Question

A well is dug in a bed of rock containing fluorspar $\left(Ca{F}_2\right)$. If the well contains $20000\text{L}$ of water, what is the amount of $F^{-}$ in it? $K_{sp}=4\times {10}^{-11}({10}^{1/3}=2.15)$

• A. 4.3 mol
• B. 6.8 mol
• C. 8.6 mol
• D. 13.6 mol

Answer 1

The correct option is C 8.6 mol

$\begin{array}{ccccc}Ca{F}_2\left(s\right)& \rightleftharpoons & C{a}^{2+}& +& 2F^{-}\\ & & s& & 2s\end{array}$
$4s^3=K_{sp}=4\times {10}^{-11}\Rightarrow s=2.15\times {10}^{-4}M$
So, amount of $F^{-}$ in $20000L$ of water $=2s\times 20000=8.6\text{mol}$