To find the pH at the equivalence point , you always start from the same core idea: at equivalence, the original acid and base have exactly neutralized each other, and the pH is controlled by the salt that is present in solution.

Key idea in one line

At the equivalence point, ignore the original acid and base and treat the solution as a pure salt solution, then do a regular weak-acid/weak-base (or neutral) pH calculation on that salt.

Step 1: Identify the titration type

Different titrations give different behaviors at equivalence.

  • Strong acid + strong base
    • The salt does not hydrolyze.
    • pH ≈ 7 at 25°C.
    • Example: HCl + NaOH → NaCl (neutral solution).
  • Weak acid + strong base
    • Salt is the conjugate base of a weak acid → acts as a weak base.
    • pH > 7 at equivalence.
    • Example: HA + NaOH → NaA (A⁻ is a weak base).
  • Weak base + strong acid
    • Salt is the conjugate acid of a weak base → acts as a weak acid.
    • pH < 7 at equivalence.
    • Example: NH₃ + HCl → NH₄Cl (NH₄⁺ is a weak acid).
  • Weak acid + weak base
    • Both ions can hydrolyze.
    • pH depends on Ka and Kb; you compare their strengths.
    • Often use pH≈7+12(pKa−pKb)\text{pH}\approx 7+\tfrac{1}{2}(\text{p}K_a-\text{p}K_b)pH≈7+21​(pKa​−pKb​) when applicable.

Step 2: Find the salt concentration at equivalence

At equivalence, moles acid = moles base.

  1. Use stoichiometry:
    • For monoprotic acid/base:
      • nacid=Cacid×Vacidn_\text{acid}=C_\text{acid}\times V_\text{acid}nacid​=Cacid​×Vacid​
      • nbase=Cbase×Vbase,eqn_\text{base}=C_\text{base}\times V_\text{base,eq}nbase​=Cbase​×Vbase,eq​
      • Set them equal to get the equivalence volume or check that you are at equivalence.
  1. All original acid/base has been converted into salt.
  2. Concentration of the salt:
    • Csalt=nsaltVtotalC_\text{salt}=\dfrac{n_\text{salt}}{V_\text{total}}Csalt​=Vtotal​nsalt​​,
      where Vtotal=Vacid+Vbase,eqV_\text{total}=V_\text{acid}+V_\text{base,eq}Vtotal​=Vacid​+Vbase,eq​.

Example structure (weak acid + strong base):

  • Start with 0.100 M HA, 25.0 mL.
  • Titrate with 0.100 M NaOH; equivalence at 25.0 mL.
  • Moles HA = 0.100 × 0.0250 = 2.50×10⁻³ mol.
  • At equivalence, moles A⁻ = 2.50×10⁻³ mol in 0.0500 L → 0.0500 M A⁻.

Step 3: Do the hydrolysis pH calculation

Now forget the titration and treat the salt like any weak acid or weak base in water.

Case A: Weak acid + strong base (salt behaves as weak base)

  • Species in water: A⁻ (conjugate base).
  • Reaction: A−+H2O⇌HA+OH−\text{A}^-+\text{H}_2\text{O}\rightleftharpoons \text{HA}+\text{OH}^-A−+H2​O⇌HA+OH−.
  • Use Kb=KwKaK_b=\dfrac{K_w}{K_a}Kb​=Ka​Kw​​.
  • Set up: Kb=[HA][OH−][A−]K_b=\dfrac{[\text{HA}][\text{OH}^-]}{[\text{A}^-]}Kb​=[A−][HA][OH−]​.
  • With initial [A⁻] = C and x = [OH⁻], approximate:
    • Kb≈x2C⇒x=KbCK_b\approx \dfrac{x^2}{C}\Rightarrow x=\sqrt{K_bC}Kb​≈Cx2​⇒x=Kb​C​.
    • pOH = −log[OH⁻], pH = 14 − pOH.

Many worked examples (like acetate or formate with NaOH) use exactly this method.

Case B: Weak base + strong acid (salt behaves as weak acid)

  • Species in water: BH⁺ (conjugate acid).
  • Reaction: BH++H2O⇌B+H3O+\text{BH}^++\text{H}_2\text{O}\rightleftharpoons \text{B}+\text{H}_3\text{O}^+BH++H2​O⇌B+H3​O+.
  • Use Ka=KwKbK_a=\dfrac{K_w}{K_b}Ka​=Kb​Kw​​.
  • Set up: Ka=[B][H3O+][BH+]K_a=\dfrac{[\text{B}][\text{H}_3\text{O}^+]}{[\text{BH}^+]}Ka​=[BH+][B][H3​O+]​.
  • With initial [BH⁺] = C and x = [H₃O⁺], approximate:
    • Ka≈x2C⇒x=KaCK_a\approx \dfrac{x^2}{C}\Rightarrow x=\sqrt{K_aC}Ka​≈Cx2​⇒x=Ka​C​.
    • pH = −log[H₃O⁺].

Case C: Strong acid + strong base

  • Salt does not hydrolyze appreciably (e.g., NaCl, KNO₃).
  • pH ≈ 7.00 at 25°C, assuming no extra strong acid or base and reasonable dilution.

Case D: Weak acid + weak base

  • Both cation and anion hydrolyze.
  • If Ka and Kb known, a common approximation is:

pH≈7+12(pKa−pKb)\text{pH}\approx 7+\frac{1}{2}(\text{p}K_a-\text{p}K_b)pH≈7+21​(pKa​−pKb​)

for the salt solution, when simplifying assumptions hold.

Common shortcut formulas you might see

For a salt of a weak acid and strong base (like CH₃COONa):

pH=7+12(pKw+log⁡Ka+log⁡C)\text{pH}=7+\frac{1}{2}(\text{p}K_w+\log K_a+\log C)pH=7+21​(pKw​+logKa​+logC)

or written in various algebraically equivalent forms, which come from combining KwK_wKw​, KaK_aKa​ and the hydrolysis expression.

For a salt of a weak base and strong acid (like NH₄Cl), similar forms exist using Kb and concentration C to directly compute pH at equivalence.

How this shows up on exam / homework questions

Typical question pattern:

  1. “Titrate X mL of 0.100 M weak acid HA (Ka given) with 0.100 M strong base. What is the pH at the equivalence point?”
  2. You:
    • Do stoichiometry to find equivalence volume and salt concentration.
    • Decide if the salt is acidic, basic, or neutral.
    • Do a weak-base or weak-acid pH calculation on the salt.

Many step‑by‑step worked examples follow this exact recipe for both weak- acid/strong-base and weak-base/strong-acid titrations.

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