Question

The solubility of CaF2 in water is 2 *10^-4mol per liter.The solubility od CaF2 in0.01M NaF is

Answer 1

Explanation:

The dissociation equilibrium of calcium fluoride is as given below.

$CaF_{2} \: ⇌ \: \binom{ {Ca}^{2 + } + 2 {F}^{ - } }{x \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2x}$

The dissociation of NaF is as shown below.

$NaF→ \binom{ {Na}^{ + } + {F}^{ - } }{0.1 \: \: \: \: \: \: \: \: \: 0.1}$

Thus total fluoride ion concentration is

2x+0.1M, but 2x<<<0.1

Hence, 2x is neglected and the total fluoride ion concentration is 0.1 M.

The expression for the solubility product is as follows.

$k_{sp} = {[ {Ca}^{2 + } ][ {F}^{ - } ]}^{2}$

Substitute values in the above expression:

$2 \times {10}^{ - 4} = x( {0.1})^{2} \\ x = 2 \times {10}^{ - 2}$