$B i_{2} S_{3}$ has to precipitated from a solution containing $0.1 M B i^{3 +}$ ion by saturating it with 0.1 M $H_{2} S$ solution. In order to just start the precipitation the pH of solution has to be maintained at ‘x’ by adding $10^{- x} M H C l$ . The same HCl was used to titrate a 20 ml ‘y’ M $N a_{2} C O_{3}$ solution using phenolphthalein as an indicator, 4000 ml of HCl was used to reach end point. Find the value of $x + y$ . $K_{s p} o f B i_{2} S_{3} = 1 \times 10^{- 56},$ $K a f o r H_{2} S = 10^{- 21}$ .

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Detailed Solution

$B i_{2} S_{3} \overset{}{⇌} 2 B i^{+ 3} + 3 S^{- 2} K_{s p} = {[B i^{+ 3}]}^{2} {[S^{- 2}]}^{3} 10^{- 56} = {(0.1)}^{2} {[S^{- 2}]}^{3} 10^{- 54} = {[S^{- 2}]}^{3} [S^{- 2}] = 10^{- 18} m o l e / l H_{2} S \overset{}{⇌} 2 H^{+} + S^{- 2} [K a] K a = \frac{{[H^{+}]}^{2} [S^{- 2}]}{[H_{2} S]} 10^{- 21} = \frac{{[H^{+}]}^{2} \times 10^{- 18}}{0.1} [H^{+}] = 10^{- 2} M o l a r [∴ x = 2] M . e q o f H C l = \frac{1}{2} \times m . e s o f N a_{2} C O_{3} 10^{- 2} \times 1 \times 4000 = \frac{1}{2} \times y \times 2 \times 20 y = 2 (x + y) = 4$