Question

The solubility product of $BaCr{{o}_{4}}$is $2.4\times {{10}^{-10}}M$. The maximum concentration of $Ba(N{{O}_{3}})$; possible without, precipitation in a $6\times {{10}^{-4}}M\,{{K}_{2}}Cr{{O}_{4}}$solution is:
A. $4\times {{10}^{-7}}M$
B. $1.2\times {{10}^{10}}M$
C. $6\times {{10}^{-4}}M\,$
D. $3\times {{10}^{-4}}M\,$

Answer 1

Hint:Solubility product of any ionic solution tells us the activity of the dissolved ions. It is the product of the cations and anions present in the solution raised to their stoichiometric coefficients. Solubility products are denoted by${{K}_{sp}}$.

Formula used: Solubility product, ${{K}_{sp}}$= ${{[A]}^{a}}{{[B]}^{b}}$, where a and b are stoichiometric coefficients of electrolytes A and B.

Complete step-by-step answer:We have been given the solubility product of $BaCr{{o}_{4}}$to be $2.4\times {{10}^{-10}}M$. So, this can be written as,
Solubility product, \[{{K}_{sp}}=[B{{a}^{2+}}][Cr{{O}_{4}}^{2-}]=2.4\times {{10}^{-10}}{{M}^{2}}\]
Now we have to find the maximum concentration of barium ions $B{{a}^{2+}}$without the precipitation in $6\times {{10}^{-4}}M\,{{K}_{2}}Cr{{O}_{4}}$solution. Precipitation does not occur, so potassium will not be of significance, we will only take the given concentration to be of $Cr{{O}_{4}}$ions, which is $6\times {{10}^{-4}}M\,$. Now the solubility product will be written with $Cr{{O}_{4}}$ions as $6\times {{10}^{-4}}M\,$, and the concentration of Barium ions will be detected as,
\[[B{{a}^{2+}}]\times 6\times {{10}^{-4}}=2.4\times {{10}^{-10}}\]
So, barium ion concentration = \[[B{{a}^{2+}}]=4\times {{10}^{-7}}M\]

Hence, the maximum concentration of barium is $4\times {{10}^{-7}}M$. So, option A is correct.

Note:Solubility product is calculated for sparingly soluble salts in aqueous solutions. Solubility product is the same as the ionic product as both have the product of concentrations raised to the number of ions. The difference in both is that solubility product is applied only for sparingly soluble salts, while ionic product for all types (saturated and unsaturated) solutions. Also solubility products depend upon temperature.