Calculating pH of Strong Bases

Spec Mapping — OCR H432 Module 5.1.3 — Acids, bases and buffers, sub-topic on strong bases: definition of a strong base as one that undergoes complete ionisation to release OH⁻, pH calculation for strong monobasic bases ($[\text{OH}^-] = c$[OH−]=c) and strong dibasic bases ($[\text{OH}^-] = 2c$[OH−]=2c), use of $K_w$Kw​ to convert $[\text{OH}^-]$[OH−] to $[\text{H}^+]$[H+], dilution and mixing calculations, and excess-species calculations for strong-acid / strong-base mixtures (refer to the official OCR H432 specification document for exact wording). This lesson completes the strong-acid / strong-base calculational pair — the bridge $[\text{H}^+] = K_w / [\text{OH}^-]$[H+]=Kw​/[OH−] is the only new piece of machinery beyond lesson 3.

The strong-base pH calculation is a two-step routine: get $[\text{OH}^-]$[OH−] from the base and its stoichiometry, then use $K_w$Kw​ to convert to $[\text{H}^+]$[H+], then take the negative log. The conceptual content is light, but the bookkeeping demands care — especially the factor of 2 for dibasic bases like Ba(OH)₂, the volume conversions in mixing problems, and the excess-species reasoning when a strong acid and strong base react together but don't fully neutralise each other. This lesson also previews the equivalence point logic of lesson 9 (where moles of acid equal moles of base, leaving only water self-ionisation to set the pH).

Key Equation: $[\text{OH}^-] = b \cdot c \quad \text{(strong base)} \qquad [\text{H}^+] = \frac{K_w}{[\text{OH}^-]}$[OH−]=b⋅c(strong base)[H+]=[OH−]Kw​​ $\text{pH} = -\log_{10}[\text{H}^+] = 14 + \log_{10}[\text{OH}^-] \text{ at 298 K}$pH=−log10​[H+]=14+log10​[OH−] at 298 K where $b$b is the number of OH⁻ per formula unit (1 for NaOH/KOH, 2 for Ba(OH)₂).

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What is a strong base?

A strong base is one that undergoes essentially complete ionisation in aqueous solution to release hydroxide ions. The most important examples at A-Level are the Group 1 hydroxides (LiOH, NaOH, KOH) and the heavier Group 2 hydroxides (Ca(OH)₂, Ba(OH)₂).

Base	Formula	Ionisation	OH⁻ per formula	OCR-examined?
Lithium hydroxide	LiOH	Li⁺ + OH⁻	1	Yes (background)
Sodium hydroxide	NaOH	Na⁺ + OH⁻	1	Yes
Potassium hydroxide	KOH	K⁺ + OH⁻	1	Yes
Calcium hydroxide	Ca(OH)₂	Ca²⁺ + 2OH⁻	2 (sparingly soluble — limewater)	Yes (synoptic)
Barium hydroxide	Ba(OH)₂	Ba²⁺ + 2OH⁻	2	Yes

Group 1 hydroxides are very soluble strong bases — NaOH dissolves to >19 mol dm⁻³ in cold water. Ca(OH)₂ is only sparingly soluble (saturated solution is ~0.020 mol dm⁻³, traditionally called limewater), but what does dissolve is fully ionised — so within the solubility limit, limewater behaves as a strong dibasic alkali.

Important contrast — NH₃ is not a strong base. Ammonia is a weak base; only a small fraction (~1 % for 0.1 mol dm⁻³ NH₃) reacts with water to produce NH₄⁺ + OH⁻. NH₃ requires its own $K_b$Kb​ treatment and is beyond OCR Module 5.1.3 calculational examination (OCR uses NH₃ qualitatively in buffer discussions and in the basic-buffer extension, but does not ask for pH from $K_b$Kb​ in routine calculation).

Why Group 1 hydroxides are strong but Group 1 fluorides are inert. Both NaOH and NaF dissolve fully in water, but only NaOH produces a strong base. The difference is in the conjugate-acid strength: OH⁻ is the conjugate base of H₂O (a very weak acid, $K_a \approx 10^{-14}$Ka​≈10−14), so OH⁻ is a moderately strong Brønsted base in water; F⁻ is the conjugate base of HF (a weak acid, $K_a \approx 6 \times 10^{-4}$Ka​≈6×10−4), so F⁻ is a very weak Brønsted base. The general rule from lesson 4 ($K_a \cdot K_b = K_w$Ka​⋅Kb​=Kw​) explains this asymmetry: the weaker the conjugate acid, the stronger the base. Water has the weakest conjugate acid of any common neutral species — which is why hydroxide is such an exceptional base.

