ACIDS AND BASES Chapter 14.

ACIDS AND BASES Chapter 14

AQUEOUS EQUILIBRIA: ACIDS AND BASES
Arrhenius Acids and Bases Acids cause [H+] to increase, bases cause [OH-] to increase Bronsted-Lowry Acids and Bases H+ /proton Donor (acid) and H+/proton Acceptor (base). The definition we will use. Lewis Acid and Bases Acids accept electron pair Bases donate electron pair

BRONSTED-LOWRY Using this theory, you should be able to write weak acid/base dissociation equations and identify acid, base, conjugate acid and conjugate base. Conjugate acid-base pair - A pair of compounds that differ by the presence of one H+ unit.

BRONSTED-LOWRY HNO3 + H2O → H3O+ + NO3− neutral compound as an acid
acid base CA CB neutral compound as an acid NH4+ + H2O → H3O+ + NH3 acid base CA CB cation as an acid H2PO4− + H2O → H3O+ + HPO42− acid base CA CB anion as an acid

HYDRONIUM ION

BRONSTED-LOWRY Notice the formation of H3O+ - called the hydronium ion. This species indicates that the solution is acidic. hydronium, H3O+ - H+ riding piggy-back on a water molecule; water is polar and the + charge of the “naked” proton is greatly attracted to the oxygen.

BRONSTED-LOWRY NH3 + H2O → NH4+ + OH− neutral compound
base acid CA CB neutral compound CO32− + H2O → HCO3− + OH− base acid CA CB anion PO43− + H2O → HPO42− + OH− base acid CA CB

BRONSTED-LOWRY Notice the formation of OH− in each of the alkaline examples. This species is named the hydroxide ion. It indicates that the resulting solution is basic.

PRACTICE ONE Identify the acid on the left and its conjugate base (CB) on the right and identify the base on the left and its conjugate acid (CA) on the right. HBr + NH3 → NH4+ + Br− What is the conjugate base of H2S? What is the conjugate acid of NO3-?

BRONSTED-LOWRY ACIDS DONATE ONLY ONE PROTON AT A TIME.
Monoprotic - acids donating one H+ (ex. HC2H3O2) Diprotic - acids donating two H+'s (ex. H2C2O4) Polyprotic - acids donating many H+'s (ex. H3PO4) polyprotic bases - accept more than one H+; anions with −2 and −3 charges (ex. PO43− ; HPO42−) Amphiprotic or amphoteric - molecules or ions that can behave as EITHER acids or bases; water, anions of weak acids (look at the examples above - sometimes water was an acid, sometimes it acted as a base)

PRACTICE TWO Write the simple dissociation reaction (omitting water) for each of the following: a. Hydrochloric acid (HCl) b. Acetic acid (HC2H3O2) c. The ammonium ion (NH4+)

PRACTICE TWO d. The anilinium ion (C6H5NH3+)
e. The hydrated aluminum(III) ion [Al(H2O)6]3+

ACID AND BASE STRENGTHS
Based on extent of dissociation. Strong Acids/Bases Dissociate nearly 100% Very large dissociation constant or K values EQ lies far to the right HA  H+ + A- Weak acids/bases dissociate very little dissociation constant is very small Equilibrium lies far to the left HA ↔ H+ + A-

Memorize these Hydrohalic acids: HCl, HBr, HI—note HF is missing. Nitric: HNO3 Sulfuric: H2SO4 Perchloric: HClO4 Permanganic: HMnO4 Chromic: H2CrO4 The more oxygen present in the ion, the stronger the acid

Taken from State University of West Georgia Chemistry Dept.

Strong Bases Certain bases are known as strong bases. These are bases that fully ionize when placed in water. LiOH, lithium hydroxide NaOH, sodium hydroxide KOH, potassium hydroxide Ca(OH)2, calcium hydroxide Sr(OH)2, strontium hydroxide Ba(OH)2, barium hydroxide Alkaline earth oxides. Lime (CaO)

RULE The stronger the acid, the weaker its conjugate base.
HCl → H+ + Cl- The weaker the acid, the stronger its conjugate base. H2O ↔ H+ + OH-

The vast majority of acid/bases are weak. That means an equilibrium is established and it lies far to the left (reactant favored). Equilibrium expression for acids = Ka (the acid dissociation constant). Set up the same way as any other EQ expression.

