Chemical and Ionic Equilibrium 2 Question 40
40. The solubility of a salt of weak acid $(A B)$ at $\mathrm{pH} 3$ is $Y \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$. The value of $Y$ is _ (Given that the value of solubility product of $A B\left(K_{\mathrm{sp}}\right)=2 \times 10^{-10}$ and the value of ionisation constant of HB $\left(K_{a}\right)=1 \times 10^{-8}$
(2018 Adv.)
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Solution:
- Key Idea Solubility of salt of weak acid (AB) in presence of $\mathrm{H}^{+}$ions from buffer solution can be calculated with the help of following formula.
Solubility $=\sqrt{K_{\mathrm{sp}}\left[\frac{\left[\mathrm{H}^{+}\right]}{k_{a}}+1\right]}$
Given, $\mathrm{pH}=3$, so $\left[\mathrm{H}^{+}\right]=10^{-3}$
$$ K_{a}=1 \times 10^{-8} \Rightarrow K_{\mathrm{sp}}=2 \times 10^{-10} $$
after putting the values in above formula
Solubility $=\sqrt{2 \times 10^{-10}\left(\frac{10^{-3}}{10^{-8}}+1\right)} \approx \sqrt{2 \times 10^{-5}}=4.47 \times 10^{-3} \mathrm{M}$
Hence, the value of $y=4.47$
