Understanding Acid Dissociation Constants and pH

Expert reviewed • 22 November 2024 • 5 minute read

Introduction

Acid dissociation constants (Ka) and their logarithmic form (pKa) are fundamental concepts that help us understand and quantify acid strength. These values provide crucial insights into acid-base chemistry and equilibrium calculations.

The Acid Dissociation Constant (Ka)

When a weak acid (HA) dissolves in water, it establishes an equilibrium:

HA(aq)+H2O(l)⇌A(aq)−+H3O(aq)+HA_{(aq)} + H_2O_{(l)} \rightleftharpoons A^-_{(aq)} + H_3O^+_{(aq)}

The acid dissociation constant (Ka) for this equilibrium is:

Ka=[A−][H3O+][HA]K_a = \frac{[A^-][H_3O^+]}{[HA]}

Example Values for Common Acids

Here are typical Ka values for some common weak acids:

Weak AcidKa
Hydrofluoric acid5.6 × 10⁻⁴
Methanoic acid1.6 × 10⁻⁴
Ethanoic acid1.7 × 10⁻⁵
Hydrogen sulfide8.9 × 10⁻⁸

Understanding pKa

The pKa is the negative logarithm of Ka:

pKa=−log⁡10(Ka)pK_a = -\log_{10}(K_a)

Conversely, Ka can be calculated from pKa:

Ka=10−pKaK_a = 10^{-pK_a}

Relationship Between Ka and pKa Values

  • Stronger acids have larger Ka values and smaller pKa values
  • A one-unit change in pKa represents a ten-fold change in Ka
  • Most weak acids have pKa values between 0 and 14
  • Strong acids typically have negative pKa values

The pH-pKa Relationship

At equilibrium, when pH = pKa:

  • The concentrations of acid [HA] and conjugate base [A⁻] are equal
  • The acid is 50% dissociated
  • [HA]=[A−][HA] = [A^-]

Effects of Dilution

Adding water to an acid solution affects several parameters:

  • pH increases
  • Degree of ionization increases
  • Ka remains constant

The degree of ionization (α) is given by:

α=[H+][HA]initial×100%\alpha = \frac{[H^+]}{[HA]_{initial}} \times 100\%

Calculating pH from pKa

Practice Question 1

Calculate the pH of a 0.0800 M hypochlorous acid (HOCl) solution with pKa = 7.46.
1. Calculate Ka:

Ka=10−7.46=3.5×10−8K_a = 10^{-7.46} = 3.5 \times 10^{-8}

  • Set up ICE table and solve: 3.5×10−8=x20.08003.5 \times 10^{-8} = \frac{x^2}{0.0800} x=[H+]=5.3×10−5x = [H^+] = 5.3 \times 10^{-5}

  • Calculate pH: pH=−log⁡(5.3×10−5)=4.28pH = -\log(5.3 \times 10^{-5}) = 4.28

Key Points to Remember

  • pKa is independent of concentration
  • Temperature affects Ka values
  • Strong acids have negative pKa values
  • Weak acids have positive pKa values
  • pH = pKa at 50% dissociation

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