Question

A well is dug in a bed of rock containing fluorspar $\left(Ca{F}_2\right)$. If the well contains $20000L$ 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

Given:
$Ca{F}_2\left(s\right)\rightleftharpoons \underset{s}{C{a}^{2+}}+\underset{2s}{2F^{-}}$
Given $K_{sp}=4\times {10}^{-11}$
We know $Q_{sp}=\left[C{a}^{+2}\right][F^{-}{]}^2$
$=\left(s\right)(2s{)}^2$
$\because K_{sp}=4\times {10}^{-11}$
So, $\left(s\right)(2s{)}^2=4\times {10}^{-11}$
$s=2.15\times {10}^{-4}$
$\left[F^{-}\right]=2s$
$=2\times 2.15\times {10}^{-4}$
$=4.3\times {10}^{-4}$
One liter water contains $4.3\times {10}^{-4}$ moles of fluoride ions.
So $2000L$ of water will contain
$=4.3\times 20000\times {10}^{-4}$
$=8.6\text{mole}$