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.