Chapter 16 Aqueous Ionic Equilibrium Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro Chapter 16 Aqueous Ionic Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall

Buffers buffers are solutions that resist changes in pH when an acid or base is added they act by neutralizing the added acid or base but just like everything else, there is a limit to what they can do, eventually the pH changes many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion Tro, Chemistry: A Molecular Approach

Making an Acid Buffer Tro, Chemistry: A Molecular Approach

How Acid Buffers Work HA(aq) + H2O(l)  A−(aq) + H3O+(aq) buffers work by applying Le Châtelier’s Principle to weak acid equilibrium buffer solutions contain significant amounts of the weak acid molecules, HA – these molecules react with added base to neutralize it you can also think of the H3O+ combining with the OH− to make H2O; the H3O+ is then replaced by the shifting equilibrium the buffer solutions also contain significant amounts of the conjugate base anion, A− - these ions combine with added acid to make more HA and keep the H3O+ constant Tro, Chemistry: A Molecular Approach

Common Ion Effect HA(aq) + H2O(l)  A−(aq) + H3O+(aq) adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts the position of equilibrium to the left this causes the pH to be higher than the pH of the acid solution lowering the H3O+ ion concentration Tro, Chemistry: A Molecular Approach

Common Ion Effect Tro, Chemistry: A Molecular Approach

Ex 16. 1 - What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0 Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? 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 HC2H3O2 + H2O  C2H3O2 + H3O+ [HA] [A-] [H3O+] initial 0.100 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach

Ex 16. 1 - What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0 Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? HC2H3O2 + H2O  C2H3O2 + 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 0.100 change equilibrium x +x +x 0.100 x 0.100 + x x Tro, Chemistry: A Molecular Approach

Ex 16. 1 - What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0 Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? Ka for HC2H3O2 = 1.8 x 10-5 determine the value of Ka since Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init solve for x [HA] [A-] [H3O+] initial 0.100 ≈ 0 change -x +x equilibrium x 0.100 x 0.100 +x Tro, Chemistry: A Molecular Approach

Ex 16. 1 - What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0 Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? [HA] [A-] [H3O+] initial 0.100 ≈ 0 change -x +x equilibrium 1.8E-5 substitute [H3O+] into the formula for pH and solve Tro, Chemistry: A Molecular Approach

Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3 Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Tro, Chemistry: A Molecular Approach

Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3 Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? 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 HF + H2O  F + H3O+ [HA] [A-] [H3O+] initial 0.14 0.071 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach

Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3 Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? HF + H2O  F + 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 0.14 0.071 change equilibrium x +x +x 0.14 x 0.071 + x x Tro, Chemistry: A Molecular Approach

Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3 Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Ka for HF = 7.0 x 10-4 determine the value of Ka since Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init solve for x [HA] [A-] [H3O+] initial 0.14 0.071 ≈ 0 change -x +x equilibrium 0.012 0.100 x 0.14 x 0.071 +x Tro, Chemistry: A Molecular Approach

Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3 Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Ka for HF = 7.0 x 10-4 check if the approximation is valid by seeing if x < 5% of [HC2H3O2]init [HA] [A2-] [H3O+] initial 0.14 0.071 ≈ 0 change -x +x equilibrium x x = 1.4 x 10-3 the approximation is valid Tro, Chemistry: A Molecular Approach

Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3 Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? [HA] [A2-] [H3O+] initial 0.14 0.071 ≈ 0 change -x +x equilibrium 0.072 1.4E-3 substitute x into the equilibrium concentration definitions and solve 0.14 x 0.071 + x x x = 1.4 x 10-3 Tro, Chemistry: A Molecular Approach

Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3 Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? [HA] [A-] [H3O+] initial 0.14 0.071 ≈ 0 change -x +x equilibrium 0.072 1.4E-3 substitute [H3O+] into the formula for pH and solve Tro, Chemistry: A Molecular Approach

Henderson-Hasselbalch Equation calculating the pH of a buffer solution can be simplified by using an equation derived from the Ka expression called the Henderson-Hasselbalch Equation the equation calculates the pH of a buffer from the Ka and initial concentrations of the weak acid and salt of the conjugate base as long as the “x is small” approximation is valid Tro, Chemistry: A Molecular Approach

