Aim #13: How can we use a base (acid) to determine the concentration of an acid (base) (continued)? H.W. # 13a Study pp. 740-747 Ans. ques. p. 753 # 60 (plot the titration curve), 65,69,74 (for problem 58 only), 77,79

V A. Titrating a weak base with a strong acid Problem: For the titration of 50.00 mL of 0.050 M NH3 with 0.10 M HCl, calculate the pH of the resulting solution a) after the addition of 0 mL HCl b) after the addition of 10.00 mL HCl c) at the equivalence point d) after the addition of 50.00 mL HCl Ans: a) Kb = 1.8 x 10-5 = [NH4+][OH-] ≈ x2 ͘ [NH3] .050 x2 = (.050)(1.8 x 10-5) = 9.0 x 10-7 [OH-] = x = 9.5 x 10-4 pOH = 3.02 and pH = 10.98

b) neutralization: (.050mmol/mL)(50.00 mL) = 2.5 mmol NH3 (0.10 mmol/mL)(10.00 mL) = 1.0 mmol HCl mol NH3 remaining = 2.5 mmmol – 1.0 mmol = 1.5 mmol [NH3] = 1.5 mmol = 2.5 x 10-2 M (50.00 + 10.00) mL [NH4+] = 1.0 mmol = 1.7 x 10-2 M (50.00 + 10.00) mL equilibrium: NH3(aq) + H2O(ℓ) NH4+(aq) + OH-(aq) Kb = 1.8 x 10-5 = [NH4+][OH-] = (.017 + x)(x) [NH3] (.025 – x) 1.8 x 10-5 ≈ (.017)(x) and [OH-] = x = 2.6 x 10-5 (.025) pOH = -log(2.6 x 10-5) = 4.58 and pH = 9.42

C) at the equivalence point we need 2 C) at the equivalence point we need 2.5 mmol HCl to neutralize the NH3 completely. [NH4+] = 2.5 mmol = 3.3 x 10-2 M equilibrium: NH4+(aq) + H2O(ℓ) NH3(aq) + H3O+(aq) Ka = Kw = 1.0 x 10-14 = 5.56 x 10-10 = x2 ͘ Kb 1.8 x 10-5 3.3 x 10-2 – x 5.56 x 10-10 ≈ x2 ͘ 3.3 x 10-2 x2 = 1.85 x 10-11 [H+] = x = 4.3 x 10-6 M pH = 5.37

neutralization: 5.0 mmool HCl – 2.5 mmol NH3 = 2.5 mmol HCl remaining d) The [H+] is due to the excess of HCl. (0.10 mmol/mL HCl)(50.00 mL) = 5.0 mmol HCl neutralization: 5.0 mmool HCl – 2.5 mmol NH3 = 2.5 mmol HCl remaining NH3(aq) + HCl(aq) → NH4+(aq) + Cl-(aq) Because NH4+ is a weak acid and HCl is a strong acid [H+] = [HCl] = 2.5 mmol = 2.5 x 10-2 M (50.00 + 50.00)mL pH = -log(2.5 x 10-2) = 1.60

B. Graphing the titration curve 14.00 10.00 7.00 pH 5.00 equivalence pt. 2.00 0.00 0 10 20 30 40 50 60 70 Volume of HCl added (mL)

VI Titrations of polyprotic acids – produce more than one equivalence point. Problem: Sketch the curve of the titration of 25.0 mL 0.1 M H2CO3 with 0.10 M NaOH. Ans: The polyprotic acid dissociates in 2 distinctive steps. step 1: H2CO3(aq) + OH-(aq) → HCO3-(aq) + H2O(ℓ) Ka1 = 4.3 x 10-7 step 2: HCO3-(aq) + OH-(aq) → CO32-(aq) + H2O(ℓ) Ka2 = 5.6 x 10-11

When VNaOH = ½ VH2CO3 , [H2CO3] = [HCO3-] and pH = pKa use 1st VNaOH = VH2CO3 , pH = 1st equivalence point equation When VNaOH = 3/2 VH2CO3 , [HCO3-] = [CO3-] and pH = pKa2 use 2nd VNaOH = 2VH2CO3 , pH = 2nd equivalence point equation 14.00 12.00 11.00 10.00 pH 7.00 6.00 3.00 0.00 0 10 20 30 40 50 60 70 Volume NaOH added (mL)

VII Acid-Base Indicators Endpoint = Equivalence Point Choose an indicator that minimizes the error Acid-base indicators are themselves weak acids. (Acid is one color, its conjugate base is another.) HIn(aq) H+(aq) + In-(aq) red blue If Ka = [H+][In-] = 1.0 x 10-8 [HIn] Ka = [In-] If pH = 1.0 , [H+] [HIn]

Ka = 1.0 x 10-8 = 10-7 = 1 = [In-] ͘ [H+] 1.0 x 10-1 10,000,000 [HIn] the solution will be red. As we add OH-, a color change will be noticeable when [In-] = 1 purple [HIn] 10 [H+] = Ka [HIn] = 1.0 x 10-8 (10) = 1.0 x 10-7 [In-] (1) OR pH = 7.00 As we add more OH-, pH ↑ If [In-] >> 1 , HI H+ + In- shifts to the right [HIn] 10 the solution will be blue.

For [In-] = 1 ͘ [HIn] 10 pH = pKa + log(1/10) = pKa – 1 For [In-] = 10 [HIn] 1 pH = pKa + log(10/1) = pKa + 1 Therefore, choose an indicator whose pKa equals pH (of the equivalence point) ± 1.