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC6H4NO2? (Ka = 1.4 x 10−5 @ 25 °C) Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the percent ionization of a 0 Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC6H4NO2? write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H3O+] from water is ≈ 0 HC6H4NO2 + H2O  C6H4NO2 + H3O+ [HA] [A−] [H3O+] initial 0.012 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the percent ionization of a 0 Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC6H4NO2? represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression HC6H4NO2 + H2O  C6H4NO2 + H3O+ [HA] [A−] [H3O+] initial 0.012 change equilibrium x +x +x x x 0.012 x Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the percent ionization of a 0 Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC6H4NO2? Ka = 1.4 x 10−5 HC6H4NO2 + H2O  C6H4NO2 + H3O+ determine the value of Ka because Ka is very small, approximate the [HA]eq = [HA]init and solve for x [HA] [A−] [H3O+] initial 0.012 ≈ 0 change −x +x equilibrium x 0.012 x Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the percent ionization of a 0 Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC6H4NO2? substitute x into the equilibrium concentration definitions and solve [HA] [A−] [H3O+] initial 0.012 ≈ 0 change −x +x equilibrium 0.012−x x x = 4.1 x 10−4 Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the percent ionization of a 0 Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC6H4NO2? apply the definition and compute the percent ionization [HA] [A−] [H3O+] initial 0.012 ≈ 0 change −x +x equilibrium 4.1E-04 because the percent ionization is < 5%, the “x is small” approximation is valid Tro, Chemistry: A Molecular Approach, 2/e

Relationship Between [H3O+]equilibrium & [HA]initial Increasing the initial concentration of acid results in increased [H3O+] at equilibrium Increasing the initial concentration of acid results in decreased percent ionization This means that the increase in [H3O+] concentration is slower than the increase in acid concentration Tro, Chemistry: A Molecular Approach, 2/e

Why doesn’t the increase in H3O+ keep up with the increase in HA? The reaction for ionization of a weak acid is HA(aq) + H2O(l)  A−(aq) + H3O+(aq) According to Le Châtelier’s Principle, if we reduce the concentrations of all the (aq) components, the equilibrium should shift to the right to increase the total number of dissolved particles we can reduce the (aq) concentrations by using a more dilute initial acid concentration The result will be a larger [H3O+] in the dilute solution compared to the initial acid concentration This will result in a larger percent ionization Tro, Chemistry: A Molecular Approach, 2/e

Finding the pH of Mixtures of Acids Generally, you can ignore the contribution of the weaker acid to the [H3O+]equil For a mixture of a strong acid with a weak acid, the complete ionization of the strong acid provides more than enough [H3O+] to shift the weak acid equilibrium to the left so far that the weak acid’s added [H3O+] is negligible For mixtures of weak acids, generally only need to consider the stronger for the same reasons as long as one is significantly stronger than the other, and their concentrations are similar Tro, Chemistry: A Molecular Approach, 2/e

Example 15. 10: Find the pH of a mixture of 0 Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq) write the reactions for the acids with water and determine their Kas if the Kas are sufficiently different, use the strongest acid to construct an ICE table for the reaction enter the initial concentrations – assuming the [H3O+] from water is ≈ 0 HF + H2O  F + H3O+ Ka = 3.5 x 10−4 HClO + H2O  ClO + H3O+ Ka = 2.9 x 10−8 H2O + H2O  OH + H3O+ Kw = 1.0 x 10−14 [HF] [F−] [H3O+] initial 0.150 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

Example 15. 10: Find the pH of a mixture of 0 Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq) represent the change in the concentrations in term of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HF] [F-] [H3O+] initial 0.150 change equilibrium x +x +x 0.150 x x x Tro, Chemistry: A Molecular Approach, 2/e

Example 15. 10: Find the pH of a mixture of 0 Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq) Ka for HF = 3.5 x 10−4 because Ka is very small, approximate the [HF]eq = [HF]init and solve for x [HF] [F-] [H3O+] initial 0.150 ≈ 0 change −x +x equilibrium x 0.150 x Tro, Chemistry: A Molecular Approach, 2/e

Example 15. 10: Find the pH of a mixture of 0 Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq) Ka for HF = 3.5 x 10−4 check if the approximation is valid by seeing if x < 5% of [HF]init [HF] [F-] [H3O+] initial 0.150 ≈ 0 change −x +x equilibrium x x = 7.2 x 10−3 the approximation is valid Tro, Chemistry: A Molecular Approach, 2/e

Example 15. 10: Find the pH of a mixture of 0 Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq) Ka for HF = 3.5 x 10−4 substitute x into the equilibrium concentration definitions and solve [HF] [F-] [H3O+] initial 0.150 ≈ 0 change −x +x equilibrium 0.143 0.0072 [HF] [F-] [H3O+] initial 0.150 ≈ 0 change −x +x equilibrium 0.150-x x x = 7.2 x 10−3 Tro, Chemistry: A Molecular Approach, 2/e

