Name__________________________________________ Date____________________ Chapter 15 – Buffers and Titration 1. Calculate the pH of the each of the following solutions given that for ammonia, Kb= 1.8x10–5. a. 0.100 M NH3 NH3 + H2O NH4+ + OH– 0.100 0 0 –x +x +x 0.100 x x 2 –5 x / 0.100 = 1.8 x 10 x = [OH–] = 1.3 x 10–3 M pH = 11.13 b. 0.100 M NH4Cl NH4+ NH3 + H+ 0.100 0 0 –x +x +x 0.100 x x 2 –14 x = 1.0 x 10 x = [H+] = 7.5 x 10–6 M pH = 5.13 –5 0.100 1.8 x 10 c. A solution containing 0.100 M NH3 and 0.100 M NH4Cl NH3 + H2O NH4+ + OH– 0.100 0.100 0 –x +x +x 0.100 0.100 x 0.100 x / 0.100 = 1.8 x 10–5 x = [OH–] = 1.8 x 10–5 M pH = 9.26 2. Calculate the pH of solution 1c after the addition of 0.035 moles of NaOH to 1 L of solution. Is this solution still a buffer at equilibrium? OH– + NH4+ NH3 + H2O 0.035 0.100 0.100 –.035 –.035 +.035 0 0.065 0.135 still has both base and conjugate acid – it’s still a buffer for pH, do ICE chart: NH3 + 0.135 H2O 0.065 x / 0.135 = 1.8 x 10–5 NH4+ + OH– 0.065 x x = [OH–] = 3.7 x 10–5 M pH = 9.57 3. Calculate the pH of solution 2 after the addition of 0.200 moles of HCl to 1 L of solution. Is this solution still a buffer at equilibrium? H+ + NH3 NH4+ 0.200 0.135 0.065 –.135 –.135 +.135 0.065 0 0.200 has only acid left – not still a buffer for pH, use –log [H+] = 1.19 4. A 25.0-mL sample of 0.100 M lactic acid (HC3H5O3, pKa = 3.86) is titrated with 0.100 M NaOH solution. Calculate the pH after the following volumes of NaOH have been added. a. 0.0 mL c. 12.5 mL e. 32.0 mL b. 8.00 mL d. 25.0 mL pKa = 3.86 so Ka = 1.4 x 10–4 a. pH of weak acid: HC3H5O3 ⇄ H+ + C3H5O3– 0.100 0 0 –x +x +x 0.100 x x x2 / 0.100 = 1.4 x 10–4 x = [H+] = 3.7 x 10–3 M pH = 2.43 b. 8.00 mL base added: calculate new molarities [HC3H5O3] = (25.0 mL)(0.100 M) / 33.0 mL = 0.0758 M [OH–] = (8.00 mL)(0.100 M) / 33.0 mL = 0.0242 M one thing gets used up: OH– + HC3H5O3 → C3H5O3– + H2O 0.0242 0.0758 0 –.0242 –.0242 +0.0242 0 0.0516 0.0242 + – do an ICE chart: HC3H5O3 ⇄ H + C3H5O3 0.0516 x 0.0242 0.0242x / 0.0516 = 1.4 x 10–4 x = [H+] = 2.9 x 10–4 M pH = 3.53 c. 12.5 mL base added: use method from part b, or realize that this is the half-equiv. point pH = pKa = 3.86 d. 25.0 mL base added: equivalence point; [HC3H5O3] = [OH–] calculate new molarity: [HC3H5O3] = (25.0 mL)(0.100 M) / 50.0 mL = 0.0500 M both get used up, OH– + HC3H5O3 → C3H5O3– + H2O conj. base left 0.0500 0.0500 0 –.0500 –.0500 +0.0500 0 0 0.0500 do an ICE chart: C3H5O3– + H2O ⇄ HC3H5O3 + OH– 0.0500 x x x2 = 1.0 x 10–14 0.0500 1.4 x 10–4 x = [OH–] = 1.9 x 10–6 M pH = 8.28 e. 32.0 mL base added: calculate new molarities [HC3H5O3] = (25.0 mL)(0.100 M) / 57.0 mL = 0.0439 M [OH–] = (32.0 mL)(0.100 M) / 57.0 mL = 0.0561 M OH– + HC3H5O3 → C3H5O3– + H2O 0.0561 0.0439 0 -.0439 -.0439 +0.0439 0.0122 0 0.0439 pOH = 1.914 so pH = 12.086 one thing gets used up: [OH–] = 0.0122 M 5. Consider the titration of 80.0 mL of 0.100 M Sr(OH)2 by 0.400 M HCl. Calculate the pH of the resulting solution after the following volumes of HCl have been added. a. 0.0 mL c. 40.0 mL b. 20.0 mL d. 80.0 mL Since Sr(OH)2 → Sr2+ + 2 OH– , the concentration of OH– is 2 x 0.100 = 0.200 M a. [OH–] = 0.200 M, pOH = 0.699, pH = 13.30 b. 20.0 mL of acid added to base: calculate new molarites [OH–] = (80.0 mL)(0.200 M) / 100.0 mL = 0.160 M [H+] = (20.0 mL)(0.400 M) / 100.0 mL = 0.0800 M the acid reacts with the base: one thing gets used up: H+ + .0800 –.0800 0 OH– → H2O 0.160 –.0800 0.080 pOH = 1.09, pH = 12.90 c. 40.0 mL of acid added: equivalence point; [OH–] = [H+] pH = 7.000 d. 80.0 mL of acid added: calculate new molarities [OH–] = (80.0 mL)(0.200 M) / 160.0 mL = 0.100 M [H+] = (80.0 mL)(0.400 M) / 160.0 mL = 0.200 M the acid reacts with the base: one thing gets used up: H+ + .200 –.100 0.100 OH– → H2O 0.100 –.100 0 pH = 1.000 6. A sample of an ionic compound NaA, where A– is the anion of a weak acid, was dissolved in enough water to make 100.0 mL of solution and was then titrated with 0.100 M HCl. After 500.0 mL of HCl was added, the pH was measured to be 5.00. The experimenter found that 1.00 L of 0.100 M HCl was required to reach the equivalence point of the titration. a. What is the Kb value for A–? b. Calculate the pH of the solution at the equivalence point of the titration. a. 1.00 L of HCl is the equivalence point, so 500.0 mL is the half-equivalence point. At the half-equivalence point, pOH = pKb. Since pH = 5.00, pOH = 9.00 = pKb Kb = 1.0 x 10–9 b. equivalence point: calculate molarity of A– using M1V1 = M2V2 (0.100 M)(1.00 L) = M2 (0.1000 L) M2 = 1.00 M one solution added to another: calculate new molarities (they are equivalent!) [A–] = [H+] = (0.1000 L)(1.00 M) / 1.10 L = 0.0909 M both gets used up, conj. acid left H+ + A– → HA 0 0 0.0909 do ICE chart: HA ⇄ H+ + A– 0.0909 x x x2 = 1.0 x 10–14 .0909 1.0 x 10–9 x = [H+] = 9.5 x 10–4 M pH = 3.02