# Chapter 19 – Ionic Equilibria in Aqueous Solutions

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Chapter 19 – Ionic Equilibria in Aqueous Solutions
19.1 – Equilibria of Acid-Base Buffer Systems 19.2 – Acid-base Titration Curves 19.3 – Equilibria of Slightly Soluble Ionic Compounds 19.4 – Equilibria Involving Complex Ions 19.5 – Application of Ionic Equilibria to Chemical Analysis

19.1 - Equilibria of Acid-Base Buffer Systems

19.1 - Equilibria of Acid-Base Buffer Systems
Table 19.1 The Effect of Added Acetate Ion on the Dissociation of Acetic Acid [CH3COOH]initial [CH3COO-]added % Dissociation* pH 0.10 0.00 0.050 0.15 1.3 2.89 0.036 4.44 0.018 4.74 0.012 4.92 * % Dissociation = [CH3COOH]dissoc [CH3COOH]initial x 100

19.1 - Equilibria of Acid-Base Buffer Systems
Figure 19.3 How a buffer works. Buffer after addition of H3O+ H2O + CH3COOH H3O+ + CH3COO- Buffer with equal concentrations of conjugate base and acid Buffer after addition of OH- CH3COOH + OH H2O + CH3COO- H3O+ OH-

19.1 - Equilibria of Acid-Base Buffer Systems
Sample Problem 19.1 Calculating the Effect of Added H3O+ or OH- on Buffer pH PROBLEM: Calculate the pH: (a) of a buffer solution consisting of 0.50M CH3COOH and 0.50M CH3COONa (b) after adding 0.020mol of solid NaOH to 1.0L of the buffer solution in part (a) (c) after adding 0.020mol of HCl to 1.0L of the buffer solution in part (a) Ka of CH3COOH = 1.8x10-5. (Assume the additions cause negligible volume changes. PLAN: We know Ka and can find initial concentrations of conjugate acid and base. Make assumptions about the amount of acid dissociating relative to its initial concentration. Proceed step-wise through changes in the system. SOLUTION: CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Concentration (M) (a) Initial Change Equilibrium 0.50 - 0.50 - x - + x + x 0.50-x - 0.50 +x x

19.1 - Equilibria of Acid-Base Buffer Systems
Sample Problem 19.1 Calculating the Effect of Added H3O+ and OH- on Buffer pH continued (2 of 4) [H3O+] = x [CH3COOH]equil ≈ 0.50M [CH3COO-]initial ≈ 0.50M [H3O+][CH3COO-] [CH3COOH] [H3O+] = x = Ka [CH3COO-] [CH3COOH] Ka = = 1.8x10-5M Check the assumption: 1.8x10-5/0.50 X 100 = 3.6x10-3 % (b) [OH-]added = 0.020 mol 1.0L soln = 0.020M NaOH CH3COOH(aq) + OH-(aq) CH3COO-(aq) + H2O (l) Concentration (M) Before addition 0.50 - 0.50 - Addition - 0.020 - - After addition 0.48 0.52 -

19.1 - Equilibria of Acid-Base Buffer Systems
Sample Problem 19.1 Calculating the Effect of Added H3O+ and OH- on Buffer pH continued (3 of 4) Set up a reaction table with the new values. CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Concentration (M) Initial 0.48 - 0.52 Change - x - + x + x Equilibrium 0.48 -x - 0.52 +x x [H3O+] = 1.8x10-5 0.48 0.52 = 1.7x10-5 pH = 4.77 0.020 mol 1.0L soln (c) [H3O+]added = = 0.020M H3O+ Before addition 0.50 - 0.50 - Addition - 0.020 - - After addition 0.48 0.52 -

19.1 - Equilibria of Acid-Base Buffer Systems
Sample Problem 19.1 Calculating the Effect of Added H3O+ and OH- on Buffer pH continued (4 of 4) Set up a reaction table with the new values. CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Concentration (M) Initial 0.52 - 0.48 Change - x - + x + x Equilibrium 0.52 -x - 0.48 +x x [H3O+] = 1.8x10-5 0.48 0.52 = 2.0x10-5 pH = 4.70

19.1 - Equilibria of Acid-Base Buffer Systems
The Henderson-Hasselbalch Equation HA + H2O H3O+ + A- Ka = [H3O+] [A-] [HA] Ka [HA] [A-] [H3O+] = [A-] [HA] - log[H3O+] = - log Ka + log pH = pKa + log [base] [acid]

19.1 - Equilibria of Acid-Base Buffer Systems
Buffer capacity is the ability to resist pH change. The more concentrated the components of a buffer, the greater the buffer capacity. The pH of a buffer is distinct from its buffer capacity. A buffer has the highest capacity when the component concentrations are equal. Buffer range is the pH range over which the buffer acts effectively. Buffers have a usable range within ± 1 pH unit of the pKa of its acid component.

