Question

The solubility of a sparingly soluble salt $A_x{B}_y$ in water at ${25}^{o}C$ is $1.4\times {10}^{-4}M$. The solubility product is $1.1\times {10}^{-11}$. The possible value(s) of x and y is/are:

• A. $x=1,y=2$
• B. $x=2,y=1$
• C. $x=1,y=3$
• D. $x=3,y=1$

Answer 1

Answer: (A), (B)

The correct options are
A $x=1,y=2$
B $x=2,y=1$

Let S moles per liter be the solubility of the salt $A_x{B}_y$.

The expression for the solubility product will be

$K_{sp}=\left[A^{y+}{]}^x\right[B^{x-}{]}^y=(x\times S{)}^x(y\times y{)}^y=(S{)}^{x+y}{x}^x{y}^y$

By substituting values in the above expression, we get

$K_{sp}=1.1\times {10}^{-11}=(1.4\times {10}^{-4}{)}^{x+y}{x}^x.y^y$

On comparing the values, $x+y=3$.

Hence, $x^x.y^y=\frac{1.1\times {10}^{-11}}{1.4\times 1.4\times 1.4\times {10}^{-12}}=\frac{11}{1.96\times 1.4}=4$

Thus, either $x=1,y=2$ or $y=1,x=2$.