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Model Paper 4 — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryModel Paper 45 Marks
Question
The ionization constant of benzoic acid is $6.46 \times 10^{-5}$ and $K _{\text {sp }}$ for silver benzoate is $2.5 \times 10^{-13}$. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?
✓

Answer

$
C_6 H_5 COOAg \rightarrow C_6 H_5 COO^{-}+Ag^{+}
$
Solubility of water. Suppose solubility in water $= x mol L L ^{-1}$ then
$
\begin{aligned}
& x^2=K_{sp} \text { or } x=\sqrt{K_{s p}}=\sqrt{2.5 \times 10^{-13}}
\end{aligned}
$
$
=5 \times 10^{-7} mol L^{-1}
$
Solubility in buffer of $pH =3.19$
$
\begin{aligned}
& pH=3.19 \text { means }-\log \left[H^{+}\right]=3.19 \\
& \text { or } \log \left[H^{+}\right]=-3.19=\overline{4} .81 \text { or }\left[H^{+}\right]=6.457 \times 10^{-4} M
\end{aligned}
$
$C _6 H _5 COO ^{-}$ions now combine with the $H ^{+}$ions to form benzoic acid but $\left[ H ^{+}\right]$remains almost constant because we have buffer solution. Now
$
C_6 H_5 COOH \rightleftharpoons C_6 H_5 COO^{-}+H^{+}
$
$
\therefore K_a=\frac{\left[C_6 H_5 COO^{-}\right]\left[H^{+}\right]}{\left[C_6 H_5 COOH\right]} \text { or } \frac{\left[C_6 H_5 COOH\right]}{\left[C_6 H_5 COO^{-}\right]}=\frac{\left[H^{+}\right]}{K_a}=\frac{6.457 \times 10^{-4}}{6.46 \times 10^{-5}}=10 \ldots (i)
$
Suppose solubility in the buffer solution is ' $y^{\prime} mol L ^{-1}$. Then as most of the benzoate ions are converted into benzoic acid molecules (which remain almost ionized), we have
$
\begin{aligned}
& y=\left[Ag^{+}\right]=\left[C_6 H_5 COO^{-}\right]+\left[C_6 H_5 COOH\right]=\left[C_6 H_5 COO^{-}\right]+10\left[C_6 H_5 COO^{-}\right]=11\left[C_6 H_5 COO^{-}\right] \text {using equation (i) } \\
& \therefore\left[C_6 H_5 COO^{-}\right]=\frac{y}{11} \therefore K_{s p}=\left[C_6 H_5 COO^{-}\right]\left[Ag^{+}\right] \\
& 2.5 \times 10^{-3}=\frac{y}{11} \times y \text { or } y^2=2.75 \times 10^{-12} \text { or } y=1.66 \times 10^{-6} \\
& \frac{y}{x}=\frac{1.66 \times 10^{-6}}{5 \times 10^{-7}}=3.32
\end{aligned}
$
Note that in case of salts of weak acids, the solubility is more in the acidic solution than in water. The reason, in general, may be explained as follows: Taking example of $C _6 H _5 COOAg$, we have
$
C_6 H_5 COOAg \rightleftharpoons C_6 H_5 COO^{-}+Ag^{+}
$
In acidic solution, the anions $\left( C _6 H _5 COO ^{-}\right.$in the present case) undergo protonation in presence of acid. Thus, $C _6 H _5 COO ^{-}$ions are removed. Hence, equilibrium shifts forward producing more $Ag ^{+}$ions. Alternatively, as $C _6 H _5 COO ^{-}$ions are removed, $Q _{ sp }$ decreases. In order to maintain solubility product equilibrium $\left( Q _{ sp }= K _{ sp }\right), Ag ^{+}$ion concentration must increase. Hence, solubility is more.

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