Question

A saturated solution of $C{a}_3(P{O}_4{)}_2$ contains $2.0\times {10}^{-8}$ M $C{a}^{2+}$ and $1.6\times {10}^{-5}$ M $P{O}_4^{3-}$ at a certain temperature. The solubility product $\left(K_{sp}\right)$ of $C{a}_3(P{O}_4{)}_2$ at that temperature is:

• A. $2.048\times {10}^{-34}$
• B. $2.04\times {10}^{-33}$
• C. $3.20\times {10}^{-34}$
• D. $8\times {10}^{-34}$

Answer 1

The correct option is B $2.04\times {10}^{-33}$

$C{a}_3(P{O}_4{)}_2\leftrightharpoons 3C{a}^{2+}+2P{O}_4^{3-}$
Now,

$K_{sp}=[C{a}^{2+}{]}^3\times [P{O}_4^{3-}{]}^2$

$K_{sp}=(2.0\times {10}^{-8}{)}^3\times (1.6\times {10}^{-5}{)}^2$

$K_{sp}=$ $2.04\times {10}^{-33}$