What is the pH of a solution that is 0.20 M H2S and 0.20 M HS^-? Ka=1 x 10^-7 fo

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For the first part, begin with writing the equation for the dissociation of Hydrosulfuric acid, H2S(aq) H2S(aq) -------------> HS-(aq) + H+(aq) Ka1 = 5.7 x 10–8 HS-(aq) --------------> S2-(aq) + H+(aq) Ka2 = Ka2 = 1 x 10-19 We can neglect the H+ due to the dissociation of HS-(aq) coz the ka for its dissociation is very small (10 -19). Thus, the only significant source of H+ is from H2S(aq) dissociation where the Ka is express as follows Ka = [HS-][H+]/[H2S] = 5.7 x 10–8 let x = [H+], since x = [HS-] = [H+ The initial concentration of [H2S]o = 0.10 M at equilibrium, [H2S] = [H2S]o - x. assuming x is too small enough so that [H2S]o - x is just equal to [H2S]o Thus, X^2/0.10 = 5.7 x 10–8 x = sqrt (5.7 x 10–8 x 0.1) = 7.54 × 10-5, so its B

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