Example 15. 10: Find the pH of a mixture of 0 Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq) Ka for HF = 3.5 x 10−4 substitute [H3O+] into the formula for pH and solve [HF] [F-] [H3O+] initial 0.150 ≈ 0 change −x +x equilibrium 0.143 0.0072 Tro, Chemistry: A Molecular Approach, 2/e

Example 15. 10: Find the pH of a mixture of 0 Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq) Ka for HF = 3.5 x 10−4 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HF] [F-] [H3O+] initial 0.150 ≈ 0 change −x +x equilibrium 0.143 0.0072 though not exact, the answer is reasonably close Tro, Chemistry: A Molecular Approach, 2/e

Practice – Determine the pH @ 25 ºC of a solution that is a mixture of 0.045 M HCl and 0.15 M HF Tro, Chemistry: A Molecular Approach, 2/e

Practice – Determine the pH @ 25 ºC of a solution that is a mixture of 0.045 M HCl and 0.15 M HF Given: Find: [HCl] = 4.5 x 10−2 M, [HF] = 0.15 M pH Conceptual Plan: Relationships: Because HCl is a strong acid and HF is a weak acid, [H3O+] = [HCl] [H3O+] [HCl] pH Solution: Check: pH is unitless; the fact that the pH < 7 means the solution is acidic Tro, Chemistry: A Molecular Approach, 2/e

Strong Bases The stronger the base, the more willing it is to accept H use water as the standard acid For ionic bases, practically all units are dissociated into OH– or accept H’s strong electrolyte multi-OH strong bases completely dissociated [HO–] = [strong base] x (# OH) NaOH ® Na+ + OH− Tro, Chemistry: A Molecular Approach, 2/e

Example 15. 11: Calculate the pH at 25 °C of a 0 Example 15.11: Calculate the pH at 25 °C of a 0.0015 M Sr(OH)2 solution and determine if the solution is acidic, basic, or neutral Given: Find: [Sr(OH)2] = 1.5 x 10−3 M pH Conceptual Plan: Relationships: [H3O+] [OH] pH [Sr(OH)2] [OH]=2[Sr(OH)2] Solution: [OH] = 2(0.0015) = 0.0030 M Check: pH is unitless; the fact that the pH > 7 means the solution is basic Tro, Chemistry: A Molecular Approach, 2/e 128

Example 15. 11: Calculate the pH at 25 °C of a 0 Example 15.11: Calculate the pH at 25 °C of a 0.0015 M Sr(OH)2 solution and determine if the solution is acidic, basic, or neutral Given: Find: [Sr(OH)2] = 1.5 x 10−3 M pH Conceptual Plan: Relationships: pOH [OH] pH [Sr(OH)2] [OH]=2[Sr(OH)2] pH + pOH = 14.00 Solution: [OH] = 2(0.0015) = 0.0030 M pH = 14.00 – 2.52 = 11.48 Check: pH is unitless; the fact that the pH > 7 means the solution is basic Tro, Chemistry: A Molecular Approach, 2/e 129

Practice – Calculate the pH of the following strong acid or base solutions 0.0020 M HCl 0.0015 M Ca(OH)2 [H3O+] = [HCl] = 2.0 x 10−3 M pH = −log(2.0 x 10−3) = 2.70 [OH−] = 2 x [Ca(OH)2] = 3.0 x 10−3 M pOH = −log(3.0 x 10−3) = 2.52 pH = 14.00 − pOH = 14.00 − 2.52 pH = 11.48 Tro, Chemistry: A Molecular Approach, 2/e

Weak Bases In weak bases, only a small fraction of molecules accept H’s weak electrolyte most of the weak base molecules do not take H from water much less than 1% ionization in water [HO–] << [weak base] Finding the pH of a weak base solution is similar to finding the pH of a weak acid NH3 + H2O Û NH4+ + OH− Tro, Chemistry: A Molecular Approach, 2/e

Base Ionization Constant, Kb Base strength measured by the size of the equilibrium constant when react with H2O :Base + H2O  OH− + H:Base+ The equilibrium constant is called the base ionization constant, Kb larger Kb = stronger base Tro, Chemistry: A Molecular Approach, 2/e 132

Tro, Chemistry: A Molecular Approach, 2/e

Structure of Amines Tro, Chemistry: A Molecular Approach, 2/e

Example 15.12:Find the pH of 0.100 M NH3(aq) NH3 + H2O  NH4+ + OH write the reaction for the base with water construct an ICE table for the reaction enter the initial concentrations – assuming the [OH] from water is ≈ 0 [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change equilibrium [NH3] [NH4+] [OH] initial change equilibrium because no products initially, Qc = 0, and the reaction is proceeding forward Tro, Chemistry: A Molecular Approach, 2/e

