   Chapter 14, Problem 108E

Chapter
Section
Textbook Problem

# Arsenic acid (H3AsO4) is a triprotic acid with K a 1 = 5.5 × 10−3, K a 2 = 1.7 × 10−7, and K a 3 = 5.1 × 10−12. Calculate [H+], [OH−], [H3AsO4], [H2AsO4−], [HAsO42−], and [AsO43−] in a 0.20-M arsenic acid solution.

Interpretation Introduction

Interpretation: The value of [H+],[OH],[H2AsO4],[HAsO42] and [AsO43] for a 0.20M solution of arsenic acid is to be calculated.

Concept introduction: Arsenic acid is a weak acid; therefore, it dissociates up to a very small extent and slowly in an aqueous solution.

To determine: The value of [H+],[OH],[H3AsO4],[HAsO42] and [AsO43] for a 0.20M solution of arsenic acid.

Explanation

Explanation

Given

The standard value of Ka1 for arsenic acid is 5.5×103.

The standard value of Ka2 for arsenic acid is 1.7×107.

The standard value of Ka3 for arsenic acid is 5.1×1012.

The dissociated amount of [H+] and [H2AsO4] from the arsenic acid (H3AsO4) is assumed to be x. The concentration of [H+] and [H2AsO4] is calculated by using the ICE (Initial Change Equilibrium) table.

H3AsO4H++H2AsO4Initial(M):0.2000Change(M):xxxEquilibrium(M):0.20xxx

The value of acid dissociation equilibrium constant Ka is calculated by the formula,

Ka=[Concentrationofproducts][Concentrationofreactants]

Thus, Ka for the above stated reaction is,

Ka1=[H+][H2AsO42][H3AsO4]

Substitute the values of Ka1, concentration of reactants and products from the ICE table in the above expression.

5.5×103=[x][x][0.20x]x2+5.5×103x1.1×103=0x=0.0305M_

Hence, the values of [H+] and [H2AsO4] at equilibrium is 0.0305M_.

The dissociated amount of [H+] and [HAsO42] from H2AsO41 is assumed to be y. The concentration of [H+] and [HAsO42] is calculated by using the ICE (Initial Change Equilibrium) table.

H2AsO4H++HAsO42Initial(M):0.030500Change(M):yyyEquilibrium(M):0.0305yyy

The value of acid dissociation equilibrium constant Ka is calculated by the formula.

Ka=[Concentrationofproducts][Concentrationofreactants]

Thus, Ka for the above stated reaction is,

Ka2=[H+][HAsO42][H2AsO41]

Substitute the values of Ka2, concentration of reactants and products from the ICE table in the above expression.

1.7×107=[x][x][0.0305x]x2+1.7×107x5.18×109=0x=7.18×10-5M_

Hence, the values of [H+] and [HAsO42] at equilibrium is 7

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