For weak acid reactions: HA + H2O ↔ H3O+ + A− Ka = [H3O+][A−] < < 1 [HA]

HA  H+ + A- Extent of dissociation depends on H-A bond strength and Electronegativity (or stability of negative charge) on A. This explanation works for Halogen acids, Organic oxoacids, or Inorganic oxoacids. Oxidation # of the Halide is not necessary.

PRACTICE THREE Write the Ka expression for acetic acid.
(Note: Water is a pure liquid and is thus, left out of the equilibrium expression.)

Weak bases (bases without OH−) react with water to produce a hydroxide ion. Common examples of weak bases are ammonia (NH3), methylamine (CH3NH2), and ethylamine (C2H5NH2). The lone pair on N forms a bond with a H+. Most weak bases involve N. The equilibrium expression for bases is known as the Kb.

For weak base reactions: B + H2O → HB+ + OH− Kb = [HB+][OH−] << 1 [B]

PRACTICE FOUR Write the Kb expression for ammonia.

Notice that Ka and Kb expressions look very similar. The difference is that a base produces the hydroxide ion in solution, while the acid produces the hydronium ion in solution. H+ and H3O+ are both equivalent terms here. This has become an accepted practice.

PRACTICE FIVE Using table 14.2, arrange the following species according to their strength as bases: H2O, F−, Cl−, NO2−, and CN−.

DISSOCIATION OF WATER H2O + H2O ↔ H3O+ + OH-
The equilibrium expression is products over reactants. The molarity for the water is a constant at any specific temperature. Kw = [H3O1+] [OH1-] The quantity on the right hand side of the equation is formally defined as Kw.

DISSOCIATION OF WATER Knowing this value allows us to calculate the OH− and H+ concentration for various situations. [OH−] = [H+] solution is neutral (in pure water, each of these is 1.0 × 10−7) [OH−] > [H+] solution is basic [OH−] < [H+] solution is acidic

2H2O(l) → H3O+(aq) + OH−(aq)
PRACTICE SIX At 60°C, the value of Kw is 1 × 10−13. a. Using Le Chatelier’s principle, predict whether the reaction below is exothermic or endothermic. 2H2O(l) → H3O+(aq) + OH−(aq) b. Calculate [H+] and [OH−] in a neutral solution at 60°C.

DISSOCIATION OF WATER Equilibrium constants exist then for both acid dissociation and base. (Ka and Kb) The higher the Ka, the stronger the acid and the higher the Kb, the stronger the base. Ka and Kb are related. Kw = KaKb

DISSOCIATION OF WATER As Ka gets larger the strength of the acid gets higher, but Kb must fall. Therefore the stronger the acid, the weaker the conjugate base. It can now be said that the conjugate base (acid) of a weak acid (base) is a weak base (acid) and the conjugate base (acid) of a strong acid (base) is a worthless base (acid).

THE pH SCALE

Power of Hydronium (Hydrogen)
THE pH SCALE pH = -log [H+] Power of Hydronium (Hydrogen) pOH = - log [OH-] pH + pOH = 14 [H+][OH-] = 1 x 10-14

THE pH SCALE Similar log scales are used for representing other quantities: pK = -logK Will be using this in the lab Monday A weak acid has a pKa value in the approximate range −2 to 12 in water. Acids with a pKa value of less than about −2 are said to be strong acids

THE pH SCALE Reporting the correct number of sig. figs on a pH is problematic since it is a logarithmic scale. The rule is to report as many decimal places on a pH as there are in the least accurate measurement you are given. Example: The problem states a 1.15 M solution. That is your cue to report a pH with 3 decimal places. If the problem had stated a 1.2 M solution, then you would report your calculated pH to 2 decimal places.

