Examples of Calculating pH of Strong Bases: 3 Practical Scenarios (Plus More)

Strong bases are the low-drama characters of acid–base chemistry. In water, they fully dissociate, so you don’t need equilibrium tables or ICE charts. You just:

• Get the hydroxide ion concentration, \([OH^-]\)

• Calculate pOH using:

  \[\text{pOH} = -\log[OH^-]\]

• Convert to pH using:

  \[\text{pH} = 14.00 - \text{pOH}\]
  (at 25 °C, where \(K_w = 1.0 \times 10^{-14}\))

The best examples of calculating pH of strong bases all follow this same backbone, but the twist is in the details: mono- vs. polyhydroxide bases, very concentrated vs. very dilute, and mixtures.

Example of calculating pH of a simple strong base: 0.010 M NaOH

Let’s kick off with the kind of problem you’d see in the first week of acid–base work.

Problem: A solution is prepared by dissolving NaOH to make a 0.010 M solution. NaOH is a strong base. Find the pH.

Step 1 – Write the dissociation.
\(\text{NaOH} \rightarrow \text{Na}^+ + OH^-\)

Every mole of NaOH gives one mole of OH⁻, so:

\[ [OH^-] = 0.010\,\text{M} \]

Step 2 – Calculate pOH.

\[ \text{pOH} = -\log(0.010) = 2.00 \]

Step 3 – Convert to pH.

\[ \text{pH} = 14.00 - 2.00 = 12.00 \]

That’s it. This is one of the cleanest examples of calculating pH of strong bases: 3 practical examples we’ll see later are just this logic with more moving parts.

Examples of calculating pH of strong bases: 3 practical examples you’ll actually see

Now let’s walk through three core scenarios that keep showing up in classes, exams, and lab work. These are the backbone examples of calculating pH of strong bases: 3 practical examples that cover most of what you need for strong base calculations.

Practical Example 1: Polyhydroxide base – 0.020 M Ca(OH)₂

Problem: What is the pH of a 0.020 M solution of calcium hydroxide, Ca(OH)₂, at 25 °C?

Step 1 – Recognize it’s a strong base with two OH⁻ per formula unit.
Dissociation:

\[ \text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\,OH^- \]

For each 1 mole of Ca(OH)₂, you get 2 moles of OH⁻. So:

\[ [OH^-] = 2 \times 0.020\,\text{M} = 0.040\,\text{M} \]

Step 2 – Calculate pOH.

\[ \text{pOH} = -\log(0.040) \approx 1.40 \]

Step 3 – Convert to pH.

\[ \text{pH} = 14.00 - 1.40 = 12.60 \]

This is a classic example of calculating pH when the base contributes more than one hydroxide ion per formula unit. Many of the best examples of calculating pH of strong bases are just this pattern with different numbers.

Practical Example 2: Very dilute NaOH – when water matters

Problem: Estimate the pH of a 1.0 × 10⁻⁸ M NaOH solution at 25 °C.

If you mindlessly apply the usual method:

\[ [OH^-]_{\text{from NaOH}} = 1.0 \times 10^{-8}\,\text{M} \]

\[ \text{pOH} = -\log(1.0 \times 10^{-8}) = 8.00 \]

\[ \text{pH} = 14.00 - 8.00 = 6.00 \]

This says a basic solution has pH 6.00, which is acidic. That’s a red flag.

At this concentration, the OH⁻ from NaOH is on the same order of magnitude as the OH⁻ from water’s own autoionization. You can’t ignore water anymore.

Let \(x\) be the contribution of water to \([H^+]\) and \([OH^-]\). Then:

• From NaOH: \([OH^-] = 1.0 \times 10^{-8}\,\text{M}\)
• From water: \([H^+] = x\), \([OH^-] = x\)

Total \([OH^-] = x + 1.0 \times 10^{-8}\)

Use \(K_w = [H^+][OH^-] = 1.0 \times 10^{-14}\):

\[ x(x + 1.0 \times 10^{-8}) = 1.0 \times 10^{-14} \]

Solving this quadratic gives \(x \approx 9.5 \times 10^{-8}\,\text{M}\). Then:

\[ [H^+] \approx 9.5 \times 10^{-8}\,\text{M} \]

\[ \text{pH} = -\log(9.5 \times 10^{-8}) \approx 7.02 \]

So the pH is slightly basic, just above 7. This is one of the more subtle examples of calculating pH of strong bases: 3 practical examples rarely include this twist, but modern textbooks and exams increasingly do.

