Question

$50litre$ of a solution containing ${10^{ - 5}}mole$ of $A{g^ + }$ is mixed with $50litre$ of a $2 \times {10^{ - 7}}M$ $HBr$ solution. $[A{g^ + }]$ in resultant solution is: $[Given:{K_{sp}}(AgBr) = 5 \times {10^{ - 13}}]$
A. ${10^{ - 5}}M$
B. ${10^{ - 6}}M$
C. ${10^{ - 7}}M$
D. None of these

Answer 1

Hint: The ionic product is defined as the product of the concentrations of ions, each raised to the power specified by its stoichiometric coefficient in a solution of a salt. The solubility product of a compound in any solution is the product of the concentration of the ions in the solution.

Complete step by step answer:

The reaction of the solution containing silver univalent cation with the hydrogen bromide, when it is added to the acid will be:
$A{g^ + } + HBr \to AgBr + {H^ + }$
The silver bromide precipitate is formed in the reaction. The solubility product of the silver bromide is given as: ${K_{sp}}(AgBr) = 5 \times {10^{ - 13}}$
Thus, in a reaction, for the formation of the precipitate, the most necessary condition is that the ionic product of the ions of the salt in the solution should be greater than the solubility product of the salt.
$Q > {K_{sp}}$
Where, $Q = $ ionic product
Thus, the ionic product for silver bromide salt will be:
$Q = [A{g^ + }][B{r^ - }]$
As per the question, the molarity of the solution containing $A{g^ + }$ as well as $HBr$ ions will be:
$Molarity = \dfrac{{moles}}{{volume(l)}}$
$[A{g^ + }] = \dfrac{{{{10}^{ - 5}}}}{{50 + 50}} = {10^{ - 7}}M$
Similarly, the concentration of $B{r^ - }$ in the solution will be:
The number of moles of $B{r^ - }$ can be calculated by:
$n = molarity \times volume$

Thus, the molarity of the bromide ions will be$ = \dfrac{{{n_{B{r^ - }}}}}{{{V_{total}}}}$
$[B{r^ - }] = \dfrac{{2 \times {{10}^{ - 7}} \times 50}}{{50 + 50}} = {10^{ - 7}}M$
Thus, the ionic product will be = $Q = {10^{ - 7}} \times {10^{ - 7}} = {10^{ - 14}}$
As mentioned earlier, for the formation of precipitate, $Q > {K_{sp}}$.
Here, ${10^{ - 14}} < 5 \times {10^{ - 13}}$ , i.e. $Q < {K_{sp}}$.
Hence, the precipitate does not form. This implies that the concentration of silver ion at equilibrium will be: $[A{g^ + }] = {10^{ - 7}}M$
Thus, the correct option is C. ${10^{ - 7}}M$.

Note:
The ionic product and the solubility product play an important role in determining the concentration of any ion in the solution. If the ionic product is greater than the solubility product of the salt, it will form a precipitate and then the dissociated moles need to be determined. But if the case is reversed, then the concentration of the ion will remain as such because it will exist in the same ionic form and can undergo any dissociation as the precipitate is unable to form.