Quantitative Chemical Analysis
Quantitative Chemical Analysis
9th Edition
ISBN: 9781464135385
Author: Daniel C. Harris
Publisher: W. H. Freeman
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Chapter 10, Problem 10.8P

(a)

Interpretation Introduction

Interpretation:

The equation 10-11 for potassium hydrogen phthalate has to be derived.

Concept introduction:

[H+]=K1K2F+K1KwK1+F

The pH of solution can be calculated as,

pH=-log[H+]γH-

With the above equation, the negative logarithm of activity of Hydrogen ion can be measured.

(a)

Expert Solution
Check Mark

Explanation of Solution

Chargebalance:[K+]+[H+]=[OH-]+[HP-]+2[P2-](1)Massbalance:[K+]=[H2P]+[HP-]+[P2-](2)Equilibria:K1=[H+]γH+[HP-]γHP-_(3)[H2P]γH2PK2=[H+]γH+[P2-]γP2-_(4)[H2P]γH2PKw=[H+]γH+[OH-]γOH-(5)

[K+] Value is substituted in equation (1) and (2)

[H2P]+[H+]-[P2-]-[OH-]=0

Given values apply to equation (3), (4), and (5) we get

[H+]γH+[HP-]γHP-K1γH2P+[H+]-K2[HP-]γHP-[H+]γH+γP2--Kw[H+]γH+γOH-

That can be rearranged the equation we get

[H+]K1K2[HP-]γHP-γH2PγH+γP2-+K1KwγH2PγH+γOH-K1γH2P+[HP-]γH+γHP-

(b)

Interpretation Introduction

Interpretation:

The pH of given solution has to be found.

Concept introduction:

[H+]=K1K2F+K1KwK1+F

The pH of solution can be calculated as,

pH=-log[H+]γH-

With the above equation, the negative logarithm of activity of Hydrogen ion can be measured.

(b)

Expert Solution
Check Mark

Answer to Problem 10.8P

The pH of given solution is 4.03

Explanation of Solution

The pH of given solution can be calculated as,

The strength of the ions is

[HP-]0.050M,γHP-=0.835,γP2-=0.485,γH2P1.00γH+=0.86,γOH-=0.81

[H+]=1.09×10-4=pH=-log[H+]γH+=4.03

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