PRACTICE SEVEN a. 1.0 × 10−5M OH− b. 1.0 × 10−7 M OH− c. 10.0M H+
Calculate either the [H+] or [OH−] from the information given for each of the following solutions at 25°C, and state whether the solution is neutral, acidic, or basic. a × 10−5M OH− b × 10−7 M OH− c M H+

PRACTICE EIGHT Calculate pH and pOH for each of the following solutions at 25°C. a. 1.0 × 10−3 M OH− b. 1.0 M H+

PRACTICE NINE The pH of a sample of human blood was measured to be 7.41 at 25°C. Calculate pOH, [H+], and [OH−] for the sample.

pH WITH STRONG ACIDS Need to focus on major species in solution – meaning those in relatively large amounts. A 1.0M of HCl contains no HCl. It contains H+ and Cl- ions. Also contains H2O. Amount of OH- will be minimal (minor species).

pH WITH STRONG ACIDS When attacking acid-base problems:
Write the major species in the solution. Example: Calculate the pH of 1.0M HCl. Major Species: H+, Cl-, H2O Want to calculate pH, so focus on those that can produce H+. That would be H+ from HCl and from H2O.

PRACTICE TEN a. Calculate the pH of 0.10M HNO3.
b. Calculate the pH of 1.0 × 10−10M HCl.

pH OF WEAK ACIDS Calculating pH of weak acids involves setting up an equilibrium. Always start by writing the balanced equation, setting up the acid equilibrium expression (Ka), defining initial concentrations, changes, and final concentrations in terms of x, substituting values and variables into the Ka expression and solving for x. Use the ICE TABLE method you learned in general equilibrium.

pH OF WEAK ACIDS EXAMPLE
Calculate the pH of a 1.00 × 10−4M solution of acetic acid. The Ka of acetic acid is 1.8 × 10−5. HC2H3O2 ↔ H C2H3O2− Initial × 10− Change −x +x x Equilibrium (1.00 × 10−4) – x x x

pH OF WEAK ACIDS Ka = [H+][C2H3O2-] = 1.8 x 10-5 = x2 [HC2H3O2] 1 x x You should determine that x = [H+] = 4.24 × 10-5 and that the pH = − log (4.24 × 10−4) = 4.37 (2 SF). x is negligible since they are not going to ask you to do the quadratic formula.

pH OF WEAK ACIDS Write the major species involved.
Choose the species that can produce H+ and write balanced equations for the reactions producing H+. Use Ka values to determine which one is dominant at producing H+. Write the K expression for this one. List the [initial] of the species participating. Define the change needed – define X

pH OF WEAK ACIDS 7. Write the [EQ] n terms of X.
Substitute the [EQ] into the EQ expression. Solve for X the easy way (-x is negligible). Calculate [H+] and pH

PRACTICE TWELVE The hypochlorite ion (OCl−) is a strong oxidizing agent and forms the weakly acidic hypochlorous acid (HOCl, Ka = 3.5 × 10-8). Calculate the pH of a M aqueous solution of hypochlorous acid. H+ producers: Equations: Ka values:

PRACTICE TWELVE K Expression: 5., 6., 7. ICE:

PRACTICE TWELVE 8., 9. Put in expression and solve: 10. Calculate pH

PRACTICE THIRTEEN H+ producers: Equations: Ka values:
Calculate the pH of a solution that contains 1.00 M HCN (Ka = 6.2 × 10−10) and 5.00 M HNO2 (Ka = 4.0 × 10−4). Also calculate the concentration of cyanide ion (CN−) in this solution at equilibrium. H+ producers: Equations: Ka values:

PRACTICE THIRTEEN K Expression: 5., 6., 7. ICE:

