Question

at 25°c, ( k_{sp} ) for calcium fluoride (mm = 78.08 g/mol) is ( 4.0 \times 10^{-11} ). how many grams of this salt should dissolve in 0.75 l of water. report your answer to 3 decimal places. assume that there is complete dissociation of the salt and that there are no other important equilibria that can affect the salt’s solubility.

Explanation:

Step1: Write the dissolution equation

Calcium fluoride (\(CaF_2\)) dissolves as \(CaF_2(s)
ightleftharpoons Ca^{2 +}(aq)+2F^{-}(aq)\). Let the solubility of \(CaF_2\) be \(s\) mol/L. Then, \([Ca^{2+}]=s\) and \([F^{-}]=2s\).

Step2: Write the \(K_{sp}\) expression

\(K_{sp}=[Ca^{2+}][F^{-}]^{2}\). Substituting the values, \(K_{sp}=(s)(2s)^{2}=4s^{3}\).
Given \(K_{sp}=4.0\times 10^{-11}\), we have \(4s^{3}=4.0\times 10^{-11}\).
Solving for \(s\):

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Step3: Calculate the moles of \(CaF_2\) in \(0.75\) L

Moles of \(CaF_2=n = s\times V\). Substituting \(s = 2.154\times 10^{-4}\text{ mol/L}\) and \(V = 0.75\space L\), \(n=(2.154\times 10^{-4}\text{ mol/L})\times0.75\space L=1.616\times 10^{-4}\text{ mol}\)

Step4: Calculate the mass of \(CaF_2\)

Mass \(m=n\times MM\). Given \(MM = 78.08\space g/mol\), \(m=(1.616\times 10^{-4}\text{ mol})\times78.08\space g/mol\)

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Answer:

\(0.013\)