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Question: Solubility product of silver bromide (\(AgBr\)) \(5 \times {10^{ - 13}}\). The quantity of potassium...

Solubility product of silver bromide (AgBrAgBr) 5×10−135 \times {10^{ - 13}}. The quantity of potassium bromide (KBrKBr) to be added to 11litre of 0.05M0.05M solution of silver nitrate to start the precipitation of AgBrAgBr is:
A) 6.2×10−5g6.2 \times {10^{ - 5}}g
B) 5.0×10−8g5.0 \times {10^{ - 8}}g
C) 1.2×10−10g1.2 \times {10^{ - 10}}g
D) 1.2×10−9g1.2 \times {10^{ - 9}}g

Explanation

Solution

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:
MxXy⇌xMp+(aq)+yXq−(aq){M_x}{X_y} \rightleftharpoons x{M^{p + }}_{(aq)} + y{X^{q - }}_{(aq)}
(x.p+=y.q−x.{p^ + } = y.{q^ - } )
Its solubility product is given as
Ksp=[Mp+]x[Xq−]y{K_{sp}} = {[{M^{p + }}]^x}{[{X^{q - }}]^y}
In the above case the reaction is:
AgBr⇌Ag++Br−AgBr \rightleftharpoons A{g^ + } + B{r^ - }
Its solubility product is given as:
Ksp{K_{sp}} of AgBrAgBr =[Ag+][Br−] = [A{g^ + }][B{r^ - }]
Given Ksp{K_{sp}} of AgBrAgBr =5×10−3 = 5 \times {10^{ - 3}}
Also given
[Ag+]=0.05M[A{g^ + }] = 0.05M
By substituting the given values
⇒5×10−3=0.05[Br−]\Rightarrow 5 \times {10^{ - 3}} = 0.05[B{r^ - }]
⇒[Br−]=5×10−30.05=1×10−11M\Rightarrow [B{r^ - }] = \frac{{5 \times {{10}^{ - 3}}}}{{0.05}} = 1 \times {10^{ - 11}}M
It means that 10−11{10^{ - 11}} moles of KBrKBrmoles in 11 litre of solution
Molecular mass of KBr=120gmol−1KBr = 120gmo{l^{ - 1}}
Therefore 1.2×10−9g1.2 \times {10^{ - 9}}g of KBrKBr is to be added to 11 litre of 0.05M0.05Msolution of silver nitrate in order to start the precipitation silver bromide (AgBrAgBr)
∴\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 Ksp{K_{sp}} higher will be the solubility. Solubility products represent the maximum extent a solid can dissolve in a solution.