PRACTICE THIRTEEN 8., 9. Put in expression and solve: 10. Calculate pH

DO NOW Pick up handout – due Wed. Get out homework handout and notes

PERCENT DISSOCIATION Sometimes helpful to specify the amount of weak acid that has dissociated in achieving equilibrium. Percent Dissociation = [HA] amount dissociated (M) x 100 [HA] Initial concentration (M)

PRACTICE FOURTEEN Calculate the percent dissociation of acetic acid (Ka = 1.8 × 10−5) in each of the following solutions. a. 1.00M HC2H3O2

PRACTICE FOURTEEN Calculate the percent dissociation of acetic acid (Ka = 1.8 × 10−5) in each of the following solutions. b M HC2H3O2

PRACTICE FIFTEEN Lactic acid (HC3H5O3) is a waste product that accumulates in muscle tissue during exertion, leading to pain and a feeling of fatigue. In a 0.100M aqueous solution, lactic acid is 3.7% dissociated. Calculate the value of Ka for this acid.

pH OF WEAK BASES Determination of the pH of a weak base is very similar to the determination of the pH of a weak acid. Follow the same steps. Remember, however, that x is the [OH−] and taking the negative log of x will give you the pOH and not the pH!

PRACTICE SIXTEEN Calculate the pH for a 15.0M solution of NH3 (Kb = 1.8 × 10−5).

PRACTICE SEVENTEEN Calculate the pH of a 1.0M solution of methylamine (Kb = 4.38 × 10−4).

POLYPROTIC ACIDS Acids with more than one ionizable hydrogen will ionize in steps. Each dissociation has its own Ka value. The first dissociation will be the greatest and subsequent dissociations will have much smaller equilibrium constants. As each H+ is removed, the remaining acid gets weaker and therefore has a smaller Ka. As the negative charge on the acid increases, it becomes more difficult to remove the positively charged proton.

POLYPROTIC ACIDS Example: Consider the dissociation of H3PO4.
H3PO4(aq) + H2O(l) ↔ H3O+(aq) + H2PO4−(aq) Ka1 = 7.5 × 10-3 H2PO4-(aq) + H2O(l) ↔ H3O+(aq) + HPO4−2(aq) Ka2 = 6.2 × 10-8 HPO4-2(aq) + H2O(l) ↔ H3O+(aq) + PO4−3(aq) Ka3 = 4.8 × 10-13 Looking at the Ka values, it is obvious that only the first dissociation will be important in determining the pH of the solution.

In general, Ka1 >> Ka2 >> Ka3.
POLYPROTIC ACIDS In general, Ka1 >> Ka2 >> Ka3.

POLYPROTIC ACIDS Except for H2SO4, polyprotic acids have Ka2 and Ka3 values so much weaker than their Ka1 value that the 2nd and 3rd dissociation can be ignored. The [H+] obtained from this 2nd and 3rd dissociation is negligible compared to the [H+] from the 1st dissociation. Because H2SO4 is a strong acid in its first dissociation and a weak acid in its second, we need to consider both if the concentration is more dilute than 1.0M. The quadratic equation is needed to work this type of problem.

PRACTICE EIGHTEEN Calculate the pH of a 5.0M H3PO4 solution and the equilibrium concentrations of the species H3PO4, H2PO4-, HPO42-, and PO43-.

PRACTICE NINETEEN Calculate the pH of a 1.0 M H2SO4 solution.

PRACTICE TWENTY Calculate the pH of a 1.0 × 10−2M H2SO4 solution.

ACID-BASE PROPERTIES OF SALTS
Salts are produced when an acid and base react. Salts are not always neutral. Some hydrolyze with water to produce aqueous solutions with pHs other than 7.00. Neutral Salts - Salts that are formed from the cation of a strong base reacting with the anion of a strong acid are neutral. Beware of solubility issues! One salt such is NaNO3. Think about which acid reacted with which base to form the salt…if both the acid and base are strong, then the salt is neutral.

