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Question: The solubility product of \[A{g_2}Cr{O_4}\] is \[32 \times {10^{ - 12}}\]. What is the concentration...

The solubility product of Ag2CrO4A{g_2}Cr{O_4} is 32×10−1232 \times {10^{ - 12}}. What is the concentration of CrO42−Cr{O_4}^{2 - } ions in that solution?
A)2×10−4MA)2 \times {10^{ - 4}}M
B)16×10−4MB)16 \times {10^{ - 4}}M
C)8×10−4MC)8 \times {10^{ - 4}}M
D)8×10−8MD)8 \times {10^{ - 8}}M

Explanation

Solution

Solubility is a property useful to determine the solute solubility in the solvent, it can be written as the product of concentrations of a positive ion i.e., cation and negative ion i.e., anion. The concentration of cation or ion can be determined by using the solubility product of a salt.
Formula used:
Ksp=[A+]a[B−]b{K_{sp}} = {\left[ {{A^ + }} \right]^a}{\left[ {{B^ - }} \right]^b}
Where Ksp{K_{sp}} is solubility product of a salt
[A+]\left[ {{A^ + }} \right] is concentration of cation
[B−]\left[ {{B^ - }} \right] is the concentration of anion
a,ba,b are relative concentrations

Complete answer:
Given salt is silver chromate with the molecular formula of Ag2CrO4A{g_2}Cr{O_4}, it is an inorganic salt. It can easily dissociate into silver ions and chromate ions.
Silver ion is a cation and a chromate ion is an anion.
Given that the solubility product of silver chromate Ksp{K_{sp}} is 32×10−1232 \times {10^{ - 12}}.
The chemical dissociation of silver chromate will be as follows:
Ag2CrO4→2Ag++CrO42−A{g_2}Cr{O_4} \to 2A{g^ + } + Cr{O_4}^{2 - }
Two moles of silver ion and one mole of chromate ion were formed, Then the solubility product can be written as
Ksp=(2s)2s=4s3{K_{sp}} = {\left( {2s} \right)^2}s = 4{s^3}
The value of s can be determined by
s=(Ksp1)1/3s = {\left( {\dfrac{{{K_{sp}}}}{1}} \right)^{1/3}}
Substitute the value of solubility product in the above equation,
s=(32×10−121)1/3=2×10−4Ms = {\left( {\dfrac{{32 \times {{10}^{ - 12}}}}{1}} \right)^{1/3}} = 2 \times {10^{ - 4}}M
Thus, the concentration of chromate ion CrO42−Cr{O_4}^{2 - } will be 2×10−4M2 \times {10^{ - 4}}M

So, the correct answer is “Option A”.

Note:
Since the dissociation of silver chromate produces one chromate ion and two silver ions, the silver ion concentration can be equal to the double of the concentration of chromate ion. Thus, silver ion concentration can also be determined from the solubility product of a silver chromate.