19.1 - Equilibria of Acid-Base Buffer Systems
Figure 19.4 The relation between buffer capacity and pH change.

19.1 - Equilibria of Acid-Base Buffer Systems
Sample Problem 19.2 Preparing a Buffer PROBLEM: An environmental chemist needs a carbonate buffer of pH to study the effects of the acid rain on limsetone-rich soils. How many grams of Na2CO3 must she add to 1.5L of freshly prepared 0.20M NaHCO3 to make the buffer? Ka of HCO3- is 4.7x10-11. PLAN: We know the Ka and the conjugate acid-base pair. Convert pH to [H3O+], find the number of moles of carbonate and convert to mass. SOLUTION: Ka = [CO32-][H3O+] [HCO3-] HCO3-(aq) + H2O(l) CO32-(aq) + H3O+(aq) 4.7x10-11 = [CO32-](0.20) 1.0x10-10 pH = 10.00; [H3O+] = 1.0x10-10 [CO32-] = 0.094M moles of Na2CO3 = (1.5L)(0.094mols/L) = 0.14 0.14 moles 105.99g mol = 15 g Na2CO3

19.2 – Acid-base Titration Curves
Colors and approximate pH range of some common acid-base indicators. Figure 19.5 pH

Curve for a strong acid-strong base titration
Figure 19.7 Curve for a strong acid-strong base titration

19.2 – Acid-base Titration Curves
Titration of 40.00mL of M HPr with M NaOH Figure 19.8 Curve for a weak acid-strong base titration pH = 8.80 at equivalence point pKa of HPr = 4.89 methyl red [HPr] = [Pr-]

19.2 – Acid-base Titration Curves
Sample Problem 19.3 Calculating the pH During a Weak Acid- Strong Base Titration PROBLEM: Calculate the pH during the titration of mL of M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of M NaOH: (a) 0.00mL (b) mL (c) mL (d) mL PLAN: The amounts of HPr and Pr- will be changing during the titration. Remember to adjust the total volume of solution after each addition. SOLUTION: (a) Find the starting pH using the methods of Chapter 18. Ka = [Pr-][H3O+]/[HPr] [Pr-] = x = [H3O+] [Pr-] = x = [H3O+] x = 1.1x10-3 ; pH = 2.96 (b) HPr(aq) + OH-(aq) Pr-(aq) + H2O (l) Amount (mol) Before addition - - Addition - - - After addition -

19.2 – Acid-base Titration Curves
Sample Problem 19.3 Calculating the pH During a Weak Acid- Strong Base Titration continued [H3O+] = 1.3x10-5 mol mol = 4.3x10-6M pH = 5.37 (c) When 40.00mL of NaOH are added, all of the HPr will be reacted and the [Pr -] will be ( mol) ( L) + ( L) = M Ka x Kb = Kw Kb = Kw/Ka = 1.0x10-14/1.3x10-5 = 7.7x10-10 [H3O+] = Kw / = 1.6x10-9M pH = 8.80 (d) 50.00mL of NaOH will produce an excess of OH-. mol XS base = (0.1000M)( L L) = mol [H3O+] = 1.0x10-14/ = 9.0x10-11M M = pH = 12.05

Curve for a weak base-strong acid titration
Titration of 40.00mL of M NH3 with M HCl Figure 19.9 pKa of NH4+ = 9.25 Curve for a weak base-strong acid titration pH = 5.27 at equivalence point

Titration of 40.00mL of 0.1000M H2SO3 with 0.1000M NaOH
Figure 19.10 Curve for the titration of a weak polyprotic acid. pKa = 7.19 pKa = 1.85 Titration of 40.00mL of M H2SO3 with M NaOH

