Acids & Bases Unit 13.

Acids & Bases Unit 13

Overview Acid/base properties Definitions Acid-Base Reactions
Arrhenius Bronsted-Lowry Lewis Acid-Base Reactions Neutralization Sulfides Carbonates pH/pOH Scale Acid/base strength Factor affecting Ka, Kb, Kw Percent ionization Vocabulary Polyprotic Acids Amphoteric Anhydrides Acids/Bases & Salts Determine acidity Calculations Common Ion Effect Buffers Henderson-Hasselbalch Titration Indicators 4 types of curves

Acids Sour taste React with active metals to produce hydrogen gas
Change the color of acid-base indicators React with bases to produce salt and water Conduct an electric current (electrolytes) Turn litmus paper red

Common Acids Sulfuric Acid Nitric Acid Phosphoric acid
Car batteries; production of metals, paints, dyes, detergents Nitric Acid Explosives, pharmaceuticals, rubber, plastics, dyes Phosphoric acid Soda, fertilizers, animal feed, detergents Hydrochloric Acid Stomach acid, cleaning metals, found in hardware stores (muriatic acid) Acetic Acid Vinegar, food supplements, fungicide Citric Acid Fruit juices

Acids Binary acids Oxyacids Contain only two different elements
Name as “hydro - ic acid” Example: HCl (hydrochloric acid) Oxyacids Acid consisting of hydrogen and a polyatomic anion that contains oxygen (oxyanion) To name, drop ending of polyatomic ion and ad “- ic acid” Example: HNO3 (nitric acid)

Bases Taste bitter Feel slippery
Change the color of acid-base indicators React with bases to produce salt and water Conduct an electric current (electrolytes) Turn litmus paper blue

Common Bases Ammonium hydroxide Ammonia
Household cleaners, window cleaner Ammonia (Gas) inhalant to revive unconscious person Sodium bicarbonate (baking soda) Acid neutralizers in acid spills Antacids for upset stomachs Sodium hydroxide Drain cleaner (drano), oven cleaner, production of soap Magnesium hydroxide Antacids, milk of magnesia, laxatives

Definition: Arrhenius
Acid Substance that ionizes in water and produces H+ ions Example: HCl  H+ + Cl- Base Substance that ionizes in water and produces OH- ions Example: NaOH  Na+ + OH-

Definition: Bronsted-Lowry
Acid Substance that is capable of donating a proton (H+ ion) Base Substance that is capable of accepting a proton (H+ ion)

Examples: Bronsted-Lowry
HC2H3O2 + H2O ↔ C2H3O2- + H3O+ Acids: HC2H3O2 and H3O+ Bases: H2O and C2H3O2- NH3 + H2O ↔ NH4+ + OH- Acids: H2O and NH4+ Bases: NH3 and OH- Notice that water can act as an acid or a base

Bronsted-Lowry Conjugate Pair – a BL acid/base pair(one with H+ and one without H+) Examples: HC2H3O2 and C2H3O2- H3O+ and H2O H2O and OH- NH4+ and NH3

Bronsted-Lowry The more easily a substance gives up a proton, the less easily the conjugate base accepts a proton (and vice versa) The stronger the acid, the weaker the conjugate base The stronger the base, the weaker the conjugate acid

Definition: Lewis Acid Electron pair acceptor Base Electron pair donor

Examples: Lewis Acid Base Acid Base All Bronsted-Lowry acids/bases are also Lewis acids/bases Not all Lewis acids/bases are Bronsted-Lowry acids/bases

Acid Base Reactions Neutralization Sulfides Carbonates Salt + water
Salt + sulfide gas Carbonates Salt + CO2 + H2O

HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Neutralization Solution of an acid and solution of a base are mixed Products have no characteristics of either the acid or the base Acid + Base (metal hydroxide)  salt + water Salt comes from cation of base and anion of acid HY + XOH  XY + H2O HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

