JEE Challenger
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ChemMedAdv 2009 • Paper 2

Calculation of pH of Sodium Salt of Substituted Benzoic Acid

The dissociation constant of a substituted benzoic acid at 25∘C25^\circ\text{C} is 1.0×10−41.0 \times 10^{-4}. The pH\text{pH} of a 0.01 M0.01\text{ M} solution of its sodium salt is

Official Numerical Answer8

Topics & Concepts

EquilibriumAcid-Base Equilibrium

Step-by-Step Solution

To find the pH\text{pH} of the solution of the sodium salt of a substituted benzoic acid, we recognize that this is a salt formed from a weak monobasic acid (HA\text{HA}) and a strong base (NaOH\text{NaOH}).

In an aqueous solution, the salt completely dissociates into sodium cations (Na+\text{Na}^+) and conjugate base anions (A−\text{A}^-):
NaA⟶Na++A−\text{NaA} \longrightarrow \text{Na}^+ + \text{A}^-

The anion A−\text{A}^- undergoes hydrolysis:
A−+H2O⇌HA+OH−\text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^-

The hydrolysis constant KhK_h is given by:
Kh=KwKaK_h = \frac{K_w}{K_a}

Given values:

  • Dissociation constant of the acid, Ka=1.0×10−4K_a = 1.0 \times 10^{-4}
  • Ionic product of water at 25∘C25^\circ\text{C}, Kw=1.0×10−14K_w = 1.0 \times 10^{-14}
  • Concentration of the salt, C=0.01 M=10−2 MC = 0.01\text{ M} = 10^{-2}\text{ M}

Thus, the hydrolysis constant is:
Kh=1.0×10−141.0×10−4=1.0×10−10K_h = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-4}} = 1.0 \times 10^{-10}

Since KhK_h is very small, the degree of hydrolysis h≪1h \ll 1, and the concentration of hydroxide ions [OH−][\text{OH}^-] can be approximated as:
[OH−]=Kh⋅C=Kw⋅CKa[\text{OH}^-] = \sqrt{K_h \cdot C} = \sqrt{\frac{K_w \cdot C}{K_a}}

Substituting the values:
[OH−]=1.0×10−10×10−2=1.0×10−12=1.0×10−6 M[\text{OH}^-] = \sqrt{1.0 \times 10^{-10} \times 10^{-2}} = \sqrt{1.0 \times 10^{-12}} = 1.0 \times 10^{-6}\text{ M}

From this, the pOH\text{pOH} is:
pOH=−log⁡10[OH−]=−log⁡10(1.0×10−6)=6\text{pOH} = -\log_{10}[\text{OH}^-] = -\log_{10}(1.0 \times 10^{-6}) = 6

Since pH+pOH=14\text{pH} + \text{pOH} = 14 at 25∘C25^\circ\text{C}:
pH=14−pOH=14−6=8\text{pH} = 14 - \text{pOH} = 14 - 6 = 8

Alternatively, using the standard formula for the pH\text{pH} of a salt of a weak acid and a strong base:
pH=12[pKw+pKa+log⁡10C]\text{pH} = \frac{1}{2}\left[\text{p}K_w + \text{p}K_a + \log_{10} C\right]
pKa=−log⁡10(1.0×10−4)=4\text{p}K_a = -\log_{10}(1.0 \times 10^{-4}) = 4
log⁡10C=log⁡10(10−2)=−2\log_{10} C = \log_{10}(10^{-2}) = -2
pH=12[14+4+(−2)]=12(16)=8\text{pH} = \frac{1}{2}[14 + 4 + (-2)] = \frac{1}{2}(16) = 8

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