19.3 – Equilibria of Slightly Soluble Ionic Compounds
Sample Problem 19.4 Writing Ion-Product Expressions for Slightly Soluble Ionic Compounds PROBLEM: Write the ion-product expression for each of the following: (a) Magnesium carbonate (b) Iron (II) hydroxide (c) Calcium phosphate (d) Silver sulfide PLAN: Write an equation which describes a saturated solution. Take note of the sulfide ion produced in part (d). SOLUTION: (a) MgCO3(s) Mg2+(aq) + CO32-(aq) Ksp = [Mg2+][CO32-] (b) Fe(OH)2(s) Fe2+(aq) + 2OH- (aq) Ksp = [Fe2+][OH-] 2 (c) Ca3(PO4)2(s) Ca2+(aq) + 2PO43-(aq) Ksp = [Ca2+]3[PO43-]2 (d) Ag2S(s) Ag+(aq) + S2-(aq) S2-(aq) + H2O(l) HS-(aq) + OH-(aq) Ksp = [Ag+]2[HS-][OH-] Ag2S(s) + H2O(l) Ag+(aq) + HS-(aq) + OH-(aq)

19.3 – Equilibria of Slightly Soluble Ionic Compounds
Sample Problem 19.5 Determining Ksp from Solubility PROBLEM: (a) Lead (II) sulfate is a key component in lead-acid car batteries. Its solubility in water at 250C is 4.25x10-3g/100mL solution. What is the Ksp of PbSO4? (b) When lead (II) fluoride (PbF2) is shaken with pure water at 250C, the solubility is found to be 0.64g/L. Calculate the Ksp of PbF2. PLAN: Write the dissolution equation; find moles of dissociated ions; convert solubility to M and substitute values into solubility product constant expression. SOLUTION: PbSO4(s) Pb2+(aq) + SO42-(aq) (a) Ksp = [Pb2+][SO42-] 1000mL L 4.25x10-3g 100mL soln 303.3g PbSO4 mol PbSO4 = 1.40x10-4M PbSO4 Ksp = [Pb2+][SO42-] = (1.40x10-4)2 = 1.96x10-8

19.3 – Equilibria of Slightly Soluble Ionic Compounds
Sample Problem 19.5 Determining Ksp from Solubility continued (b) PbF2(s) Pb2+(aq) + 2F-(aq) Ksp = [Pb2+][F-]2 0.64g L soln 245.2g PbF2 mol PbF2 = 2.6x10-3 M Ksp = (2.6x10-3)(5.2x10-3)2 = 7.0x10-8

19.3 – Equilibria of Slightly Soluble Ionic Compounds
Table Solubility-Product Constants (Ksp) of Selected Ionic Compounds at 250C Name, Formula Ksp Aluminum hydroxide, Al(OH)3 3 x 10-34 Cobalt (II) carbonate, CoCO3 1.0 x 10-10 Iron (II) hydroxide, Fe(OH)2 4.1 x 10-15 Lead (II) fluoride, PbF2 3.6 x 10-8 Lead (II) sulfate, PbSO4 1.6 x 10-8 Mercury (I) iodide, Hg2I2 4.7 x 10-29 Silver sulfide, Ag2S 8 x 10-48 Zinc iodate, Zn(IO3)2 3.9 x 10-6

19.3 – Equilibria of Slightly Soluble Ionic Compounds
Sample Problem 19.6 Determining Solubility from Ksp PROBLEM: Calcium hydroxide (slaked lime) is a major component of mortar, plaster, and cement, and solutions of Ca(OH)2 are used in industry as a cheap, strong base. Calculate the solubility of Ca(OH)2 in water if the Ksp is 6.5x10-6. PLAN: Write out a dissociation equation and Ksp expression; Find the molar solubility (S) using a table. SOLUTION: Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Ksp = [Ca2+][OH-]2 Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Concentration (M) Initial - Change - +S + 2S Equilibrium - S 2S Ksp = (S)(2S)2 S = = 1.2x10x-2M

19.3 – Equilibria of Slightly Soluble Ionic Compounds
Table Relationship Between Ksp and Solubility at 250C No. of Ions Formula Cation:Anion Ksp Solubility (M) 2 MgCO3 1:1 3.5 x 10-8 1.9 x 10-4 2 PbSO4 1:1 1.6 x 10-8 1.3 x 10-4 2 BaCrO4 1:1 2.1 x 10-10 1.4 x 10-5 3 Ca(OH)2 1:2 5.5 x 10-6 1.2 x 10-2 3 BaF2 1:2 1.5 x 10-6 7.2 x 10-3 3 CaF2 1:2 3.2 x 10-11 2.0 x 10-4 3 Ag2CrO4 2:1 2.6 x 10-12 8.7 x 10-5

19.3 – Equilibria of Slightly Soluble Ionic Compounds
Figure 19.12 The effect of a common ion on solubility PbCrO4(s) Pb2+(aq) + CrO42-(aq) CrO42- added PbCrO4(s) Pb2+(aq) + CrO42-(aq)