How Much Does the pH of a Buffer Change When an Acid or Base Is Added? though buffers do resist change in pH when acid or base are added to them, their pH does change calculating the new pH after adding acid or base requires breaking the problem into 2 parts a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other added acid reacts with the A− to make more HA added base reacts with the HA to make more A− an equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−] Tro, Chemistry: A Molecular Approach

Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? If the added chemical is a base, write a reaction for OH− with HA. If the added chemical is an acid, write a reaction for it with A−. Construct a stoichiometry table for the reaction HC2H3O2 + OH−  C2H3O2 + H2O HA A- OH− mols Before 0.100 mols added - 0.010 mols After Tro, Chemistry: A Molecular Approach

Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L and than has 0.010 mol NaOH added to it? HC2H3O2 + OH−  C2H3O2 + H2O Fill in the table – tracking the changes in the number of moles for each component HA A- OH− mols Before 0.100 ≈ 0 mols added - 0.010 mols After 0.090 0.110 Tro, Chemistry: A Molecular Approach

Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? 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, and using the new molarities of the [HA] and [A−] HC2H3O2 + H2O  C2H3O2 + H3O+ [HA] [A-] [H3O+] initial 0.090 0.110 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach

Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + H2O  C2H3O2 + 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 0.090 0.110 change equilibrium x +x +x 0.090 x 0.110 + x x Tro, Chemistry: A Molecular Approach

Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? Ka for HC2H3O2 = 1.8 x 10-5 determine the value of Ka since Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init solve for x [HA] [A-] [H3O+] initial 0.100 ≈ 0 change -x +x equilibrium 0.090 0.110 x 0.090 x 0.110 +x Tro, Chemistry: A Molecular Approach

the approximation is valid Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? Ka for HC2H3O2 = 1.8 x 10-5 check if the approximation is valid by seeing if x < 5% of [HC2H3O2]init [HA] [A-] [H3O+] initial 0.090 0.110 ≈ 0 change -x +x equilibrium x x = 1.47 x 10-5 the approximation is valid Tro, Chemistry: A Molecular Approach

Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? [HA] [A-] [H3O+] initial 0.090 0.110 ≈ 0 change -x +x equilibrium 1.5E-5 substitute x into the equilibrium concentration definitions and solve 0.090 x 0.110 + x x x = 1.47 x 10-5 Tro, Chemistry: A Molecular Approach

Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? [HA] [A-] [H3O+] initial 0.090 0.110 ≈ 0 change -x +x equilibrium 1.5E-5 substitute [H3O+] into the formula for pH and solve Tro, Chemistry: A Molecular Approach

Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? Ka for HC2H3O2 = 1.8 x 10-5 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 0.090 0.110 ≈ 0 change -x +x equilibrium 1.5E-5 the values match Tro, Chemistry: A Molecular Approach

Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? find the pKa from the given Ka Assume the [HA] and [A-] equilibrium concentrations are the same as the initial HC2H3O2 + H2O  C2H3O2 + H3O+ Ka for HC2H3O2 = 1.8 x 10-5 [HA] [A-] [H3O+] initial 0.090 0.110 ≈ 0 change -x +x equilibrium x Tro, Chemistry: A Molecular Approach

Ex 16. 3 – Compare the effect on pH of adding 0 Ex 16.3 – Compare the effect on pH of adding 0.010 mol NaOH to a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L to adding 0.010 mol NaOH to 1.00 L of pure water? HC2H3O2 + H2O  C2H3O2 + H3O+ pKa for HC2H3O2 = 4.745 Tro, Chemistry: A Molecular Approach

Basic Buffers B:(aq) + H2O(l)  H:B+(aq) + OH−(aq) buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B+Cl− H2O(l) + NH3 (aq)  NH4+(aq) + OH−(aq) Tro, Chemistry: A Molecular Approach

Buffer Solutions Consider HOAc/OAc- to see how buffers work. ACID USES UP ADDED OH-. We know that OAc- + H2O HOAc + OH- has Kb = 5.6 x 10-10 Therefore, the reverse reaction of the WEAK ACID with added OH- has Kreverse = 1/ Kb = 1.8 x 109 Kreverse is VERY LARGE, so HOAc completely uses up the OH- !!!!