Example 15.12: Find the pH of 0.100 M NH3(aq) represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [NH3] [NH4+] [OH] initial 0.100 change equilibrium x +x +x x x 0.100 x Tro, Chemistry: A Molecular Approach, 2/e

Example 15.12: Find the pH of 0.100 M NH3(aq) determine the value of Kb from Table 15.8 because Kb is very small, approximate the [NH3]eq = [NH3]init and solve for x Kb for NH3 = 1.76 x 10−5 [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change −x +x equilibrium x 0.100 x Tro, Chemistry: A Molecular Approach, 2/e

Example 15.12: Find the pH of 0.100 M NH3(aq) Kb for NH3 = 1.76 x 10−5 check if the approximation is valid by seeing if x < 5% of [NH3]init [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change −x +x equilibrium x x = 1.33 x 10−3 the approximation is valid Tro, Chemistry: A Molecular Approach, 2/e

Example 15.12: Find the pH of 0.100 M NH3(aq) Kb for NH3 = 1.76 x 10−5 substitute x into the equilibrium concentration definitions and solve [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change -x +x equilibrium 0.099 1.33E-3 [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change -x +x equilibrium 0.100 x x x = 1.33 x 10−3 Tro, Chemistry: A Molecular Approach, 2/e

Example 15.12: Find the pH of 0.100 M NH3(aq) Kb for NH3 = 1.76 x 10−5 use the [OH−] to find the [H3O+] using Kw substitute [H3O+] into the formula for pH and solve [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change −x +x equilibrium 0.099 1.33E−3 Tro, Chemistry: A Molecular Approach, 2/e

Example 15.12: Find the pH of 0.100 M NH3(aq) Kb for NH3 = 1.76 x 10−5 use the [OH−] to find the pOH use pOH to find pH [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change −x +x equilibrium 0.099 1.33E−3 alternate method Tro, Chemistry: A Molecular Approach, 2/e

Example 15.12: Find the pH of 0.100 M NH3(aq) Kb for NH3 = 1.76 x 10−5 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb [NH3] [NH4+] [OH] initial 0.100 ≈ 0 change −x +x equilibrium 0.099 1.33E−3 though not exact, the answer is reasonably close Tro, Chemistry: A Molecular Approach, 2/e

Practice – Find the pH of a 0. 0015 M morphine solution, Kb = 1 Practice – Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10−6 Tro, Chemistry: A Molecular Approach, 2/e

Practice – Find the pH of a 0.0015 M morphine solution write the reaction for the base with water construct an ICE table for the reaction enter the initial concentrations – assuming the [OH] from water is ≈ 0 B + H2O  BH+ + OH [B] [BH+] [OH] initial 0.0015 ≈ 0 change equilibrium because no products initially, Qc = 0, and the reaction is proceeding forward Tro, Chemistry: A Molecular Approach, 2/e

Practice – Find the pH of a 0.0015 M morphine solution B + H2O  BH+ + OH represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [B] [BH+] [OH] initial 0.0015 change equilibrium x +x +x x x 0.0015 x Tro, Chemistry: A Molecular Approach, 2/e

Practice – Find the pH of a 0.0015 M morphine solution Kb for morphine = 1.6 x 10−6 determine the value of Kb because Kb is very small, approximate the [B]eq = [B]init and solve for x [B] [BH+] [OH] initial 0.0015 ≈ 0 change −x +x equilibrium x 0.0015 x Tro, Chemistry: A Molecular Approach, 2/e

Practice – Find the pH of a 0.0015 M morphine solution Kb for morphine = 1.6 x 10−6 check if the approximation is valid by seeing if x < 5% of [B]init [B] [BH+] [OH] initial 0.0015 ≈ 0 change −x +x equilibrium x x = 4.9 x 10−5 the approximation is valid Tro, Chemistry: A Molecular Approach, 2/e

Practice – Find the pH of a 0.0015 M morphine solution Kb for morphine = 1.6 x 10−6 substitute x into the equilibrium concentration definitions and solve [B] [BH+] [OH] initial 0.0015 ≈ 0 change −x +x equilibrium 4.9E−5 [B] [BH+] [OH] initial 0.0015 ≈ 0 change −x +x equilibrium 0.0015 x x x = 4.9 x 10−5 Tro, Chemistry: A Molecular Approach, 2/e

Practice – Find the pH of a 0.0015 M morphine solution Kb for morphine = 1.6 x 10−6 use the [OH−] to find the [H3O+] using Kw substitute [H3O+] into the formula for pH and solve [B] [BH+] [OH] initial 0.0015 ≈ 0 change −x +x equilibrium 4.9E−5 Tro, Chemistry: A Molecular Approach, 2/e

Practice – Find the pH of a 0.0015 M morphine solution Kb for morphine = 1.6 x 10−6 use the [OH−] to find the pOH use pOH to find pH [B] [BH+] [OH] initial 0.0015 ≈ 0 change −x +x equilibrium 4.9E−5 alternate method Tro, Chemistry: A Molecular Approach, 2/e