HCl(aq) + Na2S(aq)  NaCl(aq) + H2S(g)
Sulfides Acid reacts with a sulfide Gaseous product (H2S) has a foul odor (rotten eggs) Acid + metal sulfide  salt + hydrogen sulfide Salt comes from cation of sulfide and anion of acid HY + XS  XY + H2S HCl(aq) + Na2S(aq)  NaCl(aq) + H2S(g)

Carbonates HY + XHCO3  XY + H2CO3
Carbonates and bicarbonates react with acids HY + XHCO3  XY + H2CO3 H2CO3 is not stable so breaks into H2O and CO2 Then HY + XHCO3  XY + H2O + CO2 HCl(aq) + NaHCO3(aq)  NaCl(aq) + H2CO3 (aq) HCl(aq) + NaHCO3(aq)  NaCl(aq) + H2O(l) + CO2(g)

pH = -log[H+] therefore [H+] = 10-pH
The pH Scale pH scale: measures concentration of hydrogen ions in solution pH = -log[H+] therefore [H+] = 10-pH Example: What is the pH of a solution with a [H+] of 1.4×10-5? pH = -log[1.4×10-5] = 4.9

pH Scale Acids: pH < 7 Neutral: pH = 7 Bases: pH > 7
Increasing [H+] means decreasing pH Increasing pH means decreasing [H+]

pOH = -log[OH-] therefore [OH-] = 10-pOH
The pOH Scale pOH scale: measures concentration of hydroxide ions in solution pOH = -log[OH-] therefore [OH-] = 10-pOH Example: What is the [OH-] of a solution with a pOH of 6.2? [OH-] = = 6.3×10-7

Comparing pH and pOH pH + pOH = 14
An acid has a pH of 4, what is the pOH? 4 + pOH = 14 pOH = 10

Measuring pH pH meter Acid-base indicators Electrodes measure [H+]
Change color in presence of acid or base (or certain pH ranges) Litmus paper, phenolphthalein, cabbage juice, methyl orange, thymol blue…

Strong Acids and Bases Completely ionize in solution (strong electrolytes) Strong Acids Strong Bases HCl Group 1 metals + OH HBr (LiOH, NaOH, KOH,…) HI HClO3 Heavy group 2 metals + OH HClO4 Ca(OH)2, Sr(OH)2, Ba(OH)2 HNO3 H2SO4 If acid/base is not on this list, it is a weak acid/base

Weak Acids and Bases Common weak acids: Common weak bases:
Do not completely ionize in water (weak electrolytes) Common weak acids: HF, acids with -COOH group Common weak bases: NH3

Factors Affecting Acid Strength
Electronegativity of element bonded to H Binary acids More electronegative bond = stronger acid Example: HCl stronger than HBr Bond Strength Stronger bonds do not allow hydrogen to dissociate as easily Reason why HF is not a strong acid (F is most electronegative, but H-F bond is strongest bond) Stability of Conjugate base More stable the conjugate base, the stronger the acid

Factors Affecting Acid Strength
For polyatomic ions, the more electronegative the nonmetal, the stronger the acid (when comparing acids with same number of O atoms) Example: HClO3 is stronger than HBrO3 For polyatomic ions, when nonmetal is the same, the more O atoms, the stronger the acid Example: HClO3 is stronger than HClO2

Initial Acid Concentration
Percent Ionization Tells us what percent of an acid (or base) is ionized in water Helps determine the strength of an acid (or base) Percent Ionization = ×100 [H+] at equilibrium Initial Acid Concentration

Percent Ionization (Example)
A M solution of HNO2 contains 3.7×10-3 M H+(aq). Calculate the percent ionization. = = 11% This means that 11% of the acid will dissociate in water. 3.7×10-3 M 0.035 M

Equilibrium Time!

