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pH of Weak Acids/Bases

pH of Weak Acids/Bases

8 min read Updated Mar 26, 2026

Calculating pH for a weak acid or base requires an equilibrium calculation. The acid does not fully dissociate, so you need to solve for the equilibrium concentration of H⁺ using Ka and an ICE table.

Weak Acid pH - The ICE Table Method

For a weak acid HA with initial concentration c and acid dissociation constant Ka:

Step 1: Write the equilibrium and set up the ICE table.

HA ⇌ H⁺ + A⁻

HAH⁺A⁻
Initialc00
Change-x+x+x
Equilibriumc - xxx

Step 2: Write the Ka expression and substitute.

Ka = [H⁺][A⁻] / [HA] = (x)(x) / (c - x) = x² / (c - x)

Step 3: Apply the x-is-small approximation if c/Ka > 100.

If c >> x (the acid barely dissociates), then c - x ≈ c:

Ka ≈ x² / c

x = √(Ka x c)

Step 4: Find pH = -log(x).

Worked Example - Weak Acid

What is the pH of 0.10 M acetic acid (Ka=1.8×10−5K_a = 1.8 \times 10^{-5})?

  1. Check the approximation: c/Ka=0.10/(1.8×10−5)=5,600c/K_a = 0.10 / (1.8 \times 10^{-5}) = 5{,}600. Since 5,600 >> 100, the approximation is valid.

  2. [H+]=Ka×c=1.8×10−5×0.10=1.8×10−6[\text{H}^+] = \sqrt{K_a \times c} = \sqrt{1.8 \times 10^{-5} \times 0.10} = \sqrt{1.8 \times 10^{-6}}

  3. 1.8×10−6=18×10−7≈4.2×10−7\sqrt{1.8 \times 10^{-6}} = \sqrt{18 \times 10^{-7}} \approx 4.2 \times 10^{-7}… wait, let us redo more carefully:

    • 1.8×10−6=1.8×10−3≈1.34×10−3\sqrt{1.8 \times 10^{-6}} = \sqrt{1.8} \times 10^{-3} \approx 1.34 \times 10^{-3}

  4. pH=−log⁡(1.34×10−3)=3−log⁡(1.34)≈3−0.13=\text{pH} = -\log(1.34 \times 10^{-3}) = 3 - \log(1.34) \approx 3 - 0.13 = 2.87

Weak Base pH - Same Method, Different Constant

For a weak base B with initial concentration c and Kb:

B + H₂O ⇌ BH⁺ + OH⁻

[OH⁻] = √(Kb x c) (same approximation, same conditions)

Then: pOH = -log[OH⁻], and pH = 14 - pOH.

Worked Example - Weak Base

What is the pH of 0.20 M ammonia (Kb=1.8×10−5K_b = 1.8 \times 10^{-5})?

  1. [OH−]=Kb×c=1.8×10−5×0.20=3.6×10−6=1.9×10−3[\text{OH}^-] = \sqrt{K_b \times c} = \sqrt{1.8 \times 10^{-5} \times 0.20} = \sqrt{3.6 \times 10^{-6}} = 1.9 \times 10^{-3}

  2. pOH=−log⁡(1.9×10−3)=3−log⁡(1.9)≈3−0.28=2.72\text{pOH} = -\log(1.9 \times 10^{-3}) = 3 - \log(1.9) \approx 3 - 0.28 = 2.72

  3. pH = 14 - 2.72 = 11.28

When the Approximation Fails

If c/Ka < 100, the approximation c - x ≈ c is not valid, and you need the quadratic formula:

Ka = x² / (c - x) → Ka(c - x) = x² → x² + Ka·x - Ka·c = 0

x = [-Ka + √(Ka² + 4Ka·c)] / 2

On the MCAT, this rarely comes up because they typically give you concentrations and KaK_a values where the approximation works. But if you are told that KaK_a is 10−210^{-2} and c is 0.10, then c/Ka=10c/K_a = 10 - the approximation fails, and you need the quadratic (or they will give you enough information to avoid it).

Checking Your Answer

Always verify that your answer makes sense:

  • Weak acid solution: pH must be between 0 and 7 (acidic but not as low as a strong acid at the same concentration)
  • Weak base solution: pH must be between 7 and 14
  • Percent dissociation should be small (less than 5% if you used the approximation)
  • Higher concentration = lower pH for acids (more acid = more H⁺)
What is the pH of a 0.040 M solution of HCN (Ka=4.0×10−10K_a = 4.0 \times 10^{-10})?
Click to reveal answer
pH = 5.4. Check approximation: 0.040/(4.0×10−10)=108>>1000.040 / (4.0 \times 10^{-10}) = 10^8 >> 100, so the approximation holds. [H+]=4.0×10−10×0.040=1.6×10−11=4.0×10−6[\text{H}^+] = \sqrt{4.0 \times 10^{-10} \times 0.040} = \sqrt{1.6 \times 10^{-11}} = 4.0 \times 10^{-6}. pH=−log⁡(4.0×10−6)=6−log⁡(4)=6−0.6=5.4\text{pH} = -\log(4.0 \times 10^{-6}) = 6 - \log(4) = 6 - 0.6 = 5.4.
Which has a lower pH: 0.10 M acetic acid (Ka=1.8×10−5K_a = 1.8 \times 10^{-5}) or 0.10 M HCN (Ka=6.2×10−10K_a = 6.2 \times 10^{-10})?
Click to reveal answer
Acetic acid has the lower pH (more acidic). At the same concentration, the acid with the larger KaK_a dissociates more, producing more H⁺. KaK_a of acetic acid (1.8×10−51.8 \times 10^{-5}) is roughly 30,000 times larger than KaK_a of HCN (6.2×10−106.2 \times 10^{-10}). More dissociation = more H⁺ = lower pH.
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