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Arsenic Acid Is a Triprotic Acid in Water,ionizing in the Following

question 161

Short Answer

Arsenic acid is a triprotic acid in water,ionizing in the following sequential steps:
H₃AsO₄ + H₂O $\leftrightarrows$ H₂AsO₄^- + H₃O⁺
$Arsenic acid is a triprotic acid in water,ionizing in the following sequential steps: H3AsO4 + H2O \leftrightarrows H2AsO4- + H3O+ = 5.5 *10 - 3 H2AsO4- + H2O \leftrightarrows HAsO42- + H3O+ = 1.7 *10-7 HAsO42- + H2O \leftrightarrows AsO43- + H3O+ = 5.1 * 10-12 What is the approximate pH of a 0.80 M Na3AsO4 solution?$ = 5.5 *10 ^- ³
H₂AsO₄^- + H₂O $\leftrightarrows$ HAsO₄²^- + H₃O⁺
$Arsenic acid is a triprotic acid in water,ionizing in the following sequential steps: H3AsO4 + H2O \leftrightarrows H2AsO4- + H3O+ = 5.5 *10 - 3 H2AsO4- + H2O \leftrightarrows HAsO42- + H3O+ = 1.7 *10-7 HAsO42- + H2O \leftrightarrows AsO43- + H3O+ = 5.1 * 10-12 What is the approximate pH of a 0.80 M Na3AsO4 solution?$ = 1.7 *10^-⁷
HAsO₄²^- + H₂O $\leftrightarrows$ AsO₄³^- + H₃O⁺
$Arsenic acid is a triprotic acid in water,ionizing in the following sequential steps: H3AsO4 + H2O \leftrightarrows H2AsO4- + H3O+ = 5.5 *10 - 3 H2AsO4- + H2O \leftrightarrows HAsO42- + H3O+ = 1.7 *10-7 HAsO42- + H2O \leftrightarrows AsO43- + H3O+ = 5.1 * 10-12 What is the approximate pH of a 0.80 M Na3AsO4 solution?$ = 5.1 * 10^-¹²
What is the approximate pH of a 0.80 M Na₃AsO₄ solution?

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