Buffer Solutions Consider HOAc/OAc- to see how buffers work. CONJUGATE BASE USES UP ADDED H+ HOAc + H2O OAc- + H3O+ has Ka = 1.8 x 10-5. Therefore, the reverse reaction of the WEAK BASE with added H+has Kreverse = 1/ Ka = 5.6 x 104 Kreverse is VERY LARGE, so OAc- completely uses up the H+ !

Buffer Solutions Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? in 1 L HOAc + H2O OAc- + H3O+ Ka = 1.8 x 10-5 [HOAc] [OAc-] [H3O+] initial change equilib 0.700 0.600 0 -x +x +x 0.700 - x 0.600 + x x

Buffer Solutions Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? HOAc + H2O OAc- + H3O+ Ka = 1.8 x 10-5 [HOAc] [OAc-] [H3O+] equilib 0.700 - x 0.600 + x x Assuming that x << 0.700 and 0.600, we have K a = 1.8 x 10 -5 [H 3 O + ](0.600) 0.700 [H3O+] = 2.1 x 10-5 and pH = 4.68

Adding an Acid to a Buffer Problem: What is the new pH when 1.00 mL of 1.00 M HCl is added to: a) 1.00 L of pure water (before HCl, pH = 7.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (a) Calculate [HCl] after adding 1.00 mL of HCl to 1.00 L of water M1 • V1 = M2 • V2 M2 = 1.00 x 10-3 M pH = 3.00

Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to: a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (b) Step 1 — do the stoichiometry H3O+ (from HCl) + OAc- (from buffer) ---> HOAc (from buffer) The reaction occurs completely because K is very large.

Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to: a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (b): Step 1—Stoichiometry [H3O+] [OAc-] [HOAc] Before rxn 0.00100 0.600 0.700 Change -0.00100 -0.00100 +0.00100 After rxn 0 0.599 0.701

Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to; a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (b): Step 2—Equilibrium HOAc + H2O H3O+ + OAc- [HOAc] [OAc-] [H3O+] Before rxn 0.701 0.599 0 Change -x +x +x After rxn 0.701-x 0.599+x x

Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (b): Step 2—Equilibrium HOAc + H2O H3O+ + OAc- [HOAc] [OAc-] [H3O+] After rxn 0.710-x 0.599+x x Because [H3O+] = 2.1 x 10-5 M BEFORE adding HCl, we again neglect x relative to 0.701 and 0.599.

Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (b): Step 2—Equilibrium HOAc + H2O OAc- + H3O+ [ H 3 O + ] = [HOAc] [OAc - • K a 0.701 0.599 ( 1 . 8 x 10 -5 ) [H3O+] = 2.1 x 10-5 M ------> pH = 4.68 The pH has not changed significantly upon adding HCl to the buffer!

REVIEW PROBLEMS Calculate the pH of 0.500 L of a buffer solution composed of 0.10 M sodium acetate and 0.15 M acetic acid, before and after adding 1.0 grams of sodium hydroxide solid (no volume change).

Sample Problem Calculate the pH of 0.500 L of a buffer solution composed of 0.10 M sodium acetate and 0.15 M acetic acid. HC2H3O2 <---> H+ + C2H3O2- 0.15 0 0.10 - x + x + x 0.15 x 0.10 [C2H3O2-] [H+] [HC2H3O2] x(0.10) (0.15) Ka = = = 1.8 x 10-5 X = 2.7 x 10-5 M = [H+] pH = 4.57

Sample Problem Calculate the pH after adding 1.0 grams of sodium hydroxide solid. HC2H3O (.500L)(.15M) = .075 mole NaC2H3O (.500L)(.10M) = .050 mole NaOH (1.0g)(40.0g/mole) = .025 mole HC2H3O2 + OH- ---> HOH + C2H3O2- 0.075 0.025 0.050 - 0.025 - 0.025 + 0.025 0.050 0 0 .075 .050 .500 .075 .500 [HC2H3O2] = = 0.10 [C2H3O2-] = = 0.15 M

