## Answers

H_3ASO_4 rightarrow H^+ + H_2ASO_4 0.1 0 0 rightarrow inti of 0.1.x x x rightarrow equallozam ka_1 = [H^+] [H_2 ASO_^-_4]/[H_3 ASO_4] 5.65 times 10^-3 = x^2/0.1 - x x^2 + 5.65 times 10^-3 x - 565 times 10^-4 = 0 x = 0.0211 [H^+] = 0.0211 M P^H = -log [H^+] = -log (0.0211) P^H = 1.68 from H_3 ASO_4, [HASO^-2_4] = ka_2 = 1.75 times 10^-7 M from NaH_2ASO_4 H_2ASO_4 rightarrow H^+ + ASO^-_4 0.1 0.0211 x ka_2 = (0.0211) x/0.1 = 1.75 times 10^-7 x = 8.29 times 10^-7 M There is higher H_2ASO^-_4 than H_3ASO_4.

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