Best examples of calculating the pH of acid-base mixtures (step-by-step)

Starting with real examples of calculating the pH of acid-base mixtures

Let’s skip the abstract talk and go straight into real examples of calculating the pH of acid-base mixtures. Then we’ll generalize the patterns you see.

We’ll assume room temperature (25 °C) so that \(K_w = 1.0 \times 10^{-14}\) and pH + pOH = 14.

Example 1 – Strong acid + strong base (net excess acid)

Mixture: 50.0 mL of 0.10 M HCl mixed with 25.0 mL of 0.10 M NaOH.

Both are strong, so they dissociate completely and just neutralize each other.

1. Moles before mixing

   • HCl: \(0.0500\,\text{L} \times 0.10\,\text{mol/L} = 5.0 \times 10^{-3}\,\text{mol}\)
   • NaOH: \(0.0250\,\text{L} \times 0.10\,\text{mol/L} = 2.5 \times 10^{-3}\,\text{mol}\)

2. Neutralization reaction
   \(\text{H}^+ + \text{OH}^- \to \text{H}_2\text{O}\)

   Limiting reagent: OH⁻.
   Leftover H⁺ moles: \(5.0 \times 10^{-3} - 2.5 \times 10^{-3} = 2.5 \times 10^{-3}\,\text{mol}\).

3. Total volume
   \(V_{\text{total}} = 50.0\,\text{mL} + 25.0\,\text{mL} = 75.0\,\text{mL} = 0.0750\,\text{L}\)

4. [H⁺] and pH
   \([\text{H}^+] = \dfrac{2.5 \times 10^{-3}}{0.0750} = 3.33 \times 10^{-2}\,\text{M}\)
   \(\text{pH} = -\log(3.33 \times 10^{-2}) \approx 1.48\)

This is one of the cleanest examples of calculating the pH of acid-base mixtures: for strong/strong, it’s just moles, subtraction, and a final concentration.

Example 2 – Strong acid + strong base (equivalence point)

Mixture: 25.0 mL of 0.10 M HCl with 25.0 mL of 0.10 M NaOH.

Moles of H⁺ and OH⁻ are equal:

• H⁺ moles = OH⁻ moles = \(2.5 \times 10^{-3}\,\text{mol}\)

They completely neutralize, leaving only water and spectator ions (Na⁺, Cl⁻). At 25 °C, pure water has \([\text{H}^+] = 1.0 \times 10^{-7}\,\text{M}\), so

• pH ≈ 7.00

This example of calculating the pH of acid-base mixtures shows that a strong acid–strong base equivalence point is neutral (pH ≈ 7), unlike weak acid titrations.

Example 3 – Strong base in excess (strong base + strong acid)

Flip Example 1.

Mixture: 25.0 mL of 0.10 M HCl with 50.0 mL of 0.10 M NaOH.

1. Moles before mixing

   • HCl: \(0.0250\,\text{L} \times 0.10 = 2.5 \times 10^{-3}\,\text{mol}\)
   • NaOH: \(0.0500\,\text{L} \times 0.10 = 5.0 \times 10^{-3}\,\text{mol}\)

2. Leftover OH⁻
   Excess OH⁻ moles: \(5.0 \times 10^{-3} - 2.5 \times 10^{-3} = 2.5 \times 10^{-3}\,\text{mol}\)

3. Total volume
   \(V_{\text{total}} = 0.0750\,\text{L}\)

4. [OH⁻], pOH, pH
   \([\text{OH}^-] = \dfrac{2.5 \times 10^{-3}}{0.0750} = 3.33 \times 10^{-2}\,\text{M}\)
   \(\text{pOH} = -\log(3.33 \times 10^{-2}) \approx 1.48\)
   \(\text{pH} = 14.00 - 1.48 \approx 12.52\)

Again, for strong/strong mixtures, examples of calculating the pH of acid-base mixtures are basically stoichiometry problems.

Example 4 – Strong acid + weak base (NH₃) before equivalence

Now it gets more interesting.

