Question

The solubility product ($\text{ }{{\text{K}}_{\text{sp}}}\text{ }$) of $\text{ AgCl }$ is equal to $\text{ 1}\text{.0 }\times \text{1}{{\text{0}}^{-10}}\text{ }$then which of the following option is incorrect?
A) $\text{ }{{\text{p}}_{\text{Ag}}}\text{ + }{{\text{p}}_{\text{Cl}}}\text{ = 10 }$
B)$\text{ }{{\text{p}}_{\text{Cl}}}\text{= 8 }$ in $\text{ 0}\text{.01 M A}{{\text{g}}^{\text{+}}}\text{ }$
C) ${{\text{p}}_{\text{Ag}}}\text{ + }{{\text{p}}_{\text{Cl}}}\text{ }<\text{ 10 }$in the presence of $\text{ A}{{\text{g}}^{\text{+}}}\text{ }$
D) $\text{ }{{\text{p}}_{\text{Ag}}}\text{ = }{{\text{p}}_{\text{Cl}}}\text{ = 5 }$ in the saturated $\text{ AgCl }$solution

Answer 1

Hint: Chemical concentrations can be expressed as the p-function. P-function is taking the negative logarithmic value of chemical quantities. Let’s consider a chemical quantity as ‘x’ then its p function is written as,
$\text{ pX = }-\text{log X }$
The solubility product of silver chloride is written as the product of the concentration of silver ion and chloride ion in the solution.it is represented as follows,
$\text{ }{{\text{K}}_{\text{sp}}}\text{ = }\left[ \text{A}{{\text{g}}^{\text{+}}} \right]\left[ \text{C}{{\text{l}}^{-}} \right]\text{ }$
Thus solubility products can be expressed in terms of the negative log of concentration.

Complete step by step answer:
The solubility of a compound is expressed as the solubility product. Let's consider a sparingly soluble salt like silver chloride is added to water. A very small amount of silver chloride dissolves and the rest of it remains in the solid-state .the dissolution of silver chloride is represented as follows,
$\text{ AgCl }\rightleftharpoons \text{ A}{{\text{g}}^{\text{+}}}\text{(aq) + C}{{\text{l}}^{-}}(\text{aq) }$
The solubility product of silver chloride is written as the product of the concentration of silver ion and chloride ion in the solution.it is represented as follows,
$\text{ }{{\text{K}}_{\text{sp}}}\text{ = }\left[ \text{A}{{\text{g}}^{\text{+}}} \right]\left[ \text{C}{{\text{l}}^{-}} \right]\text{ }$ (1)
Chemical concentrations can be expressed as the p-function. p-function is taking the negative logarithmic value of chemical quantities. let's consider a chemical quantity as ‘x’ then its p function is written as,
$\text{ pX = }-\text{log X }$
Take the p-function of the (1) we have,
$\text{ }-\log \left( {{\text{K}}_{\text{sp}}}\text{ } \right)\text{=}-\log \left( \left[ \text{A}{{\text{g}}^{\text{+}}} \right] \right)+-\log \left( \left[ \text{C}{{\text{l}}^{-}} \right] \right)\text{ }$ (1)
This would be equal to $-\log \left( {{\text{K}}_{\text{sp}}}\text{ } \right)\text{ = }{{\text{p}}_{\text{Ag}}}\text{ + }{{\text{p}}_{\text{Cl}}}\text{ }$ (2)
We have given that solubility product of $\text{ AgCl }$ is $\text{ 1}\text{.0 }\times \text{1}{{\text{0}}^{-10}}\text{ }$ .lets substitute the value of solubility product in the equation (2) we have,
$\begin{align}
& \text{ }{{\text{p}}_{\text{Ag}}}\text{ + }{{\text{p}}_{\text{Cl}}}\text{ }=-\log \left( \text{1}\text{.0}\times \text{1}{{\text{0}}^{-10}}\text{ } \right)\text{ } \\
 & \therefore \text{ }{{\text{p}}_{\text{Ag}}}\text{ + }{{\text{p}}_{\text{Cl}}}\text{ = 10 } \\
\end{align}$
Let’s consider that concentration of a silver ion $\text{ A}{{\text{g}}^{\text{+}}}\text{ }$in the silver chloride ion is equal to $\text{ 0}\text{.01 M }$ .lets substitute the value of silver ion concentration in equation (1) then we have,
$\begin{align}
& \text{ }-\log \left( 1.0\times {{10}^{-10}} \right)\text{ = }-\log \left( 0.01 \right)\text{ + }{{\text{p}}_{\text{Cl}}}\text{ } \\
& \Rightarrow {{\text{p}}_{\text{Cl}}}=\text{ 10}-2\text{ = 8 } \\
\end{align}$
The solubility of the insoluble compound decreases in the presence of common ion.This is known as the common ion effect. Silver chloride is an insoluble inorganic compound.in presence of ion which is common in between the compound and added salt shifts the equilibrium towards the left-hand side. Thus common is added in solution to decrease the solubility of the solute. Thus when the silver ion $\text{ A}{{\text{g}}^{\text{+}}}\text{ }$ is added to the silver chloride solution decreases the solubility of $\text{ AgCl }$ . Thus solubility decreases. The p-function of silver and chloride is greater than 10.
 $\text{ }{{\text{p}}_{\text{Ag}}}\text{ + }{{\text{p}}_{\text{Cl}}}\text{ }>\text{ 10 }$
$\text{ AgCl }$ Is a $\text{ 1:1 }$ salt. Thus at a saturated solution the concentration of $\text{ A}{{\text{g}}^{\text{+}}}\text{ }$ and $\text{ C}{{\text{l}}^{-}}\text{ }$ are equal. Then the equation (1) is modified as follows,
$\begin{align}
& \text{ 10 = }{{\text{p}}_{\text{Ag}}}\text{ + }{{\text{p}}_{\text{Cl}}}\text{ } \\
& \text{At equilibrium , }{{\text{p}}_{\text{Ag}}}\text{ = }{{\text{p}}_{\text{Cl}}} \\
& \Rightarrow \text{10 = 2}{{\text{p}}_{\text{Ag}}} \\
& \therefore {{\text{p}}_{\text{Ag}}}\text{ = }{{\text{p}}_{\text{Cl}}}\text{ = 5 } \\
\end{align}$
Thus option (A), (B), and (D) are correct and option (C) is incorrect.

Hence, (C) is the correct option.

Note: Note that, p-function is a negative logarithmic term of chemical quantity. Thus here the small value of solubility will have a high value in p-function. Thus due to the common ion effect p-function value of solubility product increases but solubility decreases.