Strong Acids acid water proton conjugate base
HA(aq) + H2O(l)  H+(aq) + A-(aq) acid water proton conjugate base Strong acids dissociate completely The dissociation is not reversible The acid is the only significant source of H+ ions, so pH can be calculated directly from the [H+] Example: A 0.20 M solution of HNO3 has an [H+] of 0.20 M pH = -log[H+]

Strong Bases Strong bases dissociate completely
The dissociation is not reversible The base is the only significant source of OH- ions, so pOH can be calculated directly from the [OH-] Example: A 0.30 M solution of NaOH has a [OH-] of 0.30 M pOH = -log[OH-]

K = Weak Acids and Bases [Products] [Reactants]
Do not dissociate completely Reversible reactions Need to use equilibrium to solve for [H+] K = [Products] [Reactants]

Acid Dissociation HA(aq) + H2O(l)  H+(aq) + A-(aq) acid water proton conjugate base Write the equilibrium expression for the acid dissociation constant, Ka.

Base Dissociation base water conjugate hydroxide acid ion
B(aq) + H2O(l) ↔ BH+(aq) + OH-(aq) base water conjugate hydroxide acid ion Write the equilibrium expression for the base dissociation constant, Kb.

Size of K The greater the Ka, the stronger the acid
The smaller the Ka, the weaker the acid The greater the Kb, the stronger the base The smaller the Kb, the weaker the base

Example: Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #1: Write the dissociation equation HC2H3O2  C2H3O2- + H+

What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #2: ICE HC2H3O2  C2H3O H+ I C E 0.50 - x +x +x x x x

What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #3: Set up the equilibrium expression If percent ionization is less than 5%, you can ignore using the quadratic.

What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #5: Solve for pH You can use the Kb expression to solve for pOH using the same method!

Example 2: Weak Acid Equilibrium - Solving for Ka
A student prepared a 0.10 M solution of formic acid (HCOOH) and measured its pH. The pH at 25°C was found to be Calculate the Ka for formic acid at this temperature. Step #1: Solve for [H+] from pH [H+] = = 4.2×10-3 M

A student prepared a 0.10 M solution of formic acid (HCOOH) and measured its pH. The pH at 25°C was found to be Calculate the Ka for formic acid at this temperature. Step #2: Set up ICE table HCOOH(aq)  HCOO H+ I C E 0.10 4.2×10-3

A student prepared a 0.10 M solution of formic acid (HCOOH) and measured its pH. The pH at 25°C was found to be Calculate the Ka for formic acid at this temperature. Step #3: Use stoichiometry to complete table HCOOH(aq)  HCOO H+ I C E 0.10 4.2×10-3 4.2×10-3 4.2×10-3 0.0096 4.2×10-3 4.2×10-3

A student prepared a 0.10 M solution of formic acid (HCOOH) and measured its pH. The pH at 25°C was found to be Calculate the Ka for formic acid at this temperature. Step #4: Solve for Ka using equilibrium expression Ka = 1.8×10-4

The larger the value of Ka, the stronger the acid
Rule of Thumb The larger the value of Ka, the stronger the acid

Initial Acid Concentration
Percent Ionization Tells us what percent of an acid (or base) is ionized in water Helps determine the strength of an acid (or base) Percent Ionization = ×100 [H+] at equilibrium Initial Acid Concentration

Percent Ionization (Example)
A M solution of HNO2 contains 3.7×10-3 M H+(aq). Calculate the percent ionization. = = 11% This means that 11% of the acid will dissociate in water. 0.035 M 3.7×10-3 M

(Self-) Auto-ionization of Water
According to Bronsted Lowry, H2O can act as either an acid or a base Auto-ionization: One water molecule can donate a proton to another water molecule Extremely rapid reaction and no molecule remains ionized for long At room temperature 1 out of every 109 molecule are ionized at a given instant Water is a nonelectrolyte and consists almost entirely of H2O molecules H2O(l) + H2O(l) ↔ H3O+ + OH-

Auto-ionization of Water
H2O(l) + H2O(l) ↔ H3O+ + OH- Auto-ionization of water is an equilibrium process (use Kw - ion product constant) Kw = [H3O+][OH-] Also written as Kw = [H+][OH-] At 25°C, Kw =1.4×10-14