The two-step strategy

To calculate the pH of a strong base solution:

1. Get $[\text{OH}^-]$[OH−] from the base concentration and its stoichiometry. For a monobasic base $[\text{OH}^-] = c$[OH−]=c; for a dibasic base $[\text{OH}^-] = 2c$[OH−]=2c.
2. Convert to $[\text{H}^+]$[H+] via $K_w$Kw​. $[\text{H}^+] = K_w / [\text{OH}^-]$[H+]=Kw​/[OH−].
3. Take the negative log. pH = $-\log_{10}[\text{H}^+]$−log10​[H+].

flowchart TD
    A["Strong base concentration c"] --> B{"Monobasic or dibasic?"}
    B -- "Monobasic NaOH KOH" --> C["[OH-] = c"]
    B -- "Dibasic Ba(OH)2 Ca(OH)2" --> D["[OH-] = 2 x c"]
    C --> E["[H+] = Kw / [OH-]"]
    D --> E
    E --> F["pH = -log10[H+]"]
    F --> G["Quote pH to 2 dp"]

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Worked example 1 — 0.100 mol dm⁻³ NaOH

NaOH is fully ionised, so $[\text{OH}^-] = 0.100$[OH−]=0.100 mol dm⁻³.

$[\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.00 \times 10^{-14}}{0.100} = 1.00 \times 10^{-13} \text{ mol dm}^{-3}$[H+]=[OH−]Kw​​=0.1001.00×10−14​=1.00×10−13 mol dm−3 $\text{pH} = -\log_{10}(1.00 \times 10^{-13}) = 13.00$pH=−log10​(1.00×10−13)=13.00

Quick cross-check using pH + pOH = 14: pOH = $-\log_{10}(0.100) = 1.00$−log10​(0.100)=1.00, so pH = 14.00 − 1.00 = 13.00. ✓

Worked example 2 — 0.0500 mol dm⁻³ KOH

$[\text{OH}^-] = 0.0500 \text{ mol dm}^{-3}$[OH−]=0.0500 mol dm−3 $[\text{H}^+] = \frac{1.00 \times 10^{-14}}{0.0500} = 2.00 \times 10^{-13} \text{ mol dm}^{-3}$[H+]=0.05001.00×10−14​=2.00×10−13 mol dm−3 $\text{pH} = -\log_{10}(2.00 \times 10^{-13}) = 12.70 \quad (2 \text{ dp})$pH=−log10​(2.00×10−13)=12.70(2 dp)

Worked example 3 — dilute NaOH

$5.00 \times 10^{-3}$5.00×10−3 mol dm⁻³ NaOH: $[\text{H}^+] = \frac{1.00 \times 10^{-14}}{5.00 \times 10^{-3}} = 2.00 \times 10^{-12} \text{ mol dm}^{-3}$[H+]=5.00×10−31.00×10−14​=2.00×10−12 mol dm−3 $\text{pH} = 11.70 \quad (2 \text{ dp})$pH=11.70(2 dp)

Worked example 4 — strong dibasic base Ba(OH)₂

Calculate the pH of 0.0250 mol dm⁻³ Ba(OH)₂ assuming complete dissociation.

$\text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2\text{OH}^-$Ba(OH)2​→Ba2++2OH− $[\text{OH}^-] = 2 \times 0.0250 = 0.0500 \text{ mol dm}^{-3}$[OH−]=2×0.0250=0.0500 mol dm−3 $[\text{H}^+] = \frac{1.00 \times 10^{-14}}{0.0500} = 2.00 \times 10^{-13} \text{ mol dm}^{-3}$[H+]=0.05001.00×10−14​=2.00×10−13 mol dm−3 $\text{pH} = 12.70 \quad (2 \text{ dp})$pH=12.70(2 dp)

The factor-of-2 reflex is critical: 0.0250 mol dm⁻³ Ba(OH)₂ has the same pH as 0.0500 mol dm⁻³ KOH because each Ba(OH)₂ delivers two OH⁻.

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Dilutions

The familiar $c_1 V_1 = c_2 V_2$c1​V1​=c2​V2​ relation applies to bases as well.

Worked example 5 — 10× dilution of NaOH

25.0 cm³ of 0.500 mol dm⁻³ NaOH is diluted to 250.0 cm³.

$c_2 = \frac{0.500 \times 25.0}{250.0} = 0.0500 \text{ mol dm}^{-3}$c2​=250.00.500×25.0​=0.0500 mol dm−3 $[\text{OH}^-] = 0.0500 \text{ mol dm}^{-3}$[OH−]=0.0500 mol dm−3 $[\text{H}^+] = 2.00 \times 10^{-13} \text{ mol dm}^{-3} \Rightarrow \text{pH} = 12.70$[H+]=2.00×10−13 mol dm−3⇒pH=12.70

The pH dropped by 1.00 (from 13.70 to 12.70) for a 10× dilution. The mirror image of strong acids: every 10× dilution of a strong base lowers pH by 1.00, just as 10× dilution of a strong acid raises pH by 1.00.

