Ionic Equilibrium | JEE Main Chemistry Chapter Wise Question Papers

(4)

$XS (s) ⇌ X^{2 +} (aq) + S^{2 -} (aq)$

$For precipitation of XS(s)$

$[X^{2 +}] [S^{2 -}] \geq K_{s p} (XS)$

$[S^{2 -}] \geq \frac{1 \times 10^{- 22}}{0.01} = 10^{- 20}$

$YS (s) ⇌ Y^{2 +} (aq) + S^{2 -} (aq)$

$For precipitation of YS(s)$

$[Y^{2 +}] [S^{2 -}] \geq K_{s p} (YS)$

$[S^{2 -}] \geq \frac{4 \times 10^{- 16}}{10^{- 2}} = 4 \times 10^{- 14}$

$Now, H_{2} S (aq) ⇌ 2 H^{+} (aq) + S^{2 -} (aq)$

$\frac{[S^{2 -}] {[H^{+}]}^{2}}{[H_{2} S]} = K_{a 1} \times K_{a 2} = 1 \times 10^{- 21}$

$[S^{2 -}] = \frac{1 \times 10^{- 21} \times [H_{2} S]}{{[H^{+}]}^{2}} \geq 4 \times 10^{- 14}$

${[H^{+}]}^{2} \leq \frac{1}{4} \times 10^{- 7} \times 10^{- 1}$

$[H^{+}] \leq \frac{1}{2} \times 10^{- 4} \Rightarrow pH \geq 4.3$

(100)

$\begin{matrix} H_{2} X & ⇌ & H^{+} & + & {HX}^{-} \\ 0.1 - x & x + y & x - y \end{matrix}$

$2.5 \times 10^{- 8} = \frac{(x + y) (x - y)}{0.1 - x}$

$\begin{matrix} {HX}^{-} & ⇌ & H^{+} & + & X^{2 -} \\ x - y & x + y & y \end{matrix}$

$1 \times 10^{- 13} = \frac{(x + y) (y)}{x - y}$

$Approximate : K_{a 1} ≫ K_{a 2} \Rightarrow So x ≫ y$

$x + y \approx x, x - y \approx x$

$10^{- 13} = \frac{x \cdot y}{x}$

$y = 10^{- 13}$

$[X^{2 -}] = 10^{- 13}$

$[X^{2 -}] = 100 \times 10^{- 15}$

Want Us to Email you About Special Offers & Updates?