1.4×10-14 = [H+][OH-] In basic solutions, [OH-] > [H+] In acidic solutions, [H+] > [OH-] In neutral solutions, [H+] = [OH-]

1.4×10-14 = [H+][OH-] If the concentration of one ion is known, you can solve for the concentration of the other ion Example: Calculate the concentration of H+ in a solution in which the concentration of OH- is 0.010M. 1.4×10-14 = [H+][0.010] [H+] = 1.0×10-12 M

Relating pKw to pKa and pKb
Acid or base dissociation constants are sometimes expressed as pKa and Kb. pKa = –logKa pKb = -logKb pKw = 14 = pKa+ pKb

Polyprotic Acids Acids with more than one ionizable H+ ion
The acid-dissociation constants are Ka1, Ka2, etc… The first proton is most easily removed As protons are removed, it becomes more and more difficult to remove protons Ka1>Ka2>Ka3…. H2SO3(aq) ↔ H+(aq) + HSO3-(aq) Ka1 = 1.7×10-2 HSO3-(aq) ↔ H+(aq) + SO3-2(aq) Ka2 = 6.4×10-8

Polyprotic Acids To calculate the overall K for the reaction, treat it as a multi-step equilibrium Overall reaction… H2SO3(aq) ↔ H+(aq) + HSO3-(aq) Ka1 = 1.7×10-2 HSO3-(aq) ↔ H+(aq) + SO3-2(aq) Ka2 = 6.4×10-8 H2SO3(aq) ↔ 2H+(aq) + SO3-2(aq) Ka = Ka1 × Ka2 Ka = (1.7×10-2)(6.4×10-8) = 1.1×10-9

Polyprotic Acids Strong acids (ex H2SO4) completely ionize with the first step pH can be calculated by treating the acid as if it were a monoprotic acid (one ionizable hydrogen) If Ka values differ by a factor of 103 or more, acids can be treated as monoprotic

Polyprotic Acids Monoprotic acid = 1 ionizable H+
Diprotic acid = 2 ionizable H + Triprotic acid = 3 ionizable H + Etc…

Amphoteric (amphiprotic) Substances
Substances that can act as either acids or bases Example: H2O Can give up H+ to become OH - Can receive H + to become H3O + Example: H2PO4- Can give up H + to become HPO4-2 Can receive H + to become H3PO4

Anhydrides Acid Anhydride: combines with water to form an acid
CO2 + H2O  H2CO3 SO3 + H2O  H + + HSO4- Base Anhydride: combines with water to form a base CaO + H2O  Ca(OH)2 Na2O + H2O  2Na+ + 2OH-

Acids Bases and Salts #1 If a salt is composed of the conjugates of a strong acid and strong base, the solution will be neutral. #2 If a salt is composed of the conjugates of a weak base and a strong acid, its solution will be acidic. #3 If a salt is composed of the conjugates of a strong base and a weak acid, its solution will be basic. #4 If a salt is composed of the conjugates of a weak base and a weak acid, the pH of its solution will depend on the relative strengths of the conjugate acid and base of the specific ions in the salt.

Strong Acid + Strong Base
If a salt is composed of the conjugates of a strong acid and strong base, the solution will be neutral. HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) The Na+ and Cl- ions do not further ionize in water

HCl(aq) + NH3(aq)  NH4Cl(s) NH4Cl(s)  NH4+(aq) + Cl-(aq)
Strong Acid + Weak Base If a salt is composed of the conjugates of a strong acid and weak base, the solution will be acidic. HCl(aq) + NH3(aq)  NH4Cl(s) The NH4Cl ionizes in water to produce some H+ ions (the Cl - ions do not react in water) NH4Cl(s)  NH4+(aq) + Cl-(aq) NH4+ + H2O  NH3+ + H+ Solution Is acidic