Practice – Find the pH of a 0.0015 M morphine solution Kb for morphine = 1.6 x 10−6 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb [B] [BH+] [OH] initial 0.0015 ≈ 0 change −x +x equilibrium 4.9E−5 the answer matches the given Kb Tro, Chemistry: A Molecular Approach, 2/e

Acid–Base Properties of Salts Salts are water-soluble ionic compounds Salts that contain the cation of a strong base and an anion that is the conjugate base of a weak acid are basic NaHCO3 solutions are basic Na+ is the cation of the strong base NaOH HCO3− is the conjugate base of the weak acid H2CO3 Salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are acidic NH4Cl solutions are acidic NH4+ is the conjugate acid of the weak base NH3 Cl− is the anion of the strong acid HCl Tro, Chemistry: A Molecular Approach, 2/e

Anions as Weak Bases Every anion can be thought of as the conjugate base of an acid Therefore, every anion can potentially be a base A−(aq) + H2O(l)  HA(aq) + OH−(aq) The stronger the acid is, the weaker the conjugate base is An anion that is the conjugate base of a strong acid is pH neutral Cl−(aq) + H2O(l)  HCl(aq) + OH−(aq) An anion that is the conjugate base of a weak acid is basic F−(aq) + H2O(l)  HF(aq) + OH−(aq) Tro, Chemistry: A Molecular Approach, 2/e

Example 15.13: Use the table to determine if the given anion is basic or neutral NO3− the conjugate base of a strong acid, therefore neutral NO2− the conjugate base of a weak acid, therefore basic Tro, Chemistry: A Molecular Approach, 2/e

Relationship between Ka of an Acid and Kb of Its Conjugate Base Many reference books only give tables of Ka values because Kb values can be found from them when you add equations, you multiply the K’s Tro, Chemistry: A Molecular Approach, 2/e

Example 15.14: Find the pH of 0.100 M NaCHO2(aq) solution Na+ is the cation of a strong base – pH neutral. The CHO2− is the anion of a weak acid – pH basic write the reaction for the anion with water construct an ICE table for the reaction enter the initial concentrations – assuming the [OH] from water is ≈ 0 Example 15.14: Find the pH of 0.100 M NaCHO2(aq) solution CHO2− + H2O  HCHO2 + OH [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change equilibrium

Example 15.14: Find the pH of 0.100 M NaCHO2(aq) solution represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x Calculate the value of Kb from the value of Ka of the weak acid from Table 15.5 substitute into the equilibrium constant expression [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change equilibrium x +x +x 0.100 x x x Tro, Chemistry: A Molecular Approach, 2/e

Example 15.14: Find the pH of 0.100 M NaCHO2(aq) Kb for CHO2− = 5.6 x 10−11 because Kb is very small, approximate the [CHO2−]eq = [CHO2−]init and solve for x [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change −x +x equilibrium x 0.100 x Tro, Chemistry: A Molecular Approach, 2/e

Example 15.14: Find the pH of 0.100 M NaCHO2(aq) Kb for CHO2− = 5.6 x 10−11 check if the approximation is valid by seeing if x < 5% of [CHO2−]init [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change −x +x equilibrium x x = 2.4 x 10−6 the approximation is valid Tro, Chemistry: A Molecular Approach, 2/e

Example 15.14: Find the pH of 0.100 M NaCHO2(aq) Kb for CHO2− = 5.6 x 10−11 substitute x into the equilibrium concentration definitions and solve [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change −x +x equilibrium 0.100 −x x [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change −x +x equilibrium 2.4E-6 x = 2.4 x 10−6 Tro, Chemistry: A Molecular Approach, 2/e

Example 15.14: Find the pH of 0.100 M NaCHO2(aq) Kb for CHO2− = 5.6 x 10−11 use the [OH−] to find the [H3O+] using Kw substitute [H3O+] into the formula for pH and solve [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change −x +x equilibrium 2.4E-6 Tro, Chemistry: A Molecular Approach, 2/e

Example 15.14: Find the pH of 0.100 M NaCHO2(aq) Kb for CHO2− = 5.6 x 10−11 use the [OH−] to find the pOH use pOH to find pH [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change −x +x equilibrium 2.4E−6 alternate method Tro, Chemistry: A Molecular Approach, 2/e

Example 15.14: Find the pH of 0.100 M NaCHO2(aq) Kb for CHO2− = 5.6 x 10−11 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb [CHO2−] [HCHO2] [OH] initial 0.100 ≈ 0 change −x +x equilibrium 2.4E−6 though not exact, the answer is reasonably close Tro, Chemistry: A Molecular Approach, 2/e

Practice – If a 0. 15 M NaA solution has a pOH of 5 Practice – If a 0.15 M NaA solution has a pOH of 5.45, what is the Ka of HA? Tro, Chemistry: A Molecular Approach, 2/e

