The ionization constant of benzoic acid is 6.46 × 10–5 and Ksp for silver benzoate are 2.5 × 10–13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Given:

The ionization constant of benzoic acid (Ka) is 6.46 × 10–5


Ksp for silver benzoate is 2.5 × 10–13


pH = 3.19


Ionization of silver benzoate:


C6H5COOAg C6H5COO- + Ag+


Solubility in water: Let solubility in water is x mol/l Then


[C6H5COO- ] = [Ag+] = x mol/l


As we know that,


Ksp = [A+] [B-]


Where A and B are the ions dissolved


In the above reaction,


[A+] = C6H5COO-


[B-] = Ag+


Ksp = [C6H5COO-][Ag+]


As Ksp = 2.5 × 10–13 (given)


2.5 × 10–13 = x2


x = 5 × 10-7


Solubility in buffer of pH =3.19


As we know that,


pH = –log [H+]


–log [H+] = 3.19


By taking antilog of both the sides, we get


[H+] = antilog -3.19


[H+] = 6.457 × 10-4 M


C6H5COO- ions combine with the H= ions o from benzoic acid but [H+] remains almost constant because we have buffer solution. Now,


C6H5COOH C6H5COO- + H+


As we know that,


Ka =


Ka =


=


[H+] = 6.457 × 10-4 M (calculated above)


Ka = 6.46 × 10–5 (given)


= = = 10


[C6H5COOH] = 10 [C6H5COO-]


Suppose solubility in the buffer solution is y mol/l.


Then as most of the benzoate ion is converted into benzoic acid molecules(which remains almost ionized) we have


Y = [Ag+] = [C6H5COO-] + [C6H5COOH]


As [C6H5COOH] = 10 [C6H5COO-] (calculated above)


y = [C6H5COO-] + 10 [C6H5COO-]


y = 11 [C6H5COO-]


[C6H5COO-] =


As we know that,


Ksp = [C6H5COO-][Ag+]


As Ksp = 2.5× 10-13 (given0


[Ag+] = y, [C6H5COO-] =


2.5× 10-13 = × y


Y2 = 2.75 × 10-12


Y = √ 2.75 × 10-12


Y = 1.66 × 10-6


Now,


= = = 3.32


Thus, the silver benzoate is 3.32 times more soluble in a buffer of pH compared to its solubility in pure water


Note: In case of weak acids, the solubility is more in the acidic solution than in water. The reason, in general, may be explained as below:


Taking example of C6H5COOAg, we have


C6H5COOAg C6H5COO- + Ag+


In acidic solution, the anions (C6H5COO- in the present case) undergo protonation in presence of acid. Thus, C6H5COO- ions are removed. Hence equilibrium shifts forward producing more Ag+ ions. Alternatively, as C6H5COO- are removed, Qsp decreases. In order to maintain solubility product equilibrium (Qsp= Ksp), Ag+ ion concentration must increase. Hence solubility is more.


Note: A buffer solution is defined as a solution which resists any change in its pH value even when small amount of acid or base added to it.


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