Sample Problem Calculate the pH after adding 1.0 grams of sodium hydroxide solid. HC2H3O2 <---> H+ + C2H3O2- 0.10 0.15 - x + x + x 0.15 x 0.10 [C2H3O2-][H+] [HC2H3O2] x(0.15) (0.10) Ka = = = 1.8 x 10-5 X = 1.2 x 10-5 M = [H+] pH = 4.92

Chlorous Acid, HClO2 pKa = 1.95 Ex. 16.5a – Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with pH 4.25? Chlorous Acid, HClO2 pKa = 1.95 Nitrous Acid, HNO2 pKa = 3.34 Formic Acid, HCHO2 pKa = 3.74 Hypochlorous Acid, HClO pKa = 7.54 Tro, Chemistry: A Molecular Approach

Chlorous Acid, HClO2 pKa = 1.95 Ex. 16.5a – Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with pH 4.25? Chlorous Acid, HClO2 pKa = 1.95 Nitrous Acid, HNO2 pKa = 3.34 Formic Acid, HCHO2 pKa = 3.74 Hypochlorous Acid, HClO pKa = 7.54 The pKa of HCHO2 is closest to the desired pH of the buffer, so it would give the most effective buffering range. Tro, Chemistry: A Molecular Approach

Buffering Capacity buffering capacity is the amount of acid or base that can be added to a buffer without destroying its effectiveness the buffering capacity increases with increasing absolute concentration of the buffer components as the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves buffers that need to work mainly with added acid generally have [base] > [acid] buffers that need to work mainly with added base generally have [acid] > [base] Tro, Chemistry: A Molecular Approach

Buffering Capacity a concentrated buffer can neutralize more added acid or base than a dilute buffer Tro, Chemistry: A Molecular Approach

Titration Curve a plot of pH vs. amount of added titrant the inflection point of the curve is the equivalence point of the titration prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH the pH of the equivalence point depends on the pH of the salt solution equivalence point of neutral salt, pH = 7 equivalence point of acidic salt, pH < 7 equivalence point of basic salt, pH > 7 beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH Tro, Chemistry: A Molecular Approach

Titration Curve: Unknown Strong Base Added to Strong Acid Tro, Chemistry: A Molecular Approach

Tro, Chemistry: A Molecular Approach

Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH HCHO2(aq) + NaOH(aq)  NaCHO2(aq) + H2O(aq) Initial pH: Ka = 1.8 x 10-4 [HCHO2] [CHO2-] [H3O+] initial 0.100 0.000 ≈ 0 change -x +x equilibrium 0.100 - x x Tro, Chemistry: A Molecular Approach

Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(aq) initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10-3 before equivalence added 5.0 mL NaOH HA A- OH− mols Before 2.50E-3 mols added - 5.0E-4 mols After 2.00E-3 ≈ 0 Tro, Chemistry: A Molecular Approach

Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(aq) initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10-3 at equivalence CHO2−(aq) + H2O(l)  HCHO2(aq) + OH−(aq) added 25.0 mL NaOH Kb = 5.6 x 10-11 [HCHO2] [CHO2-] [OH−] initial 0.0500 ≈ 0 change +x -x equilibrium x 5.00E-2-x HA A- OH− mols Before 2.50E-3 mols added - mols After ≈ 0 [OH-] = 1.7 x 10-6 M

Adding NaOH to HCHO2 added 10.0 mL NaOH 0.00150 mol HCHO2 pH = 3.56 0.00100 mol NaOH xs pH = 12.22 added 30.0 mL NaOH 0.00050 mol NaOH xs pH = 11.96 added 25.0 mL NaOH equivalence point 0.00250 mol CHO2− [CHO2−]init = 0.0500 M [OH−]eq = 1.7 x 10-6 pH = 8.23 added 5.0 mL NaOH 0.00200 mol HCHO2 pH = 3.14 initial HCHO2 solution 0.00250 mol HCHO2 pH = 2.37 added 12.5 mL NaOH 0.00125 mol HCHO2 pH = 3.74 = pKa half-neutralization added 40.0 mL NaOH 0.00150 mol NaOH xs pH = 12.36 added 15.0 mL NaOH 0.00100 mol HCHO2 pH = 3.92 added 50.0 mL NaOH 0.00250 mol NaOH xs pH = 12.52 added 20.0 mL NaOH 0.00050 mol HCHO2 pH = 4.34 Tro, Chemistry: A Molecular Approach