Weak Acid + Strong Base If a salt is composed of the conjugates of a weak acid and strong base, the solution will be basic. NaOH(aq) + HC2H3O2(aq)  NaC2H3O2(s) + H2O(l) The NaC2H3O2 ionizes in water NaC2H3O2(s)  Na+(aq) + C2H3O2-(aq) The Na+ ions do not react at all, but the C2H3O2- ions react to produce OH- in solution C2H3O2-(aq) + H2O  HC2H3O2 + OH- Solution Is basic

Weak Acid + Weak Base If a salt is composed of the conjugates of a weak acid and weak base, the pH of the solution will depend on the relative strengths of the conjugate acid and base of the specific ions in the salt. The ion with the larger equilibrium constant (Ka or Kb) will have the greater influence on pH)

Vocabulary Hydrolysis
When the ions react with water to produce hydrogen or hydroxide ions in solution Example: Strong Acid/Weak Base Example: Strong Acid/Strong Base

Acid Base Salt Calculations
Solve similar to weak acid/weak base problems Solve for Ka or Kb depending on if the salt makes an acidic (Ka) or basic (Kb) solution Use the ion of the strong acid or strong base for equilibrium equation

Example (Acid Base Salt)
What is the pH of a 0.10 M solution of acetic acid, NaC2H3O2. The Ka of HC2H3O2 is 1.8 x Step #1: We will use Kb because the solution will be basic. First find the Kb for HC2H3O2. C2H3O2- + H2O  HC2H3O2 + OH-

What is the pH of a 0.10 M solution of acetic acid, NaC2H3O2. The Ka of HC2H3O2 is 1.8 x C2H3O2- + H2O  HC2H3O2 + OH- Kw = KaKb 1.0 × = (1.8x10-5)(Kb) Kb = 5.6 × 10-10

Step #2: Write the equilibrium expression. C2H3O2- + H2O  HC2H3O2 + OH- [HC2H3O2][OH-] [C2H3O2-] Kb =

What is the pH of a 0.10 M solution of acetic acid, NaC2H3O2. The Ka of HC2H3O2 is 1.8 x Step #3: ICE HC2H3O C2H3O OH- I C E 0.10 - x +x +x x x x

Step #4: Solve for OH-. C2H3O2- + H2O  HC2H3O2 + OH- [x][x] [0.10-x] [x][x] [0.10] 5.6×10-10 = = = 7.5×10-6

Step #4: Solve for OH-. C2H3O2- + H2O  HC2H3O2 + OH- x = 7.5×10-6 = [OH-]

Step #5: Solve for pH. [OH-] = 7.5×10-6 pOH = -log[OH-] = -log[7.5×10-6] pOH = 5.1 14 = pH + pOH 14 = pH + 5.1 pH = 8.9

Common Ion Effect HC2H3O2  C2H3O2 - + H+
Use LeChatlier’s Principle to determine how the addition of a substance will affect the equilibrium The addition of the acetate ion (adding soluble salt) causes the equilibrium to shift to the LEFT The hydrogen concentration will DECREASE The acetic acid ionizes less with the addition of the acetate ion than it would alone in solution HC2H3O2  C2H3O H+

Buffers A solution with a very stable pH
You can add an acid or base to a buffer solution without it greatly affecting the pH of the solution Consists a weak acid-base conjugate pair Usually a weak acid or base with the salt of that acid or base Resists changes in addition of strong acid or base Example: blood (pH of 7.4)

Buffers Buffer Capacity – the amount of acid or base the buffer can neutralize before the pH begins to change to an appreciable degree pH Range (of a buffer) – the range over which the buffer acts effectively

Henderson-Hasselbalch Equation
Reminder: pKa = –logKa and pKb = -logKb Given on AP Cheat Sheet!

