To find the pH of a weak acid, you use its acid dissociation constant KaK_aKa​ and the acid’s initial concentration to estimate the hydrogen ion concentration, then convert that to pH with the usual pH formula.

Core idea

For a weak acid HA in water,

HA⇌H++A−\text{HA}\rightleftharpoons \text{H}^++\text{A}^-HA⇌H++A−

its strength is described by the equilibrium constant KaK_aKa​.

  • The pH is defined as pH=−log⁡[H+]\text{pH}=-\log[\text{H}^+]pH=−log[H+].
  • For many weak-acid problems, you first find [H+][\text{H}^+][H+] from KaK_aKa​ and the initial acid concentration, then plug into the pH formula.

Step‑by‑step method

  1. Write the equilibrium and KaK_aKa​ expression
    For HA:

Ka=[H+][A−][HA]K_a=\frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}Ka​=[HA][H+][A−]​

At equilibrium, starting from an initial acid concentration CCC and assuming no H+\text{H}^+H+ and A−\text{A}^-A− initially, the change in concentration is +x+x+x for H+\text{H}^+H+ and A−\text{A}^-A−, and −x-x−x for HA.

 * Equilibrium concentrations:
   * [H+]=x[\text{H}^+]=x[H+]=x
   * [A−]=x[\text{A}^-]=x[A−]=x
   * [HA]≈C−x[\text{HA}]\approx C-x[HA]≈C−x
  1. Use the weak‑acid approximation
    If the acid is sufficiently weak and not extremely dilute, dissociation is small, so C−x≈CC-x\approx CC−x≈C.

Then:

Ka≈x2CK_a\approx \frac{x^2}{C}Ka​≈Cx2​

So:

x=[H+]≈Ka⋅Cx=[\text{H}^+]\approx \sqrt{K_a\cdot C}x=[H+]≈Ka​⋅C​

  1. Convert[H+][\text{H}^+][H+] to pH
    Once xxx is known:

pH=−log⁡([H+])=−log⁡(x)\text{pH}=-\log([\text{H}^+])=-\log(x)pH=−log([H+])=−log(x)

When the shortcut fails

For stronger “weak” acids or very concentrated/dilute solutions, the approximation C−x≈CC-x\approx CC−x≈C may not be accurate.

  • In that case, use the exact expression:

Ka=x2C−xK_a=\frac{x^2}{C-x}Ka​=C−xx2​

which leads to a quadratic equation in xxx.

  • Solve the quadratic for x=[H+]x=[\text{H}^+]x=[H+], then calculate pH as before.

Simple worked pattern (no numbers)

  • Given: weak acid with KaK_aKa​ and concentration CCC.
  • Procedure:
    1. Assume [H+]=[A−]=x[\text{H}^+]=[\text{A}^-]=x[H+]=[A−]=x.
2. Write Ka=x2/(C−x)K_a=x^2/(C-x)Ka​=x2/(C−x).

3. If the acid is sufficiently weak, use C−x≈CC-x\approx CC−x≈C and set x≈KaCx\approx \sqrt{K_aC}x≈Ka​C​.

4. Compute pH=−log⁡x\text{pH}=-\log xpH=−logx.

Information gathered from public forums or data available on the internet and portrayed here.