Tro, Chemistry: A Molecular Approach

Solubility Equilibria all ionic compounds dissolve in water to some degree however, many compounds have such low solubility in water that we classify them as insoluble we can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water Tro, Chemistry: A Molecular Approach

Solubility Product the equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, Ksp for an ionic solid MnXm, the dissociation reaction is: MnXm(s)  nMm+(aq) + mXn−(aq) the solubility product would be Ksp = [Mm+]n[Xn−]m for example, the dissociation reaction for PbCl2 is PbCl2(s)  Pb2+(aq) + 2 Cl−(aq) and its equilibrium constant is Ksp = [Pb2+][Cl−]2 Tro, Chemistry: A Molecular Approach

Tro, Chemistry: A Molecular Approach

Molar Solubility solubility is the amount of solute that will dissolve in a given amount of solution at a particular temperature the molar solubility is the number of moles of solute that will dissolve in a liter of solution the molarity of the dissolved solute in a saturated solution for the general reaction MnXm(s)  nMm+(aq) + mXn−(aq) Tro, Chemistry: A Molecular Approach

PbCl2(s)  Pb2+(aq) + 2 Cl−(aq) Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C Write the dissociation reaction and Ksp expression Create an ICE table defining the change in terms of the solubility of the solid PbCl2(s)  Pb2+(aq) + 2 Cl−(aq) Ksp = [Pb2+][Cl−]2 [Pb2+] [Cl−] Initial Change +S +2S Equilibrium S 2S Tro, Chemistry: A Molecular Approach

Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C Substitute into the Ksp expression Find the value of Ksp from Table 16.2, plug into the equation and solve for S Ksp = [Pb2+][Cl−]2 Ksp = (S)(2S)2 [Pb2+] [Cl−] Initial Change +S +2S Equilibrium S 2S Tro, Chemistry: A Molecular Approach

CaF2(s)  Ca2+(aq) + 2 F−(aq) Ex 16.10 – Calculate the molar solubility of CaF2 in 0.100 M NaF at 25C Write the dissociation reaction and Ksp expression Create an ICE table defining the change in terms of the solubility of the solid CaF2(s)  Ca2+(aq) + 2 F−(aq) Ksp = [Ca2+][F−]2 [Ca2+] [F−] Initial 0.100 Change +S +2S Equilibrium S 0.100 + 2S Tro, Chemistry: A Molecular Approach

Ex 16. 10 – Calculate the molar solubility of CaF2 in 0 Ex 16.10 – Calculate the molar solubility of CaF2 in 0.100 M NaF at 25C Substitute into the Ksp expression assume S is small Find the value of Ksp from Table 16.2, plug into the equation and solve for S Ksp = [Ca2+][F−]2 Ksp = (S)(0.100 + 2S)2 Ksp = (S)(0.100)2 [Ca2+] [F−] Initial 0.100 Change +S +2S Equilibrium S 0.100 + 2S Tro, Chemistry: A Molecular Approach

Precipitation precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound if we compare the reaction quotient, Q, for the current solution concentrations to the value of Ksp, we can determine if precipitation will occur Q = Ksp, the solution is saturated, no precipitation Q < Ksp, the solution is unsaturated, no precipitation Q > Ksp, the solution would be above saturation, the salt above saturation will precipitate some solutions with Q > Ksp will not precipitate unless disturbed – these are called supersaturated solutions Tro, Chemistry: A Molecular Approach

a supersaturated solution will precipitate if a seed crystal is added precipitation occurs if Q > Ksp Tro, Chemistry: A Molecular Approach

