The molar solublity of CaF_ 2 (K_ sp =5.3 10^ -11 ) in 0.1 M solution of NaF will be

The molar solublity of CaF_ 2 (K_ sp =5.3 10^ -11 ) in 0.1 M solu…

MCQ

The molar solublity of $\mathrm{CaF}_{2} \left(\mathrm{K}_{\mathrm{sp}}=5.3 \times 10^{-11}\right)$ in $0.1 \mathrm{M}$ solution of NaF will be

✓

Correct option: C.

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

c
$\mathrm{CaF}_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Ca}^{+2}(\mathrm{aq})+2 \mathrm{F}(\mathrm{aq})$

$(a-s)\quad \quad \quad \quad S\quad \quad \quad \quad 2s$

$\mathrm{NaF}(\mathrm{aq}) \rightarrow \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{F}^{-}(\mathrm{aq})$

$C\quad \quad \quad \quad \quad 0\quad \quad \quad \quad 0$

$0\quad \quad \quad \quad \quad C\quad \quad \quad \quad C$

In solution- $[\mathrm{F^-}]=\left(2 \mathrm{s}+\mathrm{C}\right)$

$\left[\mathrm{F}^{-}\right] \approx \mathrm{C}$ (due to common lon effect)

$\mathrm{K}_{\mathrm{sp}\left(\mathrm{Ca} \mathrm{F}_{2}\right)}=\left[\mathrm{Ca}^{+2}\right] .\left[\mathrm{F}^{-}\right]^{2}$

$\mathrm{K}_{\mathrm{sp}\left(\mathrm{Ca} \mathrm{F}_{2}\right)}=s\cdot C^2$

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

$s=5.3 \times 10^{-9}\; \mathrm{mol} \mathrm{L}^{-1}$

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