For reference on water autoionization and \(K_w\), see resources like the NIST Chemistry WebBook (U.S. National Institute of Standards and Technology).

Practical Example 3: Strong base from mass – KOH pellets in water

Problem: You dissolve 2.80 g of KOH (solid pellets) in water and make 500.0 mL of solution. Find the pH at 25 °C.

This is a very realistic lab-style problem.

Step 1 – Convert mass to moles.
Molar mass of KOH ≈ 56.11 g/mol.

\[ n = \frac{2.80\,\text{g}}{56.11\,\text{g/mol}} \approx 0.0499\,\text{mol} \]

Step 2 – Find concentration of OH⁻.
Volume = 500.0 mL = 0.5000 L.

KOH is a strong base:

\[ \text{KOH} \rightarrow \text{K}^+ + OH^- \]

\[ [OH^-] = \frac{0.0499\,\text{mol}}{0.5000\,\text{L}} = 0.0998\,\text{M} \approx 0.100\,\text{M} \]

Step 3 – pOH and pH.

\[ \text{pOH} = -\log(0.100) = 1.00 \]

\[ \text{pH} = 14.00 - 1.00 = 13.00 \]

This is exactly the kind of example of calculating pH that connects the classroom formula to what you actually do at the bench.

More real examples of calculating pH of strong bases (beyond the “3 practical examples”)

The headline promised examples of calculating pH of strong bases: 3 practical examples, but real learning happens when you push past three. Let’s add a few more scenarios you’re likely to face.

Example 4: Strong base used in cleaning – 0.50 M NaOH

Industrial cleaners and drain openers often contain concentrated NaOH.

Problem: Estimate the pH of a 0.50 M NaOH solution.

Fully dissociated, so:

\[ [OH^-] = 0.50\,\text{M} \]

\[ \text{pOH} = -\log(0.50) \approx 0.30 \]

\[ \text{pH} = 14.00 - 0.30 = 13.70 \]

At this level, you’re dealing with a highly caustic solution. Organizations like NIOSH and OSHA emphasize protective equipment for strong base exposure because high-pH solutions can cause serious chemical burns.

Example 5: Barium hydroxide – 0.015 M Ba(OH)₂

Problem: Find the pH of 0.015 M Ba(OH)₂ at 25 °C.

Dissociation:

\[ \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2\,OH^- \]

\[ [OH^-] = 2 \times 0.015\,\text{M} = 0.030\,\text{M} \]

\[ \text{pOH} = -\log(0.030) \approx 1.52 \]

\[ \text{pH} = 14.00 - 1.52 = 12.48 \]

This is another solid example of calculating pH where paying attention to the stoichiometry (two OH⁻ per formula unit) is everything.

Example 6: Strong base mixture – NaOH + KOH in the same solution

Problem: A solution contains 0.010 M NaOH and 0.020 M KOH. Both are strong bases. What is the pH at 25 °C?

Each base fully dissociates and contributes OH⁻. Add them.

\[ [OH^-]_{\text{total}} = 0.010 + 0.020 = 0.030\,\text{M} \]

\[ \text{pOH} = -\log(0.030) \approx 1.52 \]

\[ \text{pH} = 14.00 - 1.52 = 12.48 \]

The identity of the cation (Na⁺ vs. K⁺) doesn’t matter for pH here; they’re spectator ions. This is one of the best examples of calculating pH of strong bases when more than one base is present.

