A 75.0-mL sample of 0.0500 M HCN (K_a = 6.2 ×10^-10)is titrated with 0.370 M NaOH.What is [H⁺] in the solution after 3.0 mL of 0.370 M NaOH has been added?

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Quizplus · Accepted Answer

As NaOH is added, it reacts with HCN to form NaCN and water: HCN + NaOH → NaCN + H2O This reaction consumes HCN and generates CN- ions. The equilibrium shifts to the left to partially replenish HCN. We can set up an ICE table to calculate the changes in concentration: HCN + NaOH ⇌ NaCN + H2O I 0.0500 0 0 C -0.0030 -0.0030 +0.0030 E 0.0470 0.0030 0.0030 At this point, the reaction has not reached equilibrium, and there are still HCN and CN- ions present. The NaOH added has consumed 3.0 mL/1000 mL/mL/L × 0.370 mol/L = 1.11 × 10^-3 mol of HCN. Therefore, the remaining amount of HCN is: 0.0500 mol/L × 0.0750 L - 1.11 × 10^-3 mol = 3.64 × 10^-3 mol The amount and concentration of CN- ions are equal to the amount and concentration of NaOH added, respectively: 1.11 × 10^-3 mol and 1.11 × 10^-3 mol/0.0780 L = 1.42 × 10^-2 mol/L To calculate [H+] or [OH-], we need to use the ion product constant of water Kw = [H+][OH-] = 1.0 × 10^-14 at 25°C, and the equilibrium constant for the ionization of water: H2O ⇌ H+ + OH- Kw = [H+][OH-] = 1.0 × 10^-14 Let x = [H+] = [OH-], then Kw = x^2 and x = sqrt(Kw) = 1.0 × 10^-7 M The solution is basic because [OH-] > [H+], and the amount of OH- ions added is: 1.11 × 10^-3 mol × 1 mol OH-/1 mol NaOH = 1.11 × 10^-3 mol Therefore, the total amount and concentration of OH- ions are: 1.0 × 10^-7 mol/L × 0.0750 L + 1.11 × 10^-3 mol = 8.36 × 10^-5 mol [OH-] = 8.36 × 10^-5 mol/0.0750 L = 1.12 × 10^-3 mol/L To calculate [HS+], we need to use the acid dissociation constant of HCN: HCN ⇌ H+ + CN- Ka = [H+][CN-]/[HCN] = 6.2 × 10^-10 Let x = [H+] = [CN-], then Ka = x^2/[HCN - x] and [HS+] = [H+] + [CN-] = 2x Because x << [HCN], we can approximate [HCN - x] ≈ [HCN], and solve for x: 6.2 × 10^-10 = x^2/(0.0500 - x) ≈ x^2/0.0500 x = sqrt(Ka × [HCN]) = sqrt(6.2 × 10^-10 × 0.0500) = 1.11 × 10^-6 M [HS+] = 2x = 2.22 × 10^-6 M [HS+] + [S2-] = [CN-] = 1.12 × 10^-3 M [HS+] = 1.12 × 10^-3 M - [S2-] = 1.12 × 10^-3 M - Kw/[HS+] = 1.12 × 10^-3 M - 1.0 × 10^-7 M^2/1.12 × 10^-3 M = 1.12 × 10^-3 M - 8.93 × 10^-8 M ≈ 1.12 × 10^-3 M Therefore, [HS+] ≈ 1.12 × 10^-3 M + [S2-] = 1.12 × 10^-3 M + 2.8 × 10^-10 M ≈ 1.12 × 10^-3 M Finally, we can calculate [S2-] = [CN-] - [HS+] = 1.12 × 10^-3 M - 1.12 × 10^-3 M = 0 M, because the concentration of CN- ions is much higher than that of HS+ ions, and the reaction has not reached equilibrium. The excess CN- ions can react with H2O to form HCN and OH- ions: CN- + H2O ⇌ HCN + OH- Kw/Ka = ([H+][OH-])/[HCN] = (1.0 × 10^-14)/(6.2 × 10^-10) ≈ 1.6 × 10^-5 x^2/(0.0500 - x) ≈ 1.6 × 10^-5 x ≈ sqrt(1.6 × 10^-5 × 0.0500) ≈ 2.53 × 10^-4 M [OH-] = [CN-] - x = 1.12 × 10^-3 M - 2.53 × 10^-4 M ≈ 8.65 × 10^-4 M The solution is basic because [OH-] > [H+], and the pH is: pH = -log[H+] ≈ 7.95 Therefore, [HS+] ≈ 1.12 × 10^-3 M, and the answer is (C).

A 750-ML Sample of 0