Mixture: 25.0 mL of 0.10 M NH₃ with 10.0 mL of 0.10 M HCl.

• \(K_b\) for NH₃ at 25 °C ≈ \(1.8 \times 10^{-5}\)
• Conjugate acid: NH₄⁺, with \(K_a = \dfrac{K_w}{K_b} = \dfrac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.6 \times 10^{-10}\)

1. Moles before reaction

   • NH₃: \(0.0250 \times 0.10 = 2.5 \times 10^{-3}\,\text{mol}\)
   • HCl: \(0.0100 \times 0.10 = 1.0 \times 10^{-3}\,\text{mol}\)

2. Reaction:
   \(\text{NH}_3 + \text{H}^+ \to \text{NH}_4^+\)

   H⁺ is limiting. After reaction:

   • NH₃ left: \(2.5 \times 10^{-3} - 1.0 \times 10^{-3} = 1.5 \times 10^{-3}\,\text{mol}\)
   • NH₄⁺ formed: \(1.0 \times 10^{-3}\,\text{mol}\)

3. Total volume
   \(V_{\text{total}} = 25.0 + 10.0 = 35.0\,\text{mL} = 0.0350\,\text{L}\)

4. Buffer concentrations

   • \([\text{NH}_3] = \dfrac{1.5 \times 10^{-3}}{0.0350} = 0.0429\,\text{M}\)
   • \([\text{NH}_4^+] = \dfrac{1.0 \times 10^{-3}}{0.0350} = 0.0286\,\text{M}\)

5. Use Henderson–Hasselbalch for the conjugate acid
   For the NH₄⁺/NH₃ buffer:
   \(\text{p}K_a = -\log(5.6 \times 10^{-10}) \approx 9.25\)

   \[ \text{pH} = \text{p}K_a + \log\left(\dfrac{[\text{base}]}{[\text{acid}]}\right) = 9.25 + \log\left(\dfrac{0.0429}{0.0286}\right) \]

   \(\dfrac{0.0429}{0.0286} = 1.50\), so \(\log(1.50) \approx 0.176\).

   \(\text{pH} \approx 9.25 + 0.18 = 9.43\)

This is a classic example of calculating the pH of acid-base mixtures that form a buffer: you first do stoichiometry, then use the Henderson–Hasselbalch equation.

Example 5 – Weak acid + strong base (acetic acid buffer region)

Mixture: 50.0 mL of 0.10 M acetic acid (CH₃COOH) with 25.0 mL of 0.10 M NaOH.

Data: \(K_a(\text{CH}_3\text{COOH}) \approx 1.8 \times 10^{-5}\), so pKₐ ≈ 4.74.

1. Moles before mixing

   • CH₃COOH: \(0.0500 \times 0.10 = 5.0 \times 10^{-3}\,\text{mol}\)
   • OH⁻: \(0.0250 \times 0.10 = 2.5 \times 10^{-3}\,\text{mol}\)

2. Reaction:
   \(\text{CH}_3\text{COOH} + \text{OH}^- \to \text{CH}_3\text{COO}^- + \text{H}_2\text{O}\)

   OH⁻ is limiting. After reaction:

   • CH₃COOH left: \(5.0 \times 10^{-3} - 2.5 \times 10^{-3} = 2.5 \times 10^{-3}\,\text{mol}\)
   • CH₃COO⁻ formed: \(2.5 \times 10^{-3}\,\text{mol}\)

3. Total volume
   \(V_{\text{total}} = 75.0\,\text{mL} = 0.0750\,\text{L}\)

4. Buffer concentrations

   • \([\text{CH}_3\text{COOH}] = \dfrac{2.5 \times 10^{-3}}{0.0750} = 0.0333\,\text{M}\)
   • \([\text{CH}_3\text{COO}^-] = 0.0333\,\text{M}\)

5. pH using Henderson–Hasselbalch
   \[ \text{pH} = \text{p}K_a + \log\left(\dfrac{[\text{base}]}{[\text{acid}]}\right) = 4.74 + \log(1.00) = 4.74 \]

This buffer-region case is one of the best examples of calculating the pH of acid-base mixtures because it hits a common exam favorite: when acid and conjugate base are equal, pH = pKₐ.

