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**Problem**: Arsenic acid (H3AsO4) is a triprotic acid with K a1 = 5.5 X 10 -3, Ka2 = 1.7 X 10 -7, and Ka3 = 5.1 X 10-12. Calculate [H+], [OH-], [H3AsO4], [H2AsO4-], [HAsO42-], and [AsO43-] in a 0.20-M arsenic acid solution.

###### FREE Expert Solution

###### FREE Expert Solution

Arsenic acid (H_{3}AsO_{4}) is a **weak oxyacid** with only one oxygen greater than hydrogen. H_{3}AsO_{4} is a triprotic acid, meaning it can donate three protons (H^{+}).

It has three K values and it will have three equilibrium reactions.

$\stackrel{\mathbf{a}\mathbf{c}\mathbf{i}\mathbf{d}\mathbf{i}\mathbf{c}\mathbf{}\mathbf{f}\mathbf{o}\mathbf{r}\mathbf{m}}{\stackrel{\mathbf{\u23de}}{{\mathbf{H}}_{\mathbf{3}}\mathbf{A}\mathbf{s}{\mathbf{O}}_{\mathbf{4}}}}\mathbf{}\underset{{\mathbf{K}}_{\mathbf{1}}}{\mathbf{\rightleftharpoons}}\mathbf{}\stackrel{\mathbf{i}\mathbf{n}\mathbf{t}\mathbf{e}\mathbf{r}\mathbf{m}\mathbf{e}\mathbf{d}\mathbf{i}\mathbf{a}\mathbf{t}\mathbf{e}\mathbf{}\mathbf{\#}\mathbf{1}}{\stackrel{\mathbf{\u23de}}{{\mathbf{H}}_{\mathbf{2}}\mathbf{A}\mathbf{s}{\mathbf{O}}_{\mathbf{4}}^{\mathbf{-}}}}\underset{{\mathbf{K}}_{\mathbf{2}}}{\mathbf{\rightleftharpoons}}\mathbf{}\stackrel{\mathbf{i}\mathbf{n}\mathbf{t}\mathbf{e}\mathbf{r}\mathbf{m}\mathbf{e}\mathbf{d}\mathbf{i}\mathbf{a}\mathbf{t}\mathbf{e}\mathbf{}\mathbf{\#}\mathbf{2}}{\stackrel{\mathbf{\u23de}}{\mathbf{H}\mathbf{A}\mathbf{s}{\mathbf{O}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}}}\underset{{\mathbf{K}}_{\mathbf{3}}}{\mathbf{\rightleftharpoons}}\mathbf{}\stackrel{\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{i}\mathbf{c}\mathbf{}\mathbf{f}\mathbf{o}\mathbf{r}\mathbf{m}}{\stackrel{\mathbf{\u23de}}{\mathbf{A}\mathbf{s}{\mathbf{O}}_{\mathbf{4}}^{\mathbf{3}\mathbf{-}}}}$

###### Problem Details

Arsenic acid (H_{3}AsO_{4}) is a triprotic acid with K _{a1} = 5.5 X 10 ^{-3}, K_{a2 }= 1.7 X 10 ^{-7}, and K_{a3} = 5.1 X 10^{-12}. Calculate [H^{+}], [OH^{-}], [H_{3}AsO_{4}], [H_{2}AsO_{4}^{-}], [HAsO_{4}^{2-}], and [AsO_{4}^{3-}] in a 0.20-M arsenic acid solution.

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