Ionic EquilibriumMedium
Question
Solubility of BaF2 in a solution of Ba(NO3)2 will be represented by the concentration term
Options
A.[Ba2+]
B.[F–]
C.0.5 [F–]
D.2 [NO3 –]
Solution
$BaF_{2}(s) \rightleftharpoons \underset{(S + C)\text{ M}}{Ba^{2 +}} + \underset{2S\text{ M}}{2F^{-}}$
Where S = solubility of BaF2 in CM-Be(NO3)2 solution.
Hence, Solubility, $S = \frac{\left\lbrack F^{-} \right\rbrack}{2}$
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