Example 7: Temperature-aware thinking (2024–2025 context)

Most textbook examples of calculating pH of strong bases assume 25 °C, where \(K_w = 1.0 \times 10^{-14}\) and pH + pOH = 14. In real lab or industrial settings, temperature can drift, and \(K_w\) changes with temperature.

For instance, at 50 °C, \(K_w\) is larger (around \(5.5 \times 10^{-14}\)), which means neutral pH is lower than 7. That shifts the pH scale a bit. Many modern analytical chemistry courses and AP/IB curricula (2024–2025) highlight this.

If you’re working at nonstandard temperatures, you should:

• Look up \(K_w\) for your temperature (for example, from NIST or standard physical chemistry data tables)
• Use \(K_w = [H^+][OH^-]\) directly instead of assuming pH + pOH = 14

Even though this isn’t a full numeric example, it’s an updated way of thinking about examples of calculating pH of strong bases in real-world, 2024-era lab conditions.

Strategy summary: spotting patterns in examples of calculating pH of strong bases

If you look back across all these examples of calculating pH of strong bases—3 practical examples plus several extensions—you’ll notice the same pattern repeating.

When you see a strong base problem, ask yourself:

• Is it a strong base?
  Check if it’s one of the usual suspects: NaOH, KOH, LiOH, RbOH, CsOH, Ca(OH)₂, Sr(OH)₂, Ba(OH)₂.

• How many OH⁻ ions per formula unit?
  One for NaOH, KOH, etc.; two for Ca(OH)₂, Ba(OH)₂, etc.

• Am I given molarity, mass, or something else?
  Convert everything to \([OH^-]\).

• Is the solution extremely dilute (10⁻⁷–10⁻⁹ M range)?
  If yes, consider water’s autoionization instead of blindly trusting pH + pOH = 14.

Once you can answer those questions, the examples of calculating pH of strong bases basically solve themselves.

For more structured practice and reference, many general chemistry courses (for example, MIT OpenCourseWare or university gen chem syllabi) walk through similar problems; you can browse materials from sites like MIT OpenCourseWare or Khan Academy for additional worked examples.

FAQ: common questions about examples of calculating pH of strong bases

Q1: Can you give another quick example of calculating pH for a strong base?
Sure. Suppose you have 0.0050 M LiOH.

LiOH is a strong base and gives one OH⁻ per formula unit:

\[ [OH^-] = 0.0050\,\text{M} \]

\[ \text{pOH} = -\log(0.0050) \approx 2.30 \]

\[ \text{pH} = 14.00 - 2.30 = 11.70 \]

This fits right into the family of examples of calculating pH of strong bases we’ve been using.

Q2: Why are examples of calculating pH of strong bases easier than weak bases?
Because for strong bases, you don’t need an equilibrium constant. They’re treated as fully dissociated, so \([OH^-]\) comes straight from stoichiometry. Weak bases require you to set up equilibrium expressions and sometimes solve quadratic equations.

Q3: Do I always use 14.00 in the pH + pOH = 14 equation?
Only if you’re assuming 25 °C. The relationship comes from:

\[ \text{pH} + \text{pOH} = pK_w \]

At 25 °C, \(pK_w = 14.00\). At other temperatures, \(pK_w\) shifts. For most introductory examples of calculating pH of strong bases, you’ll see 25 °C used by default unless the problem says otherwise.

Q4: Are there any safety-related examples of strong base pH I should know?
Yes. Household drain cleaners often contain concentrated NaOH or KOH, with pH values above 13. These are highly corrosive. Agencies like NIH and CDC/NIOSH provide safety guidance on handling caustic solutions, which is a real-world angle on why understanding pH isn’t just academic.

Q5: How do examples of strong base pH compare to strong acid pH problems?
They’re mirror images. For strong acids, you go directly to \([H^+]\), then pH. For strong bases, you go to \([OH^-]\), then pOH, then pH. The math is symmetric; the only catch is remembering when water’s autoionization matters at very low concentrations.

If you can comfortably work through these examples of calculating pH of strong bases—3 practical examples and the extra ones—you’re in good shape for most general chemistry problems.