Solveeit Logo
Login

Question

Question: The solubility product constant of a metal carbonate $MCO_3$ is $2 \times 10^{-12}$ at $25^\circ C$....

The solubility product constant of a metal carbonate MCO3MCO_3 is 2×10−122 \times 10^{-12} at 25∘C25^\circ C. A solution is 0.1 M in M(NO3)2M(NO_3)_2 and it is saturated with 0.01M CO2CO_2. Also the ionization constant of CO2CO_2 are : Ka1=4×10−7K_{a_1} = 4 \times 10^{-7} and Ka2=5×10−11K_{a_2} = 5 \times 10^{-11} at 25∘C25^\circ C. The minimum pH that must be maintained to start any precipitation is

Answer

4

Explanation

Solution

The precipitation of MCO3MCO_3 starts when the ionic product [M2+][CO32−][M^{2+}][CO_3^{2-}] exceeds the solubility product constant, KspK_{sp}.

Given Ksp(MCO3)=2×10−12K_{sp}(MCO_3) = 2 \times 10^{-12}. The concentration of M2+M^{2+} ions from 0.1 M M(NO3)2M(NO_3)_2 is [M2+]=0.1[M^{2+}] = 0.1 M.

For precipitation to start, the minimum concentration of CO32−CO_3^{2-} required is:

[CO32−]=Ksp[M2+]=2×10−120.1=2×10−11[CO_3^{2-}] = \frac{K_{sp}}{[M^{2+}]} = \frac{2 \times 10^{-12}}{0.1} = 2 \times 10^{-11} M.

The solution is saturated with 0.01 M CO2CO_2, which implies the concentration of dissolved CO2CO_2 (represented as H2CO3H_2CO_3) is [H2CO3]=0.01[H_2CO_3] = 0.01 M.

The ionization of carbonic acid occurs in two steps:

H2CO3⇌H++HCO3−H_2CO_3 \rightleftharpoons H^+ + HCO_3^-; Ka1=4×10−7K_{a_1} = 4 \times 10^{-7}

HCO3−⇌H++CO32−HCO_3^- \rightleftharpoons H^+ + CO_3^{2-}; Ka2=5×10−11K_{a_2} = 5 \times 10^{-11}

We can combine these two steps to relate [H2CO3][H_2CO_3], [H+][H^+], and [CO32−][CO_3^{2-}]:

H2CO3⇌2H++CO32−H_2CO_3 \rightleftharpoons 2H^+ + CO_3^{2-}

The equilibrium constant for this overall reaction is Koverall=Ka1×Ka2=(4×10−7)×(5×10−11)=20×10−18=2×10−17K_{overall} = K_{a_1} \times K_{a_2} = (4 \times 10^{-7}) \times (5 \times 10^{-11}) = 20 \times 10^{-18} = 2 \times 10^{-17}.

The expression for the overall equilibrium constant is Koverall=[H+]2[CO32−][H2CO3]K_{overall} = \frac{[H^+]^2[CO_3^{2-}]}{[H_2CO_3]}.

We need to find the pH (and thus [H+][H^+]) at which [CO32−][CO_3^{2-}] is 2×10−112 \times 10^{-11} M, given [H2CO3]=0.01[H_2CO_3] = 0.01 M.

Substitute the values into the KoverallK_{overall} expression:

2×10−17=[H+]2(2×10−11)0.012 \times 10^{-17} = \frac{[H^+]^2 (2 \times 10^{-11})}{0.01}

[H+]2=(2×10−17)×0.012×10−11[H^+]^2 = \frac{(2 \times 10^{-17}) \times 0.01}{2 \times 10^{-11}}

[H+]2=2×10−17×10−22×10−11[H^+]^2 = \frac{2 \times 10^{-17} \times 10^{-2}}{2 \times 10^{-11}}

[H+]2=2×10−192×10−11[H^+]^2 = \frac{2 \times 10^{-19}}{2 \times 10^{-11}}

[H+]2=10−19−(−11)=10−8[H^+]^2 = 10^{-19 - (-11)} = 10^{-8}

[H+]=10−8=10−4[H^+] = \sqrt{10^{-8}} = 10^{-4} M.

The pH is given by −log⁡10[H+]-\log_{10}[H^+].

pH = −log⁡10(10−4)=4-\log_{10}(10^{-4}) = 4.

This is the minimum pH required to achieve the necessary [CO32−][CO_3^{2-}] for precipitation to start. At pH values higher than 4, [H+][H^+] is lower, which shifts the carbonic acid equilibrium further to the right, increasing [CO32−][CO_3^{2-}] above 2×10−112 \times 10^{-11} M and causing precipitation. At pH values lower than 4, [H+][H^+] is higher, which shifts the equilibrium to the left, decreasing [CO32−][CO_3^{2-}] below 2×10−112 \times 10^{-11} M and preventing precipitation.