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ChemMedAdv 2025 • Paper 1

Hydrogen Ion Concentration in Weak Monobasic Acid Solution

At 25 ∘C25\text{ }^\circ\text{C}, the concentration of H+\text{H}^+ ions in 1.00×10−3 M1.00 \times 10^{-3}\text{ M} aqueous solution of a weak monobasic acid having acid dissociation constant (KaK_{\text{a}}) of 4.00×10−114.00 \times 10^{-11} is X×10−7 MX \times 10^{-7}\text{ M}. The value of XX is ______.

Use: Ionic product of water (KwK_{\text{w}}) =1.00×10−14= 1.00 \times 10^{-14} at 25 ∘C25\text{ }^\circ\text{C}

Official Numerical Answer2.2 to 2.3

Topics & Concepts

EquilibriumAcid-Base Equilibrium

Step-by-Step Solution

To find the concentration of H+\text{H}^+ ions in the solution, we must consider both the dissociation of the weak monobasic acid (HA\text{HA}) and the auto-ionization of water, since the contribution of H+\text{H}^+ from the weak acid is comparable to that from water.

The relevant equilibria in the aqueous solution are:

  1. Dissociation of the weak acid: HA(aq)⇌H+(aq)+A−(aq)\text{HA(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{A}^-\text{(aq)} Ka=[H+][A−][HA]K_{\text{a}} = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}

  2. Auto-ionization of water: H2O(l)⇌H+(aq)+OH−(aq)\text{H}_2\text{O(l)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} Kw=[H+][OH−]K_{\text{w}} = [\text{H}^+][\text{OH}^-]

Given:

  • Initial concentration of acid, C=1.00×10−3 MC = 1.00 \times 10^{-3}\text{ M}
  • Acid dissociation constant, Ka=4.00×10−11K_{\text{a}} = 4.00 \times 10^{-11}
  • Ionic product of water, Kw=1.00×10−14K_{\text{w}} = 1.00 \times 10^{-14}

Since KaK_{\text{a}} is extremely small, the degree of dissociation of HA\text{HA} is negligible, so [HA]≈C=1.00×10−3 M[\text{HA}] \approx C = 1.00 \times 10^{-3}\text{ M}.

From the equilibrium expressions: [A−]=Ka[HA][H+]≈KaC[H+][\text{A}^-] = \frac{K_{\text{a}} [\text{HA}]}{[\text{H}^+]} \approx \frac{K_{\text{a}} C}{[\text{H}^+]} [OH−]=Kw[H+][\text{OH}^-] = \frac{K_{\text{w}}}{[\text{H}^+]}

From the electroneutrality (charge balance) condition: [H+]=[A−]+[OH−][\text{H}^+] = [\text{A}^-] + [\text{OH}^-]

Substituting [A−][\text{A}^-] and [OH−][\text{OH}^-] into the charge balance equation: [H+]=KaC[H+]+Kw[H+][\text{H}^+] = \frac{K_{\text{a}} C}{[\text{H}^+]} + \frac{K_{\text{w}}}{[\text{H}^+]}

Multiplying through by [H+][\text{H}^+]: [H+]2=KaC+Kw[\text{H}^+]^2 = K_{\text{a}} C + K_{\text{w}}

Now, substituting the given values: KaC=(4.00×10−11)×(1.00×10−3)=4.00×10−14 M2K_{\text{a}} C = (4.00 \times 10^{-11}) \times (1.00 \times 10^{-3}) = 4.00 \times 10^{-14}\text{ M}^2 Kw=1.00×10−14 M2K_{\text{w}} = 1.00 \times 10^{-14}\text{ M}^2

Therefore: [H+]2=4.00×10−14+1.00×10−14=5.00×10−14 M2[\text{H}^+]^2 = 4.00 \times 10^{-14} + 1.00 \times 10^{-14} = 5.00 \times 10^{-14}\text{ M}^2 [H+]=5.00×10−14=5×10−7 M≈2.24×10−7 M[\text{H}^+] = \sqrt{5.00 \times 10^{-14}} = \sqrt{5} \times 10^{-7}\text{ M} \approx 2.24 \times 10^{-7}\text{ M}

Given that [H+]=X×10−7 M[\text{H}^+] = X \times 10^{-7}\text{ M}, the value of XX is: X=5≈2.24X = \sqrt{5} \approx 2.24

Hydrogen Ion Concentration in Weak Monobasic Acid Solution | Chemistry PYQ Solution - JEE Challenger