Example 6 – Weak acid + strong base (beyond equivalence point)

Take the same acetic acid but add more NaOH.

Mixture: 50.0 mL of 0.10 M CH₃COOH with 75.0 mL of 0.10 M NaOH.

1. Moles before reaction

   • CH₃COOH: \(5.0 \times 10^{-3}\,\text{mol}\)
   • OH⁻: \(0.0750 \times 0.10 = 7.5 \times 10^{-3}\,\text{mol}\)

2. Reaction and leftover OH⁻
   All CH₃COOH is neutralized:

   • CH₃COO⁻ formed: \(5.0 \times 10^{-3}\,\text{mol}\)
   • Excess OH⁻: \(7.5 \times 10^{-3} - 5.0 \times 10^{-3} = 2.5 \times 10^{-3}\,\text{mol}\)

3. Total volume
   \(V_{\text{total}} = 125.0\,\text{mL} = 0.1250\,\text{L}\)

4. [OH⁻] and pH
   \([\text{OH}^-] = \dfrac{2.5 \times 10^{-3}}{0.1250} = 0.0200\,\text{M}\)
   \(\text{pOH} = -\log(0.0200) \approx 1.70\)
   \(\text{pH} = 14.00 - 1.70 = 12.30\)

Here, the excess strong base completely dominates; the weak-acid origin of CH₃COO⁻ barely matters for pH.

Example 7 – Weak acid + weak base mixture

Weak–weak mixtures are where students often get stuck.

Mixture: 100.0 mL of 0.10 M acetic acid with 100.0 mL of 0.10 M ammonia.

• Both are 0.010 mol initially.
• Reaction: \(\text{CH}_3\text{COOH} + \text{NH}_3 \rightleftharpoons \text{CH}_3\text{COO}^- + \text{NH}_4^+\)

This is a salt-formation equilibrium between two weak species. The exact pH requires solving an equilibrium expression. But there’s a shortcut when the acid and base are equimolar and not extremely weak or strong relative to each other.

You can approximate pH using the average of pKₐ (acid) and pK_b (base converted to pKₐ of its conjugate acid):

• pKₐ of acetic acid ≈ 4.74
• pK_b of NH₃ ≈ 4.75 ⇒ pKₐ of NH₄⁺ = 14 − 4.75 ≈ 9.25

A widely taught approximation for this kind of example of calculating the pH of acid-base mixtures is:

\[ \text{pH} \approx \dfrac{\text{p}K_a(\text{acid}) + \text{p}K_a(\text{conjugate acid of base})}{2} \]

So:

\[ \text{pH} \approx \dfrac{4.74 + 9.25}{2} \approx \dfrac{13.99}{2} = 7.00 \]

The mixture is approximately neutral, but not exactly pure water; it’s a solution of the weak conjugate pair CH₃COO⁻/NH₄⁺.

If you want to see the full equilibrium approach, the ChemLibreTexts project has detailed derivations and more examples of calculating the pH of acid-base mixtures involving weak–weak systems:
https://chem.libretexts.org

Example 8 – Real-world style: pH of a buffer made from a salt and an acid

This mimics a common lab prep.

Mixture: 0.20 mol of sodium acetate (CH₃COONa) and 0.10 mol of acetic acid dissolved together and diluted to 1.00 L.

1. Concentrations

   • \([\text{CH}_3\text{COO}^-] = 0.20\,\text{M}\)
   • \([\text{CH}_3\text{COOH}] = 0.10\,\text{M}\)

2. Use Henderson–Hasselbalch
   pKₐ = 4.74 (as before).

   \[ \text{pH} = 4.74 + \log\left(\dfrac{0.20}{0.10}\right) = 4.74 + \log(2.0) \]

   \(\log(2.0) \approx 0.30\), so pH ≈ 5.04.