Selective Precipitation a solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others a successful reagent can precipitate with more than one of the cations, as long as their Ksp values are significantly different Tro, Chemistry: A Molecular Approach

precipitating may just occur when Q = Ksp Ex 16.13 What is the minimum [OH−] necessary to just begin to precipitate Mg2+ (with [0.059]) from seawater? precipitating may just occur when Q = Ksp Tro, Chemistry: A Molecular Approach

precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M Ex 16.14 What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater? precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M Tro, Chemistry: A Molecular Approach

precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M Ex 16.14 What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater? precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M precipitating Ca2+ begins when [OH−] = 2.06 x 10-2 M when Ca2+ just begins to precipitate out, the [Mg2+] has dropped from 0.059 M to 4.8 x 10-10 M Tro, Chemistry: A Molecular Approach

Qualitative Analysis an analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis scheme wet chemistry a sample containing several ions is subjected to the addition of several precipitating agents addition of each reagent causes one of the ions present to precipitate out Tro, Chemistry: A Molecular Approach

Qualitative Analysis Tro, Chemistry: A Molecular Approach

Ag+(aq) + 2 H2O(l)  Ag(H2O)2+(aq) Complex Ion Formation transition metals tend to be good Lewis acids they often bond to one or more H2O molecules to form a hydrated ion H2O is the Lewis base, donating electron pairs to form coordinate covalent bonds Ag+(aq) + 2 H2O(l)  Ag(H2O)2+(aq) ions that form by combining a cation with several anions or neutral molecules are called complex ions e.g., Ag(H2O)2+ the attached ions or molecules are called ligands e.g., H2O Tro, Chemistry: A Molecular Approach

Formation Constants Tro, Chemistry: A Molecular Approach

SOLUBILITY AND COMPLEX IONS If the metal cation can form a complex ion with the other species present, a new net equilibrium will exist. The process is similar to that in the previous slide. If silver bromide is treated with ammonia solution, some of the solid dissolves and the complex ion is formed. AgBr(s) + 2 NH3(aq) <=====> Ag(NH3)2+(aq) + Br- Knet = Ksp . Kf = ( 3.3 x 10-13 ) ( 1.6 x 107) = 5.3 x 10-6

Simultaneous Equilibria 1. If you add sufficient chromate ion to an aqueous suspension of PbCl2, can PbCl2 be converted to PbCrO4? PbCl2 <--> Pb2+ + 2 Cl- Pb2+ + CrO42- <--> PbCrO4 1.7 x 10-5 1/1.8 x 10-14 9.4 x 108 PbCl2 + CrO42- <--> PbCrO4 + 2 Cl- Yes!

Simultaneous Equilibria 2. Can AgCl be dissolved by adding a solution of NH3? Write the overall equation and determine the K value. AgCl <--> Ag+ + Cl- Ag+ + 2 NH3 <--> Ag(NH3)2+ 1.8 x 10-10 1.6 x 107 2.9 x 10-3 AgCl + 2 NH3 <--> Ag(NH3)2+ + Cl- No, unless very high [NH3]

Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq) Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3. What is the [Cu2+] at equilibrium? Write the formation reaction and Kf expression. Look up Kf value Determine the concentration of ions in the diluted solutions Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq) Tro, Chemistry: A Molecular Approach

Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq) Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3. What is the [Cu2+] at equilibrium? Create an ICE table. Since Kf is large, assume all the Cu2+ is converted into complex ion, then the system returns to equilibrium Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq) [Cu2+] [NH3] [Cu(NH3)22+] Initial 6.7E-4 0.11 Change -≈6.7E-4 -4(6.7E-4) + 6.7E-4 Equilibrium x

Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq) Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3. What is the [Cu2+] at equilibrium? Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq) Substitute in and solve for x confirm the “x is small” approximation [Cu2+] [NH3] [Cu(NH3)22+] Initial 6.7E-4 0.11 Change -≈6.7E-4 -4(6.7E-4) + 6.7E-4 Equilibrium x since 2.7 x 10-13 << 6.7 x 10-4, the approximation is valid Tro, Chemistry: A Molecular Approach