Question

The minimum mass of $NaBr$ which should be added in $200ml$ of $0.0004M$ $ - AgN{O_3}$ solution just to start precipitation of $AgBr$ ${K_{sp}}$ of $AgBr = 4 \times {10^{ - 13}}.$ (Atomic mass of \[Br = 80\] )
$A.$ $1.0 \times {10^{ - 9}}g$
$B.$ $2 \times {10^{10}}g$
\[C.\] $2.06 \times {10^{ - 8}}g$
$D.$ $1.03 \times {10^{ - 7}}g$

Answer 1

Hint: Before solving the given question, we get an idea about the solubility product as it is based on it. Solubility product constant is the equilibrium constant for the dissolution of solid substance into an aqueous solution. It is represented by ${K_{sp.}}$


Complete solution:
In the question it is given that the solubility product constant of $AgBr$ is $4 \times {10^{ - 13}}$ , we have to calculate the minimum mass of $NaBr$ which should be added in $200ml$ of $0.0004M$ $ - AgN{O_3}$ solution just to start precipitation of $AgBr$. Let solve the question step by step;
Given, ${K_{sp}}$ of $AgBr = 4 \times {10^{ - 13}}.$
Here is the reaction from the question,
$NaBr + AgN{O_3} \to AgBr + NaN{O_3}$
Now, $AgBr$ will dissociate into its ions \[A{g^ + }\] and $B{r^ - }$ . so the ${K_{Sp}}$ will become,
${K_{sp}} = \left[ {A{g^ + }} \right]\left[ {B{r^ - }} \right]$ $\left( 1 \right)$
From this we can calculate the concentration of $\left[ {B{r^ - }} \right]$ before we have to know the concentration of $\left[ {A{g^ + }} \right]$ .
Now, the concentration of $\left[ {A{g^ + }} \right]$ can be calculated as;
$AgN{O_3} \to A{g^ + } + NO_3^ - $
From the question the concentration of $\left[ {A{g^ + }} \right]$ is
$\left[ {A{g^ + }} \right] = 4 \times {10^{ - 4}}M$
From equation \[\left( 1 \right)\] the concentration of $\left[ {B{r^ - }} \right]$ is
$\left[ {B{r^ - }} \right] = \dfrac{{4 \times {{10}^{ - 13}}}}{{4 \times {{10}^{ - 4}}}}$
$\left[ {B{r^ - }} \right] = {10^{ - 9}}M$
Now, we have to calculate minimum mass of $NaBr$ as we know the concentration of $\left[ {B{r^ - }} \right]$ is ${10^{ - 9}}$ , therefore the concentration of $[NaBr]$ also ${10^{ - 9}}$ from the reaction,
$NaBr \to N{a^ + } + B{r^ - }$
Now, the number of moles of $NaBr$ for the $200ml$
Moles of \[NaBr = 0.2 \times {10^{ - 9}}\]
$ = 2 \times {10^{ - 10}}$
Hence, the mass of $NaBr$ from the relation;
\[mass = molar\, mass \times number \,of\, moles\]
$mass = 103 \times 2 \times {10^{ - 10}}$
\[mass = 2.06 \times {10^{ - 8}}g\]

Hence the correct option is (D).


Note: It is to be noted that the solubility product constant is dependent on the temperature. Solubility product constant usually increases with increase in temperature due to increased solubility, the presence of ion-pair also affects the value of solubility product constant and the diverse effect, that is if the ion is uncommon, the value of solubility product is high.