Question

The ionization constant of phenol is $1.0 \times 10^{-10}$. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0. 01 M in sodium phenolate?

Answer 1

"The ionization reaction of phenol is as follows.

$\quad \quad { C }_{ 6 }{ H }_{ 5 }OH\quad +\quad { H }_{ 2 }O\quad \rightarrow \quad { C }_{ 6 }{ H }_{ 5 }{ 0 }^{ - }\quad +\quad { H }_{ 3 }{ 0 }^{ + }\\ Initial\quad Concentration\quad \quad 0.05\quad \quad \quad \quad 0\quad \quad0\quad \quad \\ At\quad equilibrium\quad0.05-X\quad \quad \quad \quad \quad \quad \quad x\quad \quad \quad X$

${ K }_{ a }=\frac { [{ C }_{ 6 }{ H }_{ 5 }{ O }^{ - }][{ H }_{ 3 }{ O }^{ + }] }{ [{ C }_{ 6 }{ H }_{ 5 }OH] }$

${ K }_{ a }\quad =\quad \frac { X\quad \times \quad X }{ 0.05\quad -\quad X }$

The ionization constant is very less. Therefore, we considered 0.05

$X\quad =\quad \sqrt { 1\quad \times \quad { 10 }^{ -10 }\quad \times \quad 0.05 }$

$=\quad \sqrt { 5\quad \times \quad { 10 }^{ -12 } }$

$=\quad 2.2\quad \times \quad { 10 }^{ -6 }\quad M\quad =\quad [{ H }_{ 3 }{ O }^{ + }]$

Since,$[{ H }_{ 3 }{ O }^{ + }]\quad =\quad [{ C }_{ 6 }{ H }_{ 5 }{ O }^{ - }]$

$[{ C }_{ 6 }{ H }_{ 5 }{ O }^{ - }]\quad =\quad 2.2\quad \times \quad { 10 }^{ -6 }\quad M$

In the presence of 0.01 M ${ C }_{ 6 }{ H }_{ 5 }ONa$calculate degree of ionization of phenol

${ C }_{ 6 }{ H }_{ 5 }ONa\quad \longrightarrow \quad { C }_{ 6 }{ H }_{ 5 }{ O }^{ - }\quad +\quad { Na }^{ + }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0.1$

Also

$\quad \quad \quad \quad \quad { C }_{ 6 }{ H }_{ 5 }OH\quad +\quad { H }_{ 2 }O\quad \rightarrow \quad { C }_{ 6 }{ H }_{ 5 }{ O }^{ - }\quad +\quad { H }_{ 3 }{ O }^{ + }\\ Conc\quad \quad \quad0.05-0.05\alpha \quad \quad \quad \quad \quad \quad \quad \quad \quad 0.05\alpha \quad \quad \quad \quad 0.05\alpha$  

${ [C }_{ 6 }{ H }_{ 5 }OH]\quad =\quad 0.05\quad -\quad 0.05\alpha$

${ [C }_{ 6 }{ H }_{ 5 }{ O }^{ - }]\quad =\quad 0.01\quad +\quad 0.05\alpha$

$[{ H }_{ 3 }{ O }^{ + }]\quad =\quad 0.05\alpha$

${ K }_{ a }\quad =\quad \frac { [{ C }_{ 6 }{ H }_{ 5 }{ O }^{ - }][{ H }_{ 3 }{ O }^{ + }] }{ [{ C }_{ 6 }{ H }_{ 5 }OH] }$

$=\quad \frac { (0.01)(0.05\alpha ) }{ 0.05 }$

$1.0\quad \times \quad { 10 }^{ -10 }\quad =\quad 0.01\alpha$

$\alpha \quad =\quad 1\quad \times \quad { 10 }^{ -10 }$"

Answer 2

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the

ALPHA= 1×10(^-10)