Example (Henderson-Hasselbalch)
What is the pH of a buffer solution with concentrations of 0.20M HC2H3O2 and 0.50 M C2H3O2-? The acid dissociation constant for HC2H3O2 is 1.8×10-5. pH = -log(1.8×10-5) + log(0.50/0.20) pH = = 5.1

Example 2 (Henderson-Hasselbalch)
What is the pH of a buffer solution with concentrations of 0.20M HC2H3O2 and 0.20 M C2H3O2-? The acid dissociation constant for HC2H3O2 is 1.8×10-5. pH = -log(1.8×10-5) + log(0.20/0.20) pH = = 4.7

Choosing a Buffer When concentrations of acid and conjugate base are the same, pH = pKa (and pOH = pKb) When you want to prepare a buffer with a desired pH, choose an acid with a pKa close to the desired pH.

Example of a salt of the weak acid
Acid/Salt Buffering Pairs The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH) Weak Acid Formula of the acid Example of a salt of the weak acid Hydrofluoric HF KF – Potassium fluoride Formic HCOOH KHCOO – Potassium formate Benzoic C6H5COOH NaC6H5COO – Sodium benzoate Acetic CH3COOH NaH3COO – Sodium acetate Carbonic H2CO3 NaHCO3 - Sodium bicarbonate Propanoic HC3H5O2 NaC3H5O2 - Sodium propanoate Hydrocyanic HCN KCN - potassium cyanide

Base/Salt Buffering Pairs
The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3) Base Formula of the base Example of a salt of the weak acid Ammonia NH3 NH4Cl - ammonium chloride Methylamine CH3NH2 CH3NH2Cl – methylammonium chloride Ethylamine C2H5NH2 C2H5NH3NO3 - ethylammonium nitrate Aniline C6H5NH2 C6H5NH3Cl – aniline hydrochloride Pyridine C5H5N C5H5NHCl – pyridine hydrochloride

Titration A base of known concentration is slowly added to an acid of unknown concentration (or vice versa) to reach neutralization Indicators are added to the solution that change color to signal the equivalence point Equivalence point – point at which stoichiometrically equivalent quantities of acid and base have been brought together - neutralization

Titration Curves Titration Curve graph of the pH as acid or base is added to the solution

Indicators Indicators are added to the solution that change color to signal the equivalence point Equivalence point – point at which stoichiometrically equivalent quantities of acid and base have been brought together – neutralization Different indicators change at different pH levels

Indicators Indicator Low pH color High pH color Transition pH range
Gentian violet (Methyl violet 10B) yellow 0.0–2.0 blue-violet Leucomalachite green (first transition) green Leucomalachite green (second transition) 11.6–14 colorless Thymol blue (first transition) red 1.2–2.8 Thymol blue (second transition) 8.0–9.6 blue Methyl yellow 2.9–4.0 Bromophenol blue 3.0–4.6 purple Congo red 3.0–5.0 Methyl orange 3.1–4.4 orange Bromocresol green 3.8–5.4 Methyl red 4.4–6.2 4.5–5.2 Azolitmin 4.5–8.3 Bromocresol purple 5.2–6.8 Bromothymol blue 6.0–7.6 Phenol red 6.8–8.4 Neutral red 6.8–8.0 Naphtholphthalein colorless to reddish 7.3–8.7 greenish to blue Cresol Red 7.2–8.8 reddish-purple Phenolphthalein 8.3–10.0 fuchsia Thymolphthalein 9.3–10.5 Alizarine Yellow R 10.2–12.0 Litmus

Selection of Indicators

Titration Strong acid-strong base titration
Weak acid-strong base titration Strong acid-weak base titration Polyprotic acid-strong base titration

Strong Acid - Strong Base
Endpoint is at pH 7 A solution that is 0.10 M HCl is titrated with 0.10 M NaOH

Weak Acid - Strong Base A solution that is 0.10 M CH3COOH
Endpoint is above pH 7 A solution that is 0.10 M CH3COOH is titrated with 0.10 M NaOH

Strong Acid – Weak Base A solution that is 0.10 M HCl is titrated with
Endpoint is below pH 7 A solution that is 0.10 M HCl is titrated with 0.10 M NH3

Polyprotic Acids The titration curve will have as many bumps as there are hydrogen ions to give up This curve has 2 bumps so it represents a diprotic acid

Acids & Bases Unit 13.

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