Arsenic acid is a triprotic acid in water, ionizing in the following sequential steps: H3AsO4 +...

Arsenic acid is a triprotic acid in water, ionizing in the following sequential steps: H3AsO4 + H2O <==> H2AsO4- + H3O+ Ka 5.5 x 10^-3 H2AsO4- + H2O <==> HAsO42- + H3O+ Ka 1.7 x 10^-7 HAsO42- + H2O <==> AsO43- + H3O+ Ka 5.1 x 10^-12 What is the approximate pH of a 0.80 M H3AsO4 solution?

Expert Solution

Ka = 5.5 x 10^-3

H3AsO4 + H2O <--------------------> H2AsO4- + H3O+

0.80 0 0 -----------------> initial

0.80-x x x ---------------------> equilibrium

Ka1 = x^2 / 0.8- x

5.5 x 10^-3 = x^2 / 0.8- x

x^2 + 5.5 x 10^-3 x - 4.4 x 10^-3 = 0

x = 0.0636

x = [H3O+] = 0.0636 M

pH = -log[H3O+] = -log (0.0636)

pH = 1.2

queen_honey_blossom answered 1 month ago

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