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

Last updated date: 20th Jun 2024
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Hint: As we know that Solubility product is denoted by the symbol ${{K}_{sp}}$, which is the equilibrium constant for the dissolution of a substance into an aqueous solution. It is found that higher the value of ${{K}_{sp}}$ the more soluble the complex is.

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

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

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

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.