If a 0.0015 M NaA solution has a pOH of 5.45, what is the Ka of HA? Na+ is the cation of a strong base – pOH neutral. Because pOH is < 7, the solutinon is basic. A− is basic. write the reaction for the anion with water construct an ICE table for the reaction enter the initial concentrations – assuming the [OH] from water is ≈ 0 If a 0.0015 M NaA solution has a pOH of 5.45, what is the Ka of HA? A− + H2O  HA + OH [A−] [HA] [OH] initial 0.100 ≈ 0 change equilibrium 165

Practice – If a 0. 15 M NaA solution has a pOH of 5 Practice – If a 0.15 M NaA solution has a pOH of 5.45, what is the Ka of HA? use the pOH to find the [OH−] use [OH−] to fill in other items [A−] [HA] [OH] initial 0.15 ≈ 0 change equilibrium [A−] [HA] [OH] initial 0.15 ≈ 0 change −3.6E−6 +3.6E−6 equilibrium 3.6E-6 Tro, Chemistry: A Molecular Approach, 2/e 166

Practice – If a 0. 15 M NaA solution has a pOH of 5 Practice – If a 0.15 M NaA solution has a pOH of 5.45, what is the Ka of HA? calculate the value of Kb of A− [A−] [HA] [OH] initial 0.15 ≈ 0 change −3.6E−6 +3.6E−6 equilibrium 3.6E-6 Tro, Chemistry: A Molecular Approach, 2/e 167

Practice – If a 0. 15 M NaA solution has a pOH of 5 Practice – If a 0.15 M NaA solution has a pOH of 5.45, what is the Ka of HA? use Kb of A− to find Ka of HA [A−] [HA] [OH] initial 0.15 ≈ 0 change −3.6E−6 +3.6E−6 equilibrium 3.6E-6 Tro, Chemistry: A Molecular Approach, 2/e 168

Polyatomic Cations as Weak Acids Some cations can be thought of as the conjugate acid of a weak base others are the counter-ions of a strong base Therefore, some cations can potentially be acidic MH+(aq) + H2O(l)  MOH(aq) + H3O+(aq) The stronger the base is, the weaker the conjugate acid is a cation that is the counter-ion of a strong base is pH neutral a cation that is the conjugate acid of a weak base is acidic NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq) because NH3 is a weak base, the position of this equilibrium favors the right Tro, Chemistry: A Molecular Approach, 2/e

Metal Cations as Weak Acids Cations of small, highly charged metals are weakly acidic alkali metal cations and alkali earth metal cations are pH neutral cations are hydrated Al(H2O)63+(aq) + H2O(l)  Al(H2O)5(OH)2+ (aq) + H3O+(aq) Tro, Chemistry: A Molecular Approach, 2/e

Example 15.15: Determine if the given cation Is acidic or neutral C5N5NH2+ the conjugate acid of the weak base pyridine, therefore acidic Ca2+ the counter-ion of the strong base Ca(OH)2, therefore neutral Cr3+ a highly charged metal ion, therefore acidic Tro, Chemistry: A Molecular Approach, 2/e

Classifying Salt Solutions as Acidic, Basic, or Neutral If the salt cation is the counter-ion of a strong base and the anion is the conjugate base of a strong acid, it will form a neutral solution NaCl Ca(NO3)2 KBr If the salt cation is the counter-ion of a strong base and the anion is the conjugate base of a weak acid, it will form a basic solution NaF Ca(C2H3O2)2 KNO2 Tro, Chemistry: A Molecular Approach, 2/e

Classifying Salt Solutions as Acidic, Basic, or Neutral If the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a strong acid, it will form an acidic solution NH4Cl If the salt cation is a highly charged metal ion and the anion is the conjugate base of a strong acid, it will form an acidic solution Al(NO3)3 Tro, Chemistry: A Molecular Approach, 2/e

Classifying Salt Solutions as Acidic, Basic, or Neutral If the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a weak acid, the pH of the solution depends on the relative strengths of the acid and base NH4F because HF is a stronger acid than NH4+, Ka of NH4+ is larger than Kb of the F−; therefore the solution will be acidic Tro, Chemistry: A Molecular Approach, 2/e

Example 15.16: Determine whether a solution of the following salts is acidic, basic, or neutral SrCl2 Sr2+ is the counter-ion of a strong base, pH neutral Cl− is the conjugate base of a strong acid, pH neutral solution will be pH neutral AlBr3 Al3+ is a small, highly charged metal ion, weak acid solution will be acidic CH3NH3NO3 CH3NH3+ is the conjugate acid of a weak base, acidic NO3− is the conjugate base of a strong acid, pH neutral Tro, Chemistry: A Molecular Approach, 2/e

