Question

The precipitate of Ca$F_2$ ($K_{sp}=1.7\times {10}^{-10}$) is obtained when equal volumes of which of the following are mixed?

• A. ${10}^{-4}MC{a}^{2+}+{10}^{-4}M{F}^{-}$
• B. ${10}^{-2}MC{a}^{2+}+{10}^{-3}M{F}^{-}$
• C. Both
• D. None of these

Answer 1

Answer: (B)

${10}^{-2}MC{a}^{2+}+{10}^{-3}M{F}^{-}$

The precipitate of Ca$F_2$ ($K_{sp}=1.7\times {10}^{-10}$) is not obtained when equal volumes of ${10}^{-4}MC{a}^{2+}+{10}^{-4}M{F}^{-}$ are mixed. This is because ionic product of calcium fluoride will not exceed solubility product.

$Q=\left[C{a}^{2+}\right][F^{-}{]}^2$
$Q=0.5\times {10}^{-4}\times (0.5\times {10}^{-4}{)}^2$
$Q=1.3\times {10}^{-13}$
$K_{sp}=1.7\times {10}^{-10}$

$1.3\times {10}^{-13}<1.7\times {10}^{-10}$

$Q<K$

The precipitate of Ca$F_2$ ($K_{sp}=1.7\times {10}^{-10}$) is obtained when equal volumes of ${10}^{-2}MC{a}^{2+}+{10}^{-3}M{F}^{-}$ are mixed. This is because ionic product of calcium fluoride will exceed solubility product.

$Q=\left[C{a}^{2+}\right][F^{-}{]}^2$
$Q=0.5\times {10}^{-2}\times (0.5\times {10}^{-3}{)}^2$
$Q=1.3\times {10}^{-9}$
$K_{sp}=1.7\times {10}^{-10}$

$1.3\times {10}^{-9}>1.7\times {10}^{-10}$

$Q>K$

Note: The concentrations are multiplied with 0.5 because when two solutions are mixed in equal proportions, the concentration of each species is reduced to one half.