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Question: A well is dug in a bed of rock containing fluorspar \(\left( {{\text{Ca}}{{\text{F}}_{\text{2}}}} \r...

A well is dug in a bed of rock containing fluorspar (CaF2)\left( {{\text{Ca}}{{\text{F}}_{\text{2}}}} \right). If the well contains 20000 L20000{\text{ L}} of water, what is the amount of F−{{\text{F}}^ - } in it? (KSP=4×10−11)\left( {{{\text{K}}_{{\text{SP}}}} = 4 \times {{10}^{ - 11}}} \right)
A) 4.2 mol4.2{\text{ mol}}
B) 13.6 mol13.6{\text{ mol}}
C) 8.6 mol8.6{\text{ mol}}
D) 10 mol10{\text{ mol}}

Explanation

Solution

We know that the solubility product of any salt at any temperature is the product of the molar concentration of its constituent ions. The concentration of ions is raised to the number of ions produced on dissociation of one molecule of the salt.

Complete solution:
We know that the solubility product of any salt at any temperature is the product of the molar concentration of its constituent ions. The concentration of ions is raised to the number of ions produced on dissociation of one molecule of the salt.
We know that the solubility of a salt at any temperature is calculated from its solubility product.
Now, we are given a salt fluorspar CaF2{\text{Ca}}{{\text{F}}_{\text{2}}}. The salt dissociates as follows:
CaF2⇌Ca2++2F−{\text{Ca}}{{\text{F}}_{\text{2}}} \rightleftharpoons {\text{C}}{{\text{a}}^{2 + }} + 2{{\text{F}}^ - }
The solubility product of the salt fluorspar CaF2{\text{Ca}}{{\text{F}}_{\text{2}}} is given as follows:
KSP=[Ca2+][F−]2{{\text{K}}_{{\text{SP}}}} = [{\text{C}}{{\text{a}}^{2 + }}]{[{{\text{F}}^ - }]^2}
Where KSP{{\text{K}}_{{\text{SP}}}} is the solubility product.
For the salt fluorspar CaF2{\text{Ca}}{{\text{F}}_{\text{2}}}.
KSP=[Ca2+][F−]2=s×(2s)2=4s3{{\text{K}}_{{\text{SP}}}} = [{\text{C}}{{\text{a}}^{2 + }}]{[{{\text{F}}^ - }]^2} = s \times {\left( {2s} \right)^2} = 4{s^3}
Where ss is the solubility of the ions.
We are given that the solubility product of salt fluorspar CaF2{\text{Ca}}{{\text{F}}_{\text{2}}} is 4×10−114 \times {10^{ - 11}}. Thus,
4s3=4×10−114{s^3} = 4 \times {10^{ - 11}}
s3=4×10−114{s^3} = \dfrac{{4 \times {{10}^{ - 11}}}}{4}
s=10−113s = \sqrt[3]{{{{10}^{ - 11}}}}
s=2.15×10−4s = 2.15 \times {10^{ - 4}}
Now, the concentration of F−{{\text{F}}^ - } ions is,
[F−]=2s[{{\text{F}}^ - }] = 2s
Thus,
[F−]=2×(2.15×10−4)[{{\text{F}}^ - }] = 2 \times \left( {2.15 \times {{10}^{ - 4}}} \right)
[F−]=4.3×10−4[{{\text{F}}^ - }] = 4.3 \times {10^{ - 4}}
Thus, the concentration of F−{{\text{F}}^ - } ions in CaF2{\text{Ca}}{{\text{F}}_{\text{2}}}is 4.3×10−4 M4.3 \times {10^{ - 4}}{\text{ M}}.
Thus, one litre of water contains 4.3×10−4 mol4.3 \times {10^{ - 4}}{\text{ mol}} of F−{{\text{F}}^ - } ions. Thus, the F−{{\text{F}}^ - } ions in 20000 L20000{\text{ L}} of water is,
Amount of F− ions=20000 L×4.3×10−4 mol1 L{\text{Amount of }}{{\text{F}}^ - }{\text{ ions}} = 20000{\text{ L}} \times \dfrac{{4.3 \times {{10}^{ - 4}}{\text{ mol}}}}{{1{\text{ L}}}}
Amount of F− ions=8.6 mol{\text{Amount of }}{{\text{F}}^ - }{\text{ ions}} = 8.6{\text{ mol}}
Thus, the amount of F−{{\text{F}}^ - } ions is 8.6 mol8.6{\text{ mol}}.

Thus, the correct option is (C).

Note: The solubility product is calculated using the concentration of the ions in which the salt has dissociated. Solubility factor depends on various factors such as temperature, pressure and nature of the electrolyte. The concentration of ions is affected by these factors and thus, the solubility product gets affected.