Example 15.16: Determine whether a solution of the following salts is acidic, basic, or neutral NaCHO2 Na+ is the counter-ion of a strong base, pH neutral CHO2− is the conjugate base of a weak acid, basic solution will be basic NH4F NH4+ is the conjugate acid of a weak base, acidic F− is the conjugate base of a weak acid, basic Ka(NH4+) > Kb(F−); solution will be acidic Tro, Chemistry: A Molecular Approach, 2/e

Practice – Determine whether a solution of the following salts is acidic, basic, or neutral K+ is the counter-ion of a strong base, pH neutral KNO3 CoCl3 Ba(HCO3)2 CH3NH3NO3 NO3− is the counter-ion of a strong acid, pH neutral the solution is pH neutral Co3+ is a highly charged cation, pH acidic Cl− is the counter-ion of a strong acid, pH neutral the solution is pH acidic Ba2+ is the counter-ion of a strong base, pH neutral the solution is pH basic HCO3− is the conjugate of a weak acid, pH basic CH3NH3+ is the conjugate of a weak base, pH acidic the solution is pH acidic NO3− is the counter-ion of a strong acid, pH neutral Tro, Chemistry: A Molecular Approach, 2/e

Ionization in Polyprotic Acids Because polyprotic acids ionize in steps, each H has a separate Ka Ka1 > Ka2 > Ka3 Generally, the difference in Ka values is great enough so that the second ionization does not happen to a large enough extent to affect the pH most pH problems just do first ionization except H2SO4  use [H2SO4] as the [H3O+] for the second ionization [A2−] = Ka2 as long as the second ionization is negligible Tro, Chemistry: A Molecular Approach, 2/e

Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3? (Ka1 = 4.3 x 10−7, Ka2 = 5.6 x 10−11) Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3? H2CO3 + H2O  HCO3 + H3O+ write the reactions for the acid with water one H at a time construct an ICE table for the reaction enter the initial concentrations – assuming the second ionization is negligible HCO3− + H2O  CO32− + H3O+ [HA] [A−] [H3O+] initial 0.12 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3? H2CO3 + H2O  HCO3 + H3O+ represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HA] [A−] [H3O+] initial 0.12 change equilibrium x +x +x x x 0.12 x Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3? determine the value of Ka1 because Ka1 is very small, approximate the [HA]eq = [HA]init and solve for x Ka1 = 4.3 x 10−7, Ka2 = 5.6 x 10−11 [HA] [A−] [H3O+] initial 0.12 ≈ 0 change −x +x equilibrium 0.012 x 0.12 x Tro, Chemistry: A Molecular Approach, 2/e

the approximation is valid Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3? Ka1 for H2CO3 = 4.3 x 10−7 check if the approximation is valid by seeing if x < 5% of [H2CO3]init [HA] [A−] [H3O+] initial 0.12 ≈ 0 change −x +x equilibrium x x = 2.27 x 10−4 the approximation is valid Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3? substitute x into the equilibrium concentration definitions and solve [HA] [A−] [H3O+] initial 0.12 ≈ 0 change −x +x equilibrium 0.12−x x x = 2.3 x 10−4 Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3? substitute [H3O+] into the formula for pH and solve [HA] [A−] [H3O+] initial 0.12 ≈ 0 change −x +x equilibrium 0.00023 Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3? Ka1 for H2CO3 = 4.3 x 10−7 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HA] [A−] [H3O+] initial 0.12 ≈ 0 change −x +x equilibrium 0.00023 the values match within sig figs Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the [CO32−] in a 0 Practice – What is the [CO32−] in a 0.12 M solution of carbonic acid, H2CO3? (Ka1 = 4.3 x 10−7, Ka2 = 5.6 x 10−11) Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the [CO32−] in a 0 Practice – What is the [CO32−] in a 0.12 M solution of carbonic acid, H2CO3? write the reactions for the acid with water one H at a time construct an ICE table for the reaction enter the initial concentrations for the second ionization using the equilibrium concentrations from first ionization H2CO3 + H2O  HCO3 + H3O+ HCO3− + H2O  CO32− + H3O+ [HCO3−] [CO32−] [H3O+] initial 0.00023 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the [CO32−] in a 0 Practice – What is the [CO32−] in a 0.12 M solution of carbonic acid, H2CO3? HCO3− + H2O  CO32− + H3O+ represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HCO3−] [CO32−] [H3O+] initial 0.00023 change equilibrium x +x +x x 2.3E−4 x 2.3E−4 +x Tro, Chemistry: A Molecular Approach, 2/e

Practice – What is the [CO32−] in a 0 Practice – What is the [CO32−] in a 0.12 M solution of carbonic acid, H2CO3? Ka1 = 4.3 x 10−7, Ka2 = 5.6 x 10−11 determine the value of Ka2 because Ka2 is very small, approximate the [HA]eq = [HA]init, [H3O+]eq = [H3O+]init, and solve for x using this approximation, it is seen that x = Ka2. Therefore [CO32−] = Ka2 [HCO3−] [CO32−] [H3O+] initial 0.00023 change −x +x equilibrium 2.3E−4 x [HCO3−] [CO32−] [H3O+] initial 0.00023 change −x +x equilibrium 2.3E−4 − x x 2.3E−4 + x Tro, Chemistry: A Molecular Approach, 2/e

