Question

The ionization constant of benzoic acid is 6.46 × $10^{-5}$ and $K_{sp}$ for silver benzoate is 2.5 × $10^{-13}$. How many times is silver benzoate more soluble in a buffer of pH 3·19 compared to its solubility in pure water ?

Answer 1

From the given,

                   $pH\quad =\quad 3.19$

                   $[{ H }_{ 3 }{ O }^{ + }]\quad =\quad 6.46\quad \times \quad { 10 }^{ -4 }\quad M$

                   ${ C }_{ 6 }{ H }_{ 5 }COOH\quad +\quad { H }_{ 2 }O\quad \leftrightarrow \quad { C }_{ 6 }{ H }_{ 5 }CO{ O }^{ - }\quad +\quad { H }_{ 3 }{ O }^{ + }$

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

                   $\frac { [{ C }_{ 6 }{ H }_{ 5 }COOH] }{ [{ C }_{ 6 }{ H }_{ 5 }CO{ O }^{ - }] } \quad =\quad \frac { [{ H }_{ 3 }{ O }^{ + }] }{ { K }_{ a } } \quad =\quad \frac { 6.46\quad \times { 10 }^{ -4 } }{ 6.46\times { 10 }^{ -5 } } \quad =\quad 10$

The solubility of ${ C }_{ 6 }{ H }_{ 5 }COOAg$ be x mol/L

Then

                   $[{ Ag }^{ + }]\quad =\quad x$

                   $[{ C }_{ 6 }{ H }_{ 5 }COOH]\quad +\quad [{ C }_{ 6 }{ H }_{ 5 }CO{ O }^{ - }]\quad =\quad x$

                   $10[{ C }_{ 6 }{ H }_{ 5 }CO{ O }^{ - }]\quad +\quad [{ C }_{ 6 }{ H }_{ 5 }CO{ O }^{ - }]\quad =\quad x$

                   $[{ C }_{ 6 }{ H }_{ 5 }CO{ O }^{ - }]\quad =\quad \frac { x }{ 11 }$

                   ${ K }_{ sp }[A{ g }^{ + }][{ C }_{ 6 }{ H }_{ 5 }{ COO }^{ - }]$

                   $2.5\quad \times \quad { 10 }^{ -13 }\quad =\quad x\left( \frac { x }{ 11 }  \right)$

                   $x\quad =\quad 1.66\quad \times \quad { 10 }^{ -6 }\quad mol/L$

The solubility of silver benzoate in pH 3.19

Let's calculate the solubility of ${ C }_{ 6 }{ H }_{ 5 }COOAg$ be "x" mol/L

Then,

                    $[A{ g }^{ + }]\quad =\quad { x }^{ I }M$

                    $And\quad [C{ H }_{ 3 }CO{ O }^{ - }]\quad =\quad { x }^{ I }M$

                    ${ K }_{ sp }\quad =\quad (\quad { x }^{ I }{ ) }^{ 2 }$

                    ${ x }^{ I }\quad =\quad \sqrt { { K }_{ sp } } \quad =\quad \sqrt { 2.5\quad \times \quad { 10 }^{ -13 } } \quad =\quad 5\quad \times \quad { 10 }^{ -7 }\quad mol/L$

                    $\frac { x }{ { x }^{ I } } \quad =\quad \frac { 1.66\quad \times \quad 1{ 0 }^{ -6 } }{ 5\quad \times \quad 1{ 0 }^{ -7 } } \quad =\quad 3.32$

Therefore, ${ C }_{ 6 }{ H }_{ 5 }COOAg$ is approximately 3.317 times more soluble.

Answer 2

Mate....
IT IS APROXIMATELY 3.317 TIMES MORE SOLUBLE .....
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