Ionic EquilibriumMedium
Question
An amount of 0.1millimole of CdSO4 is present in 10 ml acid solution of 0.08 M-HCl. Now H2S is passed to precipitate all the Cd2+ ions. What would be the pH of solution after filtering precipitate, boiling o H2S and making the solution 100 ml by adding water?
Options
A.3.0
B.2.0
C.4.0
D.2.22
Solution
$Cd^{2 +} + H_{2}S \rightleftharpoons CdS(s) + 2H^{+}$
$0.1 \times 10^{- 3}\text{ mole}$ $\frac{10 \times 0.08}{1000} = 0.8 \times 10^{- 3}\text{ mole}$
Final 0 $0.8 \times 10^{- 3} + 0.2 \times 10^{- 3} = 10^{- 3}\text{ mole}$
$\therefore\left\lbrack H^{+} \right\rbrack_{\text{final}} = \frac{10^{- 3}}{100} \times 1000 = 0.01\text{ M} \Rightarrow P^{H} = 2.0$
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