Question

The solubility of a sparingly soluble salt $A$. $B_y$ in water at ${25}^o$C $1.4\times {10}^4$M. The solubility product is $1.1\times {10}^{11}$. The possibilities 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)

$x=1,y=2$

$A_x{B}_y\rightleftharpoons {xA}^{v+}+{yB}^{v-}$

$K_{sp}={\left(xs\right)}^x\times {\left(ys\right)}^y=s^{x+y}{x}^x.y^y$

now, $K_{sp}=1.1\times {10}^{-11}$ and $s=1.4\times {10}^{-4}M$

So if we consider $x=1$ and $y=2\left(A\right)$

$\therefore$ $K_{sp}={(1.4\times {10}^{-4})}^{1+2}{1}^1.2^2\approx 1.1\times {10}^{-11}$

if we consider $x=2$ and $y=1\left(B\right)$

$K_{sp}\ne 1.1\times {10}^{-11}$

again if we consider $x=1$ and $y=3\left(C\right)$ or $x=3$ and $y=1\left(D\right)$ for both cases $K_{sp}\ne 1.1\times {10}^{-11}$

$\therefore$ Possibility will be $x=1$ and $y=2\left(A\right)$