Ionization in H2SO4 The ionization constants for H2SO4 are H2SO4 + H2O  HSO4 + H3O+ strong HSO4 + H2O  SO42 + H3O+ Ka2 = 1.2 x 10−2 For most sulfuric acid solutions, the second ionization is significant and must be accounted for Because the first ionization is complete, use the given [H2SO4] = [HSO4−]initial = [H3O+]initial Tro, Chemistry: A Molecular Approach, 2/e

Example 15.18: Find the pH of 0.0100 M H2SO4(aq) solution @ 25 °C H2SO4 + H2O  HSO4 + H3O+ write the reactions for the acid with water construct an ICE table for the second ionization reaction enter the initial concentrations – assuming the [HSO4−] and [H3O+] is ≈ [H2SO4] HSO4 + H2O  SO42 + H3O+ [HSO4 ] [SO42 ] [H3O+] initial 0.0100 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

Example 15.18: Find the pH of 0.0100 M H2SO4(aq) solution @ 25 °C represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HSO4 ] [SO42 ] [H3O+] initial 0.0100 change −x +x equilibrium 0.0100 −x x Tro, Chemistry: A Molecular Approach, 2/e 194

Example 15.18: Find the pH of 0.0100 M H2SO4(aq) solution @ 25 °C Ka for HSO4− = 0.012 expand and solve for x using the quadratic formula Tro, Chemistry: A Molecular Approach, 2/e 195

Example 15.18: Find the pH of 0.0100 M H2SO4(aq) solution @ 25 °C Ka for HSO4− = 0.012 substitute x into the equilibrium concentration definitions and solve [HSO4 ] [SO42 ] [H3O+] initial 0.0100 change −x +x equilibrium 0.0055 0.0045 0.0145 [HSO4 ] [SO42 ] [H3O+] initial 0.0100 change −x +x equilibrium 0.0100 −x x x = 0.0045 Tro, Chemistry: A Molecular Approach, 2/e 196

Example 15.18: Find the pH of 0.0100 M H2SO4(aq) solution @ 25 °C Ka for HSO4− = 0.012 substitute [H3O+] into the formula for pH and solve [HSO4 ] [SO42 ] [H3O+] initial 0.0100 change −x +x equilibrium 0.0055 0.0045 0.0145 Tro, Chemistry: A Molecular Approach, 2/e 197

Example 15.18: Find the pH of 0.0100 M H2SO4(aq) solution @ 25 °C Ka for HSO4− = 0.012 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HSO4 ] [SO42 ] [H3O+] initial 0.0100 change −x +x equilibrium 0.0055 0.0045 0.0145 the answer matches Tro, Chemistry: A Molecular Approach, 2/e

Strengths of Binary Acids The more d+ H─X d− polarized the bond, the more acidic the bond The stronger the H─X bond, the weaker the acid Binary acid strength increases to the right across a period acidity: H─C < H─N < H─O < H─F Binary acid strength increases down the column acidity: H─F < H─Cl < H─Br < H─I Tro, Chemistry: A Molecular Approach, 2/e

Relationship between Bond Strength and Acidity Bond Energy kJ/mol Type of Acid HF 565 weak HCl 431 strong HBr 364 Tro, Chemistry: A Molecular Approach, 2/e

Strengths of Oxyacids, H–O–Y The more electronegative the Y atom, the stronger the oxyacid HClO > HIO acidity of oxyacids decreases down a group same trend as binary acids helps weakens the H–O bond The larger the oxidation number of the central atom, the stronger the oxyacid H2CO3 > H3BO3 acidity of oxyacids increases to the right across a period opposite trend of binary acids The more oxygens attached to Y, the stronger the oxyacid further weakens and polarizes the H–O bond HClO3 > HClO2 Tro, Chemistry: A Molecular Approach, 2/e

Relationship Between Electronegativity and Acidity H─O─Y Electronegativity of Y Ka H─O─Cl 3.0 2.9 x 10−8 H─O─Br 2.8 2.0 x 10−9 H─O─I 2.5 2.3 x 10−11 Tro, Chemistry: A Molecular Approach, 2/e

Relationship Between Number of Oxygens on the Central Atom and Acidity Tro, Chemistry: A Molecular Approach, 2/e

Practice – Order the Following By Acidity (Least to Most) H3PO4 HNO3 H3PO3 H3AsO3 HCl HBr H2S HS− By Basicity (Least to Most) CO32− NO3− HCO3− BO33− Tro, Chemistry: A Molecular Approach, 2/e 204

