Question

The solubility of silver chromate in 0.01M \[{{K}_{2}}Cr{{O}_{4}}\] is \[2\times {{10}^{-8}}mol{{m}^{-3}}\]. The solubility product of silver chromate will be:
A. $8\times {{10}^{-24}}$
B. $16\times {{10}^{-24}}$
C. $1.6\times {{10}^{-18}}$
D. $16\times {{10}^{-18}}$

Answer 1

Hint: The solubility product constant, abbreviated\[{{K}_{sp}}\], is a condensed equilibrium constant that describes the equilibrium between a solid and its associated ions in a particular solution. Its value reveals how much a substance can dissociate in water. The substance is more soluble the higher the solubility product constant.

Formula used:For a reaction say:
\[{{M}_{x}}{{A}_{y}}(s)\to x{{M}^{y+}}+y{{A}^{x-}}\]
The formula for calculating solubility product \[{{K}_{sp}}\] is given as:
\[{{K}_{sp}}=\text{x }\!\![\!\!\text{ }{{\text{M}}^{y+}}\text{ }\!\!]\!\!\text{ y }\!\![\!\!\text{ }{{\text{A}}^{x-}}\text{ }\!\!]\!\!\text{ }\]

Complete Step by Step Solution:
We need to calculate the solubility product of silver chromate in the presence of 0.01M\[{{K}_{2}}Cr{{O}_{4}}\]. And we are given that the solubility of silver chromate in 0.01M \[{{K}_{2}}Cr{{O}_{4}}\] is \[2\times {{10}^{-8}}mol{{m}^{-3}}\].

The equilibrium equation of dissociation of 0.01M \[{{K}_{2}}Cr{{O}_{4}}\] is given as:
\[Given:{{K}_{2}}Cr{{O}_{4}}2{{K}^{+}}+Cr{{O}_{4}}^{2-}\]
Conc.: $0.01$ $0.02$ $0.01$
So we can see that 1 mole of \[{{K}_{2}}Cr{{O}_{4}}\] gives 2 mol of potassium ions and 1 mole of chromate ion. So, we divided the concentration at equilibrium accordingly as shown in the equation.

Now, let’s suppose the solubility of silver chromate to be S which is given in the question equal to \[2\times {{10}^{8}}mol{{m}^{-3}}\]. Therefore, for silver chromate, the reaction at equilibrium is given as:
Given: \[A{{g}_{2}}Cr{{O}_{4}}2A{{g}^{+}}+Cr{{O}_{4}}^{2-}\]
Conc.: $S$ $2S$ $2S+0.01$

As we can see that \[Cr{{O}_{4}}^{2-}\]is the common ion and that’s why their concentration will be added. So now we need to find the solubility product of silver chromate. Now according to the formula used, \[{{K}_{sp}}\]for the given reaction will be:
\[{{K}_{sp}}={{(2S)}^{2}}(2S+0.01)\], where \[S=2\times {{10}^{-8}}mol{{m}^{-3}}\]
\[{{K}_{sp}}={{(2\times 2\times {{10}^{-8}})}^{2}}(2S+0.001)\]Now we can see that.
Therefore, \[{{K}_{sp}}={{(2\times 2\times {{10}^{-8}})}^{2}}(0.01)\]
Hence, \[{{K}_{sp}}=16\times {{10}^{-18}}\]
Hence, the correct option is D. $16\times {{10}^{-18}}$

Note: Please use caution when performing the calculations because the concentration of silver chromate is stated as 0.010 M, which is equivalent to ${{10}^{-2}}$M because the zero after 1 has no meaning. Remember the solubility product formula, which has the same dimension or unit as concentration.