Question

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

• A. $4.2$ mol
• B. $13.6$ mol
• C. $8.6$ mol
• D. $10$ mol

Answer 1

The correct option is C $8.6$ mol

$Ca(F{)}_2\to C{a}^{+2}+2F^{-}$ $K_{sp}=4\times {10}^{-11}$

$S$ $2S$

$Q_{sp}=\left[C{a}^{+2}\right][F^{-}{]}^2=S\times (2S{)}^2=K_{sp}=4\times {10}^{-11}$

$S=2.15\times {10}^{-4}$

$\left[F^{-}\right]=2S=4.3\times {10}^{-4}$

So one liter water contains $4.3\times {10}^{-4}$ mole of fluoride ions

Thus, $20000$ L of water will contain $=4.3\times 20000\times {10}^{-4}=8.6$ mole.