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Question: The Ksp for silver sulfate ( \( A{g_2}S{O_4} \) ) is \( 1.2 \times {10^{ - 5}} \) .How do you calcul...

The Ksp for silver sulfate ( Ag2SO4A{g_2}S{O_4} ) is 1.2×10−51.2 \times {10^{ - 5}} .How do you calculate the solubility of silver sulfate in each of the following: a) water b) 0.10 M AgNO3AgN{O_3} (c) 0.43 M K2SO4{K_2}S{O_4} ?

Explanation

Solution

To find out the molar solubility which is denoted by “s”, first we need to draw the ICE table which is known as initial, change, equilibrium. The molar product constant and the molar solubility are related with each other.

Complete step by step answer:
(a) It is given that the Ksp for the silver sulfate is 1.2×10−51.2 \times {10^{ - 5}} .
First write the chemical equation involved when the silver sulfate is dissolved in pure water.
Ag2SO4(s)⇄2Ag+(aq)+SO42−(aq)A{g_2}S{O_4}(s) \rightleftarrows 2A{g^ + }(aq) + SO_4^{2 - }(aq)
In this reaction, one mole of solid silver sulfate on dissolving in water dissociates to give two silver ions and one sulfate ion.
Now, draw the ICE table elaborated as initial change equilibrium for the given reaction.
When we dissolve the salt, the initial concentration of the ions are taken as zero.
Ag2SO4(s)⇄2Ag+(aq)+SO42−(aq)A{g_2}S{O_4}(s) \rightleftarrows 2A{g^ + }(aq) + SO_4^{2 - }(aq)
Initial 0 0\text{Initial 0 0}
Change +2s +s\text{Change +2s +s}
Equilibrium 2s s\text{Equilibrium 2s s}
The solubility product constant is given as shown below.
KSP=[Ag+]2.[SO42−]{K_{SP}} = {[A{g^ + }]^2}.[SO_4^{2 - }]
Substitute the values in the above expression.
⇒KSP=[2s]2.[s]\Rightarrow {K_{SP}} = {[2s]^2}.[s]
⇒KSP=4s3\Rightarrow {K_{SP}} = 4{s^3}
S is the molar solubility, so s is written as shown below.
⇒s=Ksp43\Rightarrow s = \sqrt[3]{{\dfrac{{{K_{sp}}}}{4}}}
Substitute the value of solubility product constant in the above expression.
⇒s=1.2×10−543\Rightarrow s = \sqrt[3]{{\dfrac{{1.2 \times {{10}^{ - 5}}}}{4}}}
⇒s=0.0144\Rightarrow s = 0.0144
Therefore, the solubility of silver sulfate in water is 0.0144.
(b) It is given that the molarity of the silver nitrate solution is 0.10 M.
Silver nitrate is soluble in aqueous solution, hence silver nitrate dissociates into its ions.
The reaction is shown below.
AgNO3(aq)⇌Ag+(aq)+NO3−(aq)AgN{O_3}(aq) \rightleftharpoons A{g^ + }(aq) + NO_3^ - (aq)
[Ag+]=[NO3−]=0.10M[A{g^ + }] = [NO_3^ - ] = 0.10M
Draw the ICE table for the solubility of silver sulphate.
Ag2SO4(s)⇄2Ag+(aq)+SO42−(aq)A{g_2}S{O_4}(s) \rightleftarrows 2A{g^ + }(aq) + SO_4^{2 - }(aq)
Initial 0.10 0\text{Initial 0.10 0}
Change +2s +s\text{Change +2s +s}
Equilibrium 2s+0.10 s\text{Equilibrium 2s+0.10 s}
The solubility product constant is given as shown below.
KSP=[Ag+]2.[SO42−]{K_{SP}} = {[A{g^ + }]^2}.[SO_4^{2 - }]
Substitute the values in the above expression.
⇒KSP=[(2s+0.10)2][s]\Rightarrow {K_{SP}} = [{(2s + 0.10)^2}][s]
The Ksp{K_{sp}} value is so small as compared to the initial concentration of the silver cation, so it can be written as
2s+0.10≈0.102s + 0.10 \approx 0.10
So,
⇒KSP=[(0.10)2][s]\Rightarrow {K_{SP}} = [{(0.10)^2}][s]
S is the molar solubility, so s is written as shown below.
⇒s=KSP0.010\Rightarrow s = \dfrac{{{K_{SP}}}}{{0.010}}
⇒s=1.2×10−50.010\Rightarrow s = \dfrac{{1.2 \times {{10}^{ - 5}}}}{{0.010}}
⇒s=1.2×10−3\Rightarrow s = 1.2 \times {10^{ - 3}}
Therefore, the solubility of Ag2SO4A{g_2}S{O_4} in 0.10 M AgNO3AgN{O_3} is 1.2×10−31.2 \times {10^{ - 3}} .
(c) It is given that the molarity of K2SO4{K_2}S{O_4} is 0.43 M.
The reaction is shown below.
K2SO4(aq)⇌2K+(aq)+SO42−(aq){K_2}S{O_4}(aq) \rightleftharpoons 2{K^ + }(aq) + SO_4^{2 - }(aq)
[K+]=[SO42−]=0.43M[{K^ + }] = [SO_4^{2 - }] = 0.43M
Draw the ICE table for the solubility of silver sulphate.
Ag2SO4(s)⇄2Ag+(aq)+SO42−(aq)A{g_2}S{O_4}(s) \rightleftarrows 2A{g^ + }(aq) + SO_4^{2 - }(aq)
Initial 0 0.43\text{Initial 0 0.43}
Change +2s +s\text{Change +2s +s}
Equilibrium 2s s+0.43\text{Equilibrium 2s s+0.43}
The solubility product constant is given as shown below.
KSP=[Ag+]2.[SO42−]{K_{SP}} = {[A{g^ + }]^2}.[SO_4^{2 - }]
Substitute the values in the above expression.
⇒KSP=[(2s)2][s+0.43]\Rightarrow {K_{SP}} = [{(2s)^2}][s + 0.43]
The Ksp{K_{sp}} value is so small as compared to the initial concentration of the silver cation, so it can be written as
s+0.43≈0.43s + 0.43 \approx 0.43
So,
⇒KSP=[(2×0.43)2][s+0.43]\Rightarrow {K_{SP}} = [{(2 \times 0.43)^2}][s + 0.43]
S is the molar solubility, so s is written as shown below.
⇒s=KSP(2×0.43)2−0.43\Rightarrow s = \dfrac{{{K_{SP}}}}{{{{(2 \times 0.43)}^2}}} - 0.43
⇒s=1.2×10−5(2×0.43)2−0.43\Rightarrow s = \dfrac{{1.2 \times {{10}^{ - 5}}}}{{{{(2 \times 0.43)}^2}}} - 0.43
⇒s=0.42998\Rightarrow s = 0.42998
Therefore, the solubility of Ag2SO4A{g_2}S{O_4} in 0.43 M K2SO4{K_2}S{O_4} is 0.4299.

Note: You must know that the silver sulfate salt is not soluble in aqueous solution. In option second note that the molar solubility of the silver sulfate salt decreases due to the presence of silver cation due to common ion effect.