Question

If $K_{sp}$ of a salt $A_2{B}_3$ is given by $1\times {10}^{-25}$. Then find the $s^5$ of the salt?
Here 's' is solubility.

• A. $\frac{{10}^{-2}}{108}$
• B. ${10}^{-5}$
• C. $\frac{{10}^{-5}}{108}$
• D. $\frac{{10}^{-25}}{108}$

Answer 1

The correct option is C $\frac{{10}^{-5}}{108}$

For the salt $A_x{B}_y$
$A_x{B}_y=x{A}^{y+}+y{B}^{x-}s00-xsys$
the solubility of a sparingly soluble salt is given by $K_{sp}=(xs{)}^x.(ys{)}^y,$
In the given salt $A_2{B}_3$
$\text{x=2 and y=3}$
$K_{sp}=(2s{)}^2.(3s{)}^3$
$1\times {10}^{-25}=(2s{)}^2.(3s{)}^3$
$1\times {10}^{-25}=4s^2.27s^3$
$s^5=\frac{1\times {10}^{-25}}{108}$