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Question 13 Marks
Define 'Solubility product'. Derive the relationship between solubility and solubility product for $Pbl_2.$
Answer
Solubility product:
In the saturated solution of sparingly soluble salt, the product of equilibrium concentrations of the constituent ions raised to the power equal to their respective coefficients in the balanced equilibrium expression at a given temperature is called the solubility product.
Solubility equilibrium for $PbI_2,$ is given by,
$PbI _{2( s )} \rightleftharpoons Pb ^{2 \oplus}{ }_{( aq )}+2 I _{( aq )}^{\ominus}$
$\therefore x = 1, y = 2$
$\therefore K_{sp} = (1)^1 (2)^2 S^{1+2}$
$\therefore K_{sp} = 4 S^3$
The above expression shows the relationship between solubility and solubility product for $PbI_2.$
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Question 23 Marks
Calculate the $pH$ of buffer solution composed of $0.01 M$ weak base $BOH$ and $0.02 M$ of its salt $BA.$
$[K_b= 1.8 \times 10^{-5}$ for weak base$]$
Answer
$pOH = pK _{ b }+\log _{10} \frac{[\text { Salt }]}{[\text { Base }]}$
Now, $pK_b = – log_{10} K_b$
$= 5 – log_{10}1.8$
$= 5 – 0.2553$
$pK_b = 4.7447$
Then, $pOH =4.7447+\log _{10} \frac{0.02}{0.01}$
$= 4.7447 + log_{10}2$
$= 4.7447 + 0.3010$
$pOH = 5.0457$
$\because pH + pOH = 14$
$\therefore pH = 14 – pOH$
$= 14 – 5.0457$
$= 8.9543$ or $8.9$
$pH = 8.95$
The $pH$ of the buffer solution will be $8.95.$
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