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Question: Solubility product of silver bromide is \(5.0\times {{10}^{-13}}\). The quantity of potassium bromid...

Solubility product of silver bromide is 5.0×10−135.0\times {{10}^{-13}}. The quantity of potassium bromide(molar mass taken as 120 g of mol−1mo{{l}^{-1}}) to be added to 1 litre of 0.05M solution of silver nitrate to start the precipitation of AgBr is:
A. 1.2×10−101.2\times {{10}^{-10}}
B. 1.2×10−91.2\times {{10}^{-9}}
C. 6.2×10−56.2\times {{10}^{-5}}
D. 5.0×10−85.0\times {{10}^{-8}}

Explanation

Solution

As we know that Solubility product is denoted by the symbol Ksp{{K}_{sp}}, which is the equilibrium constant for the dissolution of a substance into an aqueous solution. It is found that higher the value of Ksp{{K}_{sp}} the more soluble the complex is.

Complete Step by step solution:
As we know that when AgBr is dissociated, it will give:
AgBr→Ag++Br−AgBr\to A{{g}^{+}}+B{{r}^{-}}
Now, will also dissociate and it will give:
AgNO3→Ag++NO3−AgN{{O}_{3}}\to A{{g}^{+}}+N{{O}_{3}}^{-}
We are being provided with the value of silver nitrate 0.05M, and as we have seen it will dissociate and gives Ag++NO3−A{{g}^{+}}+N{{O}_{3}}^{-}.Now, the value of Ag+A{{g}^{+}}will also be same as that of silver nitrate that is 0.05M.
Ksp=[Ag+][Br−] 5×10−13=[0.05][Br−] [Br−]=5×10−130.05 [Br−]=10−11M \begin{aligned} & {{K}_{sp}}=\left[ A{{g}^{+}} \right]\left[ B{{r}^{-}} \right] \\\ & 5\times {{10}^{-13}}=\left[ 0.05 \right]\left[ B{{r}^{-}} \right] \\\ & \left[ B{{r}^{-}} \right]=\dfrac{5\times {{10}^{-13}}}{0.05} \\\ & \left[ B{{r}^{-}} \right]={{10}^{-11}}M \\\ \end{aligned}
Moles of KBr required for the precipitation of AgBr will be:
N = c.v
Where, c is the concentration and v is the volume.
n=[Br−]×1L n=10−11mol/L×1L n=10−11mol \begin{aligned} & n=\left[ B{{r}^{-}} \right]\times 1L \\\ & n={{10}^{-11}}mol/L\times 1L \\\ & n={{10}^{-11}}mol \\\ \end{aligned}

We will convert moles into grams, as molecular weight of one mole given as 120gm, so:
=10−11×120g =1.2×10−9g \begin{aligned} & ={{10}^{-11}}\times 120g \\\ & =1.2\times {{10}^{-9}}g \\\ \end{aligned}

Hence, we can conclude that the correct option is (b), that is 1.2×10−91.2\times {{10}^{-9}}

Note:

As we know that Solubility products are a type of equilibrium constant and it is found that its value depends on temperature. As the temperature increases, the value of Solubility products increases, due to increased solubility.

We should not get confused in between the terms solubility and solubility product. As solubility is the total amount of solute that can be dissolved in the solvent.
It describes the equilibrium between solute and its ions that dissociated in the solution.