Practice – Order the Following By Acidity (Least to Most) H3PO4 HNO3 H3PO3 H3AsO3 H3AsO3 < H3PO3 < H3PO4 < HNO3 HCl HBr H2S HS− HS− < H2S < HCl < HBr By Basicity (Least to Most) CO32− NO3− HCO3− BO33− NO3− < HCO3− < CO32− < BO33− Tro, Chemistry: A Molecular Approach, 2/e 205

Lewis Acid–Base Theory Lewis Acid–Base theory focuses on transferring an electron pair lone pair  bond bond  lone pair Does NOT require H atoms The electron donor is called the Lewis Base electron rich, therefore nucleophile The electron acceptor is called the Lewis Acid electron deficient, therefore electrophile Tro, Chemistry: A Molecular Approach, 2/e

Lewis Bases Lewis Base has electrons it is willing to give away to or share with another atom Lewis Base must have lone pair of electrons on it that it can donate Anions are better Lewis Bases than neutral atoms or molecules N: < N:− Generally, the more electronegative an atom, the less willing it is to be a Lewis Base O: < S: Tro, Chemistry: A Molecular Approach, 2/e 207

Lewis Acids Electron deficient, either from being attached to electronegative atom(s) not having an octet Must have empty orbital willing to accept the electron pair H+ has empty 1s orbital B in BF3 has empty 2p orbital and an incomplete octet Many small, highly charged metal cations have empty orbitals they can use to accept electrons Atoms that are attached to highly electronegative atoms and have multiple bonds can be Lewis Acids Tro, Chemistry: A Molecular Approach, 2/e

Lewis Acid–Base Reactions The base donates a pair of electrons to the acid Generally results in a covalent bond forming H3N: + BF3  H3N─BF3 The product that forms is called an adduct Arrhenius and Brønsted-Lowry acid–base reactions are also Lewis Tro, Chemistry: A Molecular Approach, 2/e

Examples of Lewis Acid–Base Reactions Tro, Chemistry: A Molecular Approach, 2/e

Examples of Lewis Acid–Base Reactions Ag+(aq) + 2 :NH3(aq)  Ag(NH3)2+(aq) Lewis Acid Lewis Base Adduct Tro, Chemistry: A Molecular Approach, 2/e

Practice – Identify the Lewis Acid and Lewis Base in Each Reaction Tro, Chemistry: A Molecular Approach, 2/e

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) U.S. Fuel Consumption Over 85% of the energy use in the United States comes from the combustion of fossil fuels oil, natural gas, coal Combustion of fossil fuels produces CO2 CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) Natural fossil fuels also contain small amounts of S that burn to produce SO2(g) S(s) + O2(g) → SO2(g) The high temperatures of combustion allow N2(g) in the air to combine with O2(g) to form oxides of nitrogen N2(g) + 2 O2(g) → 2 NO2(g) Tro, Chemistry: A Molecular Approach, 2/e

What Is Acid Rain? Natural rain water has a pH of 5.6 naturally slightly acidic due mainly to CO2 Rain water with a pH lower than 5.6 is called acid rain Acid rain is linked to damage in ecosystems and structures Tro, Chemistry: A Molecular Approach, 2/e

What Causes Acid Rain? Many natural and pollutant gases dissolved in the air are nonmetal oxides CO2, SO2, NO2 Nonmetal oxides are acidic CO2(g) + H2O(l)  H2CO3(aq) 2 SO2(g) + O2(g) + 2 H2O(l)  2 H2SO4(aq) 4 NO2(g) + O2(g) + 2 H2O(l)  4 HNO3(aq) Processes that produce nonmetal oxide gases as waste increase the acidity of the rain natural – volcanoes and some bacterial action man-made – combustion of fuel Tro, Chemistry: A Molecular Approach, 2/e

pH of Rain in Different Regions Tro, Chemistry: A Molecular Approach, 2/e 216

Weather Patterns The prevailing winds in the United States travel west to east Weather patterns may cause rain to be acidic in regions other than where the nonmetal oxide is produced Much of the northeast United States has rain of very low pH, even though it has very low sulfur emissions, due in part to the general weather patterns Tro, Chemistry: A Molecular Approach, 2/e

Sources of SO2 from Utilities Tro, Chemistry: A Molecular Approach, 2/e 218

Damage from Acid Rain Acids react with metals, and materials that contain carbonates Acid rain damages bridges, cars, and other metallic structures Acid rain damages buildings and other structures made of limestone or cement Acidifying lakes affects aquatic life Soil acidity causes more dissolving of minerals and leaching more minerals from soil making it difficult for trees Tro, Chemistry: A Molecular Approach, 2/e

Damage from Acid Rain Tro, Chemistry: A Molecular Approach, 2/e

Acid Rain Legislation 1990 Clean Air Act attacks acid rain forces utilities to reduce SO2 Result is acid rain in the Northeast stabilized and beginning to be reduced Tro, Chemistry: A Molecular Approach, 2/e