Question

If the solubility product of a sparingly soluble salt $L{i}_{3}P{O}_4$ is $2.7\times {10}^{-9}$. The concentration $\left(mol{L}^{-1}\right)$ of $L{i}^{+}$ ions in the saturated solution of the salt is:
Given : $\left(100\right)$${}^{\frac{1}{4}}=3.16$

• A. $3.78\times {10}^{-9}mol{L}^{-1}$
• B. $2.7\times {10}^{-9}mol{L}^{-1}$
• C. $3.16\times {10}^{-3}mol{L}^{-1}$
• D. $9.48\times {10}^{-3}mol{L}^{-1}$

Answer 1

The correct option is D $9.48\times {10}^{-3}mol{L}^{-1}$

The equilibrium between the undissolved solid $L{i}_{3}P{O}_4$ and the ions in a saturated solution can be represented by the solution.
$L{i}_{3}P{O}_4\left(s\right)\rightleftharpoons 3L{i}^{+}\left(aq\right)+P{O}_4^{3-}\left(aq\right)c00c-s3ss$
"s" is solubility.
The solubility product is given by
$K_{sp}=\left[L{i}^{+}{]}^3\right[P{O}_4^{3-}]$
$K_{sp}=(3s{)}^3\times s=27s^4$
$K_{sp}=27s^4=2.7\times {10}^{-9}$
$s^4={10}^{-10}=100\times {10}^{-12}$
$s=3.16\times {10}^{-3}\left[L{i}^{+}\right]=3\times (3.16\times {10}^{-3})=9.48\times {10}^{-3}mol{L}^{-1}$