This is exactly how buffer recipes in biochemistry and molecular biology are calculated. For example, many NIH and CDC protocols rely on acetate, phosphate, or Tris buffers computed by the same method:
https://www.ncbi.nlm.nih.gov/books/NBK21607/
https://www.cdc.gov/lab

These are very real examples of calculating the pH of acid-base mixtures that matter in current (2024–2025) biomedical and environmental labs.

How to choose the right method for pH of a mixture

Looking across these examples of calculating the pH of acid-base mixtures, a pattern emerges:

• Strong + strong → do mole accounting, find excess H⁺ or OH⁻, then pH or pOH.
• Strong + weak (buffer region) → do stoichiometry first, then Henderson–Hasselbalch on the conjugate pair.
• Weak + strong (beyond equivalence) → excess strong species dominates; treat like a strong acid or base.
• Weak + weak → if equimolar and moderate strengths, use the average-pKₐ approximation; otherwise, set up full equilibrium.

Those are the mental shortcuts that let you move quickly through new problems and spot which examples include buffer behavior, which are dominated by strong species, and which actually require more algebra.

For more formal treatments and updated data tables (pKₐ values, temperature dependence, ionic strength corrections), check recent physical chemistry references or open resources like:

• NIST Chemistry WebBook: https://webbook.nist.gov/chemistry/
• MIT OpenCourseWare chemistry notes: https://ocw.mit.edu/courses/chemistry/

These sources are regularly updated and align with what you’re likely to see in 2024–2025 college courses.

FAQ: common questions on examples of calculating the pH of acid-base mixtures

Q1. Can you give another quick example of calculating the pH of acid-base mixtures in a titration?
Yes. Suppose you titrate 40.0 mL of 0.10 M HCl with 0.20 M NaOH and you’ve added 10.0 mL of NaOH.
Moles H⁺ = 0.0400 × 0.10 = 4.0 × 10⁻³ mol.
Moles OH⁻ added = 0.0100 × 0.20 = 2.0 × 10⁻³ mol.
Leftover H⁺ = 2.0 × 10⁻³ mol in 0.0500 L ⇒ [H⁺] = 0.040 M ⇒ pH ≈ 1.40.
That’s another straightforward strong/strong case.

Q2. When are Henderson–Hasselbalch examples of calculating the pH of acid-base mixtures accurate enough?
They work well when both acid and conjugate base concentrations are at least about 100 times larger than their \(K_a\) (for acids) or \(K_b\) (for bases), and when you’re not at extreme dilutions. In other words, use it in the buffer region, not right at the very start of a titration or extremely close to equivalence.

Q3. Do temperature changes matter for these examples of pH calculations?
Yes. \(K_w\), \(K_a\), and \(K_b\) all depend on temperature, so pH shifts with temperature even for neutral water. For many classroom problems we assume 25 °C, but in real lab or environmental work (for example, measuring lake water pH in summer vs winter), you either correct for temperature or measure pH directly with a calibrated meter.

Q4. What is a real example of acid-base mixture pH that matters in medicine or biology?
Blood is buffered mainly by the carbonic acid/bicarbonate system. The Henderson–Hasselbalch equation is literally built into how clinicians interpret blood-gas measurements and acid–base disorders. You can see medically oriented explanations at Mayo Clinic and NIH resources, such as:
https://www.mayoclinic.org
https://www.ncbi.nlm.nih.gov/books/NBK305/

Q5. How do ionic strength and activity coefficients affect examples of calculating the pH of acid-base mixtures?
In concentrated solutions or in systems with many ions (like seawater), you should technically use activities instead of raw concentrations. That means introducing activity coefficients (γ). For most introductory problems the effect is small and ignored, but advanced analytical chemistry and environmental modeling do correct for it, especially when precision below ±0.05 pH units matters.

With these eight worked examples of calculating the pH of acid-base mixtures, you’ve essentially seen every major pattern that shows up in high school, AP, and early college chemistry. The trick now is practice: vary the volumes, swap in different weak acids or bases, and check whether your reasoning about “who dominates” still lines up with the numbers.