Question

Solubility product of silver bromide ($AgBr$) $5 \times {10^{ - 13}}$. The quantity of potassium bromide ($KBr$) to be added to $1$litre of $0.05M$ solution of silver nitrate to start the precipitation of $AgBr$ is:
A) $6.2 \times {10^{ - 5}}g$
B) $5.0 \times {10^{ - 8}}g$
C) $1.2 \times {10^{ - 10}}g$
D) $1.2 \times {10^{ - 9}}g$

Answer 1

Hint: The solubility product of a molecule is somewhat like equilibrium constant and its value depends on temperature. Solubility products usually increase with increase in temperature due to increased solubility. If the cation and anion of an ion or molecule combine in an aqueous solution forming insoluble solid is called precipitate.

Complete step by step solution:
Consider the general expression for solubility product:
\[{M_x}{X_y} \rightleftharpoons x{M^{p + }}_{(aq)} + y{X^{q - }}_{(aq)}\]
($x.{p^ + } = y.{q^ - }$ )
Its solubility product is given as
${K_{sp}} = {[{M^{p + }}]^x}{[{X^{q - }}]^y}$
In the above case the reaction is:
$AgBr \rightleftharpoons A{g^ + } + B{r^ - }$
Its solubility product is given as:
${K_{sp}}$ of $AgBr$$ = [A{g^ + }][B{r^ - }]$
Given ${K_{sp}}$ of $AgBr$$ = 5 \times {10^{ - 3}}$
Also given
$[A{g^ + }] = 0.05M$
By substituting the given values
$ \Rightarrow 5 \times {10^{ - 3}} = 0.05[B{r^ - }]$
$ \Rightarrow [B{r^ - }] = \frac{{5 \times {{10}^{ - 3}}}}{{0.05}} = 1 \times {10^{ - 11}}M$
It means that ${10^{ - 11}}$ moles of $KBr$moles in $1$ litre of solution
Molecular mass of $KBr = 120gmo{l^{ - 1}}$
Therefore $1.2 \times {10^{ - 9}}g$ of $KBr$ is to be added to $1$ litre of $0.05M$solution of silver nitrate in order to start the precipitation silver bromide ($AgBr$)
$\therefore $Option D is the correct answer.

Additional information:There are various factors determining the formation of a precipitate. In some reactions mostly in solutions used for buffer temperature is the dependent factor on the formation of precipitate whereas in some other cases concentration of the solution is the dependent factor. The technique of separating ions from solution based on their difference in solubilities is called fractional precipitation.


Note: The solubility product literally is the product of solubilities of ions in mole per litre. It usually is a representation of the solubility of a solute in a solution. Higher the value of ${K_{sp}}$ higher will be the solubility. Solubility products represent the maximum extent a solid can dissolve in a solution.