The molar solubility of in 0.1 M solution of NaF will be [2019]

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Q.
The molar solubility of ${CaF}_{2}$ $(K_{s p} = 5.3 \times 10^{- 11})$ in 0.1 M solution of NaF will be [2019]

1 $5.3 \times 10^{- 11}$ $mol$ $L^{- 1}$

2 $5.3 \times 10^{- 8}$ $mol$ $L^{- 1}$

3 $5.3 \times 10^{- 9}$ $mol$ $L^{- 1}$

4 $5.3 \times 10^{- 10}$ $mol$ $L^{- 1}$

Ans.
(3)

${CaF}_{2} \to \underset{s}{{Ca}^{2 +}} + \underset{2 s}{2 F^{-}}$

$NaF→ \underset{0.1 M}{{Na}^{+}} + \underset{0.1 M}{F^{-}}$

$[{Ca}^{2 +}] = s, [F^{-}] = (2 s + 0.1) \approx 0.1$ $M$

$K_{s p} = [{Ca}^{2 +}] {[F^{-}]}^{2}$

$5.3 \times 10^{- 11} = (s) {(0.1)}^{2}$

$s = \frac{5.3 \times 10^{- 11}}{{(0.1)}^{2}} = 5.3 \times 10^{- 9}$ $mol$ $L^{- 1}$

$∴$ $Molar solubility is 5.3 \times 10^{- 9}$ $mol$ $L^{- 1}$

Solution

Q.
The molar solubility of ${CaF}_{2}$ $(K_{s p} = 5.3 \times 10^{- 11})$ in 0.1 M solution of NaF will be [2019]

1 $5.3 \times 10^{- 11}$ $mol$ $L^{- 1}$

2 $5.3 \times 10^{- 8}$ $mol$ $L^{- 1}$

3 $5.3 \times 10^{- 9}$ $mol$ $L^{- 1}$

4 $5.3 \times 10^{- 10}$ $mol$ $L^{- 1}$

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Solution

Q. The molar solubility of CaF2 (Ksp=5.3×10-11) in 0.1 M solution of NaF will be [2019]

1 5.3×10-11 mol L-1

2 5.3×10-8 mol L-1

3 5.3×10-9 mol L-1

4 5.3×10-10 mol L-1

Ans. (3) CaF2→Ca2+s+2F-2s NaF→Na+0.1 M+F-0.1 M [Ca2+]=s, [F-]=(2s+0.1)≈0.1 M Ksp=[Ca2+][F-]2 5.3×10-11=(s)(0.1)2 s=5.3×10-11(0.1)2=5.3×10-9 mol L-1 ∴ Molar solubility is 5.3×10-9 mol L-1

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Q.
The molar solubility of ${CaF}_{2}$ $(K_{s p} = 5.3 \times 10^{- 11})$ in 0.1 M solution of NaF will be [2019]

1 $5.3 \times 10^{- 11}$ $mol$ $L^{- 1}$

2 $5.3 \times 10^{- 8}$ $mol$ $L^{- 1}$

3 $5.3 \times 10^{- 9}$ $mol$ $L^{- 1}$

4 $5.3 \times 10^{- 10}$ $mol$ $L^{- 1}$