Question

The ionization constant of benzoic acid is 6.46 × 10^–5 and Ksp 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

${\boxed{\underline{\tt{\orange{Required \: Answer:-}}}}}$

$\displaystyle \sf{C_6H_5COO \: \: Ag \: \: is \: \: Approximately \: \: 3.32 \: \: times \: \: more \: \: soluble \: \: in \: \: low \: \: pH \: \: Solution}$

★EXPLANATION:-

pH is 3.19,

$\displaystyle \sf{[H_3O^+]= 6.46 \times 10^ {-4}M}$

$\displaystyle \sf{C_6H_5COOH + H_2O \longrightarrow \: C_6H_5COO^- + H_3O}$

$\displaystyle \sf{K_a = \frac{[C_6H_5COO^-][ H_3 { O}^{ + } ]}{[C_6H_5COOH]} }$

$:\longrightarrow \displaystyle \sf{ \frac{[C_6H_5COOH]}{[C_6H_5COO^-]} = \frac{ H_3 {O}^{ + } }{K_a} } \\ \\ :\longrightarrow\displaystyle \sf{ \frac{6.46 \times {10}^{ - 4} }{6.46 \times {10}^{ - 5} } = 10}$

$\sf{Let \: \: The \: \: solubility \: \: of \: \: C_6H_5COO \: \: Ag \: \: be \: \: x \: \: \frac{mol}{l} }$

$: \longrightarrow\displaystyle\sf{[Ag^+] = x }\\\\ \longrightarrow\displaystyle\sf{[C_6H_5COOH] + [C_6H_5COO^-] = x} \\ \\ :\longrightarrow\displaystyle\sf{10[C_6H_5COOH] + [C_6H_5COO^-] = x} \\ \\ :\longrightarrow\displaystyle\sf{ [C_6H_5COO^-] = \frac{x}{11} }$

$: \longrightarrow \displaystyle \sf{K_{sp}[Ag^+][C_6H_5COO^-]} \\ \\ :\longrightarrow \displaystyle \sf{2.5 \times {10}^{ - 13} = x (\frac{x}{11} )} \\ \\ :\longrightarrow \displaystyle \sf{1.66 \times {10}^{ - 6} \frac{mol}{l} }$

Thus, the solubility of silver benzoate in a pH of 3.19 solution is 1.66 × 10^-6 mol/L.

let the solubility of C6 H5CooAg be x' mol/L

Then,

$: \longrightarrow \displaystyle \sf{[Ag^+] = x' M \: \: and \: \: [CH_3COO^-]=x ' M } \\ \\ :\longrightarrow \displaystyle \sf{ K_{sp} = [Ag^+][CH_3COO^-]} \\ \\ :\longrightarrow \displaystyle \sf{K_{sp} = {(x' )}^{2} } \\ \\ :\longrightarrow \displaystyle \sf{x' = \sqrt{K_{sp}} = \sqrt{2.5 \times {10}^{ - 13} } = 5 \times {10}^{ - 7} \frac{mol}{l} } \\ \\ :\longrightarrow \displaystyle \sf{ \therefore \: \frac{x}{x'} = \frac{1.66 \times {10}^{ - 6} }{5 \times {10}^{ - 7} } = \pink{3.32}}$

So, it will 3.32 times more soluble in a low pH solution.

Answer 2

Since pH=3.19,

(H3O^+)=6.46×10^-4M

C6H5COOH+H2O↔C6H5COO^-+H3O

Kα(C6H5COO^-)(H3O^+)/(C6H5COOH)

(C6H5COOH/C6H5COO^-)=(H3O^+)/Kα=6.46×10^-4/6.46×10^-5=10

Let the solubility of C6H5COOAg be x mol/L

Then,

(Ag^+)=x

(C6H5COOH)+(C6H5COO^-)+=x

10(C6H5COO^-)+(C6H5COO^-)=x

(C6H5COO^-)=X/11

Ksp(Ag^+)(C6H5COO^-)

2.5×10^-13=x(x/11)

x=1.66×10^-6mol/L

Thus the solubility of silver benzoate in a pH 3.19 solution is 1.66×10^-6mol/L

Now,let the solubility of C6H5COOAg be x'mol/L

Then,(Ag^+)=x'M & (CH3COO^-)=x'M

Ksp=(Ag^+)(CH3COO^-)

Ksp=(x')²

x'=√Ksp=√2.5×10^-13=5×10^-7mol/L

Therefore x/x=1.66×10^-6/5×10^-7=3.32

Hence,C6H5COOAg is approximately 3.317 times more soluble in low pH solution.