19.3 – Equilibria of Slightly Soluble Ionic Compounds
Sample Problem 19.7 Calculating the Effect of a Common Ion on Solubility PROBLEM: In Sample Problem 19.6, we calculated the solubility of Ca(OH)2 in water. What is its solubility in 0.10M Ca(NO3)2? Ksp of Ca(OH)2 is 6.5x10-6. PLAN: Set up a reaction equation and table for the dissolution of Ca(OH)2. The Ca(NO3)2 will supply extra [Ca2+] and will relate to the molar solubility of the ions involved. SOLUTION: Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Concentration(M) Initial - 0.10 Change - +S +2S Equilibrium - S 2S Ksp = 6.5x10-6 = ( S)(2S)2 = (0.10)(2S)2 S << 0.10 Check the assumption: S = = 4.0x10-3 0.10M 4.0x10-3 x 100 = 4.0%

19.3 – Equilibria of Slightly Soluble Ionic Compounds
Sample Problem 19.8 Predicting the Effect on Solubility of Adding Strong Acid PROBLEM: Write balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of these ionic compounds: (a) Lead (II) bromide (b) Copper (II) hydroxide (c) Iron (II) sulfide PLAN: Write dissolution equations and consider how strong acid would affect the anion component. SOLUTION: (a) PbBr2(s) Pb2+(aq) + 2Br-(aq) No effect. (b) Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) OH- is the anion of water, which is a weak acid. Therefore it will shift the solubility equation to the right and increase solubility. (c) FeS(s) Fe2+(aq) + S2-(aq) S2- is the anion of a weak acid and will react with water to produce OH-. FeS(s) + H2O(l) Fe2+(aq) + HS-(aq) + OH-(aq) Both weak acids serve to increase the solubility of FeS.

19.3 – Equilibria of Slightly Soluble Ionic Compounds
Sample Problem 19.9 Predicting Whether a Precipitate Will Form PROBLEM: A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100L of 0.30M Ca(NO3)2 is mixed with 0.200L of 0.060M NaF? PLAN: Write out a reaction equation to see which salt would be formed. Look up the Ksp valus in a table. Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < Ksp. Remember to consider the final diluted solution when calculating concentrations. SOLUTION: CaF2(s) Ca2+(aq) + 2F-(aq) Ksp = 3.2x10-11 mol Ca2+ = L(0.30mol/L) = 0.030mol [Ca2+] = 0.030mol/0.300L = 0.10M mol F- = L(0.060mol/L) = 0.012mol [F-] = 0.012mol/0.300L = 0.040M Q = [Ca2+][F-]2 = (0.10)(0.040)2 = 1.6x10-4 Q is >> Ksp and the CaF2 WILL precipitate.

19.5 – Ionic Equilibria in Chemical Analysis
Sample Problem 19.12 Separating Ions by Selective Precipitation PROBLEM: A solution consists of 0.20M MgCl2 and 0.10M CuCl2. Calculate the [OH-] that would separate the metal ions as their hydroxides. Ksp of Mg(OH)2= is 6.3x10-10; Ksp of Cu(OH)2 is 2.2x10-20. PLAN: Both precipitates are of the same ion ratio, 1:2, so we can compare their Ksp values to determine which has the greater solubility. It is obvious that Cu(OH)2 will precipitate first so we calculate the [OH-] needed for a saturated solution of Mg(OH)2. This should ensure that we do not precipitate Mg(OH)2. Then we can check how much Cu2+ remains in solution. SOLUTION: Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) Ksp = 6.3x10-10 Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) Ksp = 2.2x10-20 [OH-] needed for a saturated Mg(OH)2 solution = = 5.6x10-5M

19.5 – Ionic Equilibria in Chemical Analysis
Sample Problem 19.12 Separating Ions by Selective Precipitation continued Use the Ksp for Cu(OH)2 to find the amount of Cu remaining. [Cu2+] = Ksp/[OH-]2 = 2.2x10-20/(5.6x10-5)2 = 7.0x10-12M Since the solution was 0.10M CuCl2, virtually none of the Cu2+ remains in solution.

19.5 – Ionic Equilibria in Chemical Analysis
Figure 19.17 The general procedure for separating ions in qualitative analysis. Add precipitating ion Add precipitating ion Centrifuge Centrifuge

19.5 – Ionic Equilibria in Chemical Analysis
A qualitative analysis scheme for separating cations into five ion groups. Figure 19.18 Acidify to pH 0.5; add H2S Centrifuge Add NH3/NH4+ buffer(pH 8) Centrifuge Add (NH4)2HPO4 Centrifuge Add 6M HCl Centrifuge