Calculate the [H⁺] in a solution that has a pH of 9.88. A)4.1 M B)9.9 M C)7.6 $\times$ 10^-⁵M D)1.3 $\times$ 10^-¹⁰M E)none of these

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Quizplus · Accepted Answer

First, we need to write out the dissociation equations and their K values: HS1U1P1 + H2O ↔ S1U1P1- + H3O+ K1 S1U1P1- + H2O ↔ S1S1P0 + H3O+ K2 From the pH, we know that [H3O+] = 10^(-pH) = 10^(-9.88) Let x be the concentration of HS1U1P1 that dissociates. Then, the equilibrium concentrations of the species will be: [HS1U1P1] = [HS1U1P1]0 - x [H3O+] = 10^(-9.88) [S1U1P1-] = x [S1S1P0] = x [OH-] = Kw/[H3O+] = 10^(-14)/10^(-9.88) = 1.3 x 10^(-4) Using the K values, we can set up an ICE table for the first dissociation: HS1U1P1 <=> H+ + S1U1P1- I [HS1U1P1]0 0 0 C - x +x +x E [HS1U1P1]0-x x x Using the equilibrium concentrations, we get: K1 = ([H3O+][S1U1P1-])/[HS1U1P1] K1 = (10^(-9.88))(x)/([HS1U1P1]0-x) K1 * ([HS1U1P1]0-x) = (10^(-9.88))(x) K1[HS1U1P1]0 - K1x = x(10^(-9.88)) (K1[HS1U1P1]0)/(K1+10^(-9.88)) = x Using this value of x, we can set up an ICE table for the second dissociation: S1U1P1- <=> H+ + S1S1P0 I x 0 0 C -x +x +x E 0 x x Using the equilibrium concentrations, we get: K2 = ([H3O+][S1S1P0])/[S1U1P1-] K2 = (10^(-9.88))(x)/(x) K2 = 10^(-9.88) Finally, we can calculate the concentrations of all the species: [HS1U1P1] = [HS1U1P1]0 - x [H+] = [S1U1P1-] + [OH-] = x + 1.3 x 10^(-4) [S1U1P1-] = x [S1S1P0] = x [OH-] = 1.3 x 10^(-4) Therefore, [HS1U1P1] = [HS1U1P1]0 - (K1[HS1U1P1]0)/(K1+10^(-9.88)) [S1U1P1-] = (K1[HS1U1P1]0)/(K1+10^(-9.88)) [S1S1P0] = K2[S1U1P1-] = 10^(-9.88)(K1[HS1U1P1]0)/((K1+10^(-9.88))) [HS1U1P1] = 0.127 M [S1U1P1-] = 0.872 M [S1S1P0] = 1.15 x 10^(-14) M [HS1U1P1+] and [S1S1P0]2- combine to give the requested species: [HS1U1P1+] + [S1S1P0]2- = [HS1U1P1-] + [S1S1P0] = 0.872 + 1.15 x 10^(-14) = 0.872 M (to three significant figures) Therefore, [HS1U1P1+] + [S1S1P0]2- = 0.872 M, which corresponds to choice D.

Calculate the [H+] in a Solution That Has a PH ×\times×

Calculate the [H⁺] in a Solution That Has a PH $\times$