Chemical And Ionic Equilibrium Result Question 40
40. The solubility of a salt of weak acid $(A B)$ at 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 $A B\left(K_{\text {sp }}\right)=2 \times 10^{-10}$ and the value of ionisation constant of $\mathrm{HB}\left(K_a\right)=1 \times 10^{-8}$
(2018 Adv.)
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Answer:
Correct Answer: 40. $( 4.47 )$
Solution:
$ \begin{alignedat} &40. \\ &\begin{array}{|c|} \hline \begin{array}{l} Key Idea Solubility of salt of weak acid $(\mathrm{AB})$ in presence of its conjugate base \text { the } \mathrm{H^+} \text { ions from buffer solution can be calculated with the } \\ \text { help of the following formula. } \end{array} \\ \hline \text { Solubility } =\sqrt{K _{\mathrm{sp}} \dfrac{\left[\mathrm{H}^{+}\right]}{k _a}} \\ \hline \end{array} \end{aligned} $
$ \text { Given, } \mathrm{pH} = 3 \text {, so }\left[\mathrm{H}^{+}\right] = 10^{-3} $
$ K_a=1 \times 10^{-8} \Rightarrow K_{\mathrm{sp}}=2 \times 10^{-10} $
after substituting the values into the formula
$ Solubility $=\sqrt{2 \times 10^{-10}\left(\dfrac{10^{-3}}{10^{-8}}+1\right)} \approx \sqrt{2 \times 10^{-3}}=4.47 \times 10^{-2} $ $\mathrm{M} $
Hence, the value of $y=4.47$
BETA
