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6-2.Equilibrium-II (Ionic Equilibrium)
hard
The solubility of a sparingly soluble salt $A_x B_y$ in water at $25\,^oC = 1.4 \times 10^{-4}\ M$ . The solubility product is $1.1 \times 10^{-11}$ . The values of $x$ and $y$ may be
A
$x = 1 , y = 2$
B
$x = 2, y = 2$
C
$x = 1 , y = 3$
D
$x = 3, y = 1$
Solution
${{\text{K}}_{{\text{sp}}}} = 1.1 \times {10^{ – {\text{11}}}} = $ ${\left( {1.4 \times {{10}^{ – 4}}} \right)^{{\text{x}} + {\text{y}}}}{{\text{x}}^{\text{x}}} \cdot {{\text{y}}^{\text{y}}}$
so we have $x+y=3$ (by comparing values)
so, ${x^x} \cdot {y^y} = \frac{{1.1 \times {{10}^{ – 11}}}}{{1.4 \times 1.4 \times 1.4 \times {{10}^{ – 12}}}}$ $ = \frac{{110}}{{1.96 \times 1.4}} = 4$
Hence
$x=1,\,\,y=2$ or
$y=1,\,\,x=2$
Standard 11
Chemistry