Basic Salts - Salts that are formed from the cation of a strong base reacting with the anion of a weak acid are basic. Again, beware of solubility issues! The anion hydrolyzes the water molecule to produce hydroxide ions and thus a basic solution. K2C2H3O2 should be basic since C2H3O2− is the CB of the weak acid HC2H3O2, while K+ does not hydrolyze appreciably. C2H3O2− H2O ↔ OH− HC2H3O2 strong base weak acid

Acid Salts - Salts that are formed from the cation of a weak base reacting with the anion of a strong acid are acidic. The cation hydrolyzes the water molecule to produce hydronium ions and thus an acidic solution. NH4Cl should be weakly acidic, since NH4+ hydrolyzes to give an acidic solution, while Cl− does not hydrolyze. NH H2O ↔ H3O NH3 strong acid weak base

Acid Strength Strong Weak Neutral Basic Acidic ? Strong Base Strength Weak Resulting Salt Solution

PRACTICE TWENTY-TWO Calculate the pH of a 0.30M NaF solution. The Ka value for HF is 7.2 × 10−4.

PRACTICE TWENTY-THREE
Calculate the pH of a 0.10M NH4Cl solution. The Kb value for NH3 is 1.8 × 10−5.

METAL IONS MAKING ACIDS
Highly charged metal ions can produce an acidic solution. Al+ not an acid, but Al(H2O)6+3 is a weak acid. Al(H2O)6+3 ↔ [Al(OH)(H2O)5]+2 + H+

PRACTICE TWENTY-FOUR Calculate the pH of a 0.010M AlCl3 solution. The Ka value for Al(H2O)63+ is 1.4 × 10−5.

If both the cation and the anion contribute to the pH situation, compare Ka to Kb. If Kb is larger, basic! The converse is also true.

Cation Anion Solution Example Neutral NaCl Conj. Base of a Weak Acid Basic NaF Conj. Acid of a Weak Base Acidic NH4Cl Depends on Ka and Kb Values

STEPS TO SOLVE Look at the salt and ask yourself which acid and which base reacted to form it? Ask yourself “strong or weak?” for each. Embrace the fact that “strong wins” and predict whether the salt is acidic or basic based on that victory. If you predict basic, write ↔ OH−. If you predict acidic, write ↔ H+

Relish in the fact that “strong is a spectator”. Which means the remaining ion of the salt is the reactant along with water. Write water as HOH to make it easier to see how the hydroxide or hydrogen ion was formed.

Example #1 NaOH(aq) HCl(aq)  NaCl(aq) H2O Strong Base Strong Acid Neutral Salt Example #2 NaOH(aq) HF(aq)  NaF(aq) H2O Strong Base Weak Acid Basic Salt

Example #3 NH3(aq) + HCl(aq)  NH4Cl(aq) Weak Base Strong Acid Acidic Salt

Example #4 NH3(aq) + CH3COOH(aq)  NH4OOCCH3(aq) Weak Base Weak Acid ? Salt Compare Ka to Kb Ka = 5.6 x of NH4+ Kb = 5.7 x of -OOCCH3 Salt is Neutral

Example #5 2NH3(aq) + H2CO3(aq)  (NH4)2CO3(aq) Weak Base Weak Acid ? Salt Compare Ka to Kb Ka = 5.6 x of NH4+ Kb = 1.8 x 10-4 of CO32- Salt is Basic

EXAMPLE Question: What is the qualitative pH of Fe(NO3)3? Which acid reacted? Nitric Strong or weak? Strong Which base reacted? Iron(III) hydroxide Strong or weak? Weak Strong wins! It is acidic so we write: HOH ↔ H+ to get started Strong is also a spectator so, cross out the spectator and the remaining ion is the other reactant; Fe(NO3)3 Complete hydrolysis reaction: Fe3+ + 3HOH ↔ 3H+ + Fe(OH)3

PRACTICE TWENTY-ONE Predict whether an aqueous solution of each of the following salts will be acidic, basic, or neutral. Prove with appropriate equations. 1. NaC2H3O2 2. NH4NO3 3. Al2(SO4)3

ACIDS AND BASES Chapter 14.

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