EBK EXPLORING CHEMICAL ANALYSIS
EBK EXPLORING CHEMICAL ANALYSIS
5th Edition
ISBN: 9781319416942
Author: Harris
Publisher: VST
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Chapter 8, Problem 8.35P
Interpretation Introduction

Interpretation:

The pH of Cr(ClO4)3 and Fraction of chromium is in the form Cr(H2O)w-1(OH)2+ have to be calculated.

Expert Solution & Answer
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Explanation of Solution

The given compound Cr(ClO4)3 is dissociates into Cr3+andClO4-. Cr3+ ion is acidic in nature and when it reacts with water gives Cr(H2O)63+.

The reaction as follows,

  Cr(H2O)63+Cr(H2O)5(OH)2++H+Initial(M):0.01000Change:x+x+xEquilibrium:0.010xxx

Acid dissociation constant Ka is represented as follows,

  Ka=[Cr(H2O)5(OH)2+][H+][Cr(H2O)63+]

As known that,

  Ka=10pKa,substitutepKa=3.80,Ka=103.80=1.6×1041.6×104=x20.010x

By solving the above expression, the value of ‘x’ is 0.00119.

Hence,

  [Cr(H2O)5(OH)2+]=[H+]=x=0.00119MpH=log[H+]=log(0.00119)=2.92

Therefore, the value of pH is 2.92.

The fraction of chromium in the form Cr(H2O)5(OH)2+ as follows,

  =[Cr(H2O)5(OH)2+][Cr(H2O)63+]initial=0.001190.010=0.12.

Therefore, the fraction of chromium in the form Cr(H2O)5(OH)2+ is 0.12.

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