Question

Solubility product of silver bromide $5\times {10}^{-13}$ . the quantity of potassium bromide (molar mass is taken as $120g/mol$ ) 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 constant $K_{sp}$ is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. The more soluble substance means it has a higher solubility product value. Solubility products were only applicable for sparingly soluble ionic compounds.

Complete step by step solution:
Given the value solubility product of silver bromide is,
$$K_{sp}[AgBr]=[A{g}^{+}][B{r}^{-}]=5X{10}^{-13}--(1)$$
Given, the concentration of silver nitrate = 0.05M
$$[AgN{O}_3]=0.05M$$
Silver nitrate is completely dissociated into silver ions and nitrate ions because it is a completely soluble substance. i.e, the concentration of silver nitrate is equal to concentration of silver ions usually.
$$\begin{array}{rl}& AgN{O}_3\to A{g}^{+}+N{O_3}^{-}\\ & \therefore [AgN{O}_3]=[A{g}^{+}]=0.05M--(2)\end{array}$$
From equation 1 and 2, substitute the value of $[A{g}^{+}]$ in equation (1), then
$$[B{r}^{-}]={\displaystyle \frac{5\times {10}^{-13}}{[A{g}^{+}]}}={\displaystyle \frac{5\times {10}^{-13}}{0.05}}=1\times {10}^{-11}M--(3)$$
Like silver nitrate, potassium bromide(KBr) salt which is dissociated completely. Hence, from the result equation (3), the concentration of bromide ion will equal to the concentration of KBr.
$$[B{r}^{-}]=[KBr]=1\times {10}^{-11}M--(4)$$
Given the molar mass of KBr (m)= 120 g/mole,
The volume of the solution = 1L
The quantity of potassium bromide= concentration of KBr volume of solution X m
= $1\times {10}^{-11}moles/L\times 1L\times 120g/mole$
= $1.2\times {10}^{-9}g$
Therefore the amount of KBr required to precipitate AgBr is $1.2\times {10}^{-9}g$.

The correct answer is option D.

Note: Solubility products cannot be used for normally soluble compounds like sodium chloride, silver nitrate, etc. interactions between the ions in the solution interfere with the simple equilibrium. The solubility product is a value that you get when the solution is saturated. If there is any solid present, cannot dissolve any more solid than there is in a saturated solution.