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Mixing strong bases

Sum the moles of OH⁻ from each, then divide by total volume.

Worked example 6 — NaOH + Ba(OH)₂

25.0 cm³ of 0.200 mol dm⁻³ NaOH mixed with 75.0 cm³ of 0.0500 mol dm⁻³ Ba(OH)₂.

$n(\text{OH}^-)_{\text{NaOH}} = 0.200 \times 0.0250 = 5.00 \times 10^{-3} \text{ mol}$n(OH−)NaOH​=0.200×0.0250=5.00×10−3 mol $n(\text{OH}^-)_{\text{Ba(OH)}_2} = 2 \times 0.0500 \times 0.0750 = 7.50 \times 10^{-3} \text{ mol}$n(OH−)Ba(OH)2​​=2×0.0500×0.0750=7.50×10−3 mol $n(\text{OH}^-)_{\text{total}} = 1.25 \times 10^{-2} \text{ mol}$n(OH−)total​=1.25×10−2 mol $V_{\text{total}} = 0.100 \text{ dm}^3 \Rightarrow [\text{OH}^-] = 0.125 \text{ mol dm}^{-3}$Vtotal​=0.100 dm3⇒[OH−]=0.125 mol dm−3 $[\text{H}^+] = \frac{1.00 \times 10^{-14}}{0.125} = 8.00 \times 10^{-14} \text{ mol dm}^{-3}$[H+]=0.1251.00×10−14​=8.00×10−14 mol dm−3 $\text{pH} = 13.10 \quad (2 \text{ dp})$pH=13.10(2 dp)

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Mixing strong acid and strong base — the excess-species method

When strong acid and strong base are mixed, they neutralise mole-for-mole:

$\text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{H}_2\text{O(l)}$H+(aq)+OH−(aq)→H2​O(l)

If one is in excess, the excess species sets the pH. The protocol is:

1. Calculate moles of H⁺ from the acid (× 2 if dibasic).
2. Calculate moles of OH⁻ from the base (× 2 if dibasic).
3. Subtract — find which is in excess and by how much.
4. Divide the excess by the total volume.
5. If H⁺ excess: pH = $-\log_{10}[\text{H}^+]$−log10​[H+]. If OH⁻ excess: convert via $K_w$Kw​.
6. If exactly equal: pH = 7.00 (equivalence — water self-ionisation only).

Worked example 7 — OH⁻ excess

50.0 cm³ of 0.100 mol dm⁻³ HCl mixed with 40.0 cm³ of 0.150 mol dm⁻³ NaOH.

$n(\text{H}^+) = 0.100 \times 0.0500 = 5.00 \times 10^{-3} \text{ mol}$n(H+)=0.100×0.0500=5.00×10−3 mol $n(\text{OH}^-) = 0.150 \times 0.0400 = 6.00 \times 10^{-3} \text{ mol}$n(OH−)=0.150×0.0400=6.00×10−3 mol

OH⁻ in excess: $n_{\text{excess}} = 1.00 \times 10^{-3}$nexcess​=1.00×10−3 mol. Total volume 90.0 cm³ = 0.0900 dm³.

$[\text{OH}^-]_{\text{excess}} = \frac{1.00 \times 10^{-3}}{0.0900} = 1.111 \times 10^{-2} \text{ mol dm}^{-3}$[OH−]excess​=0.09001.00×10−3​=1.111×10−2 mol dm−3 $[\text{H}^+] = \frac{1.00 \times 10^{-14}}{1.111 \times 10^{-2}} = 9.00 \times 10^{-13} \text{ mol dm}^{-3}$[H+]=1.111×10−21.00×10−14​=9.00×10−13 mol dm−3 $\text{pH} = 12.05 \quad (2 \text{ dp})$pH=12.05(2 dp)

Worked example 8 — H⁺ excess

25.0 cm³ of 0.200 mol dm⁻³ NaOH mixed with 50.0 cm³ of 0.150 mol dm⁻³ HCl.

$n(\text{OH}^-) = 5.00 \times 10^{-3} \text{ mol}; \quad n(\text{H}^+) = 7.50 \times 10^{-3} \text{ mol}$n(OH−)=5.00×10−3 mol;n(H+)=7.50×10−3 mol $n(\text{H}^+)_{\text{excess}} = 2.50 \times 10^{-3} \text{ mol}$n(H+)excess​=2.50×10−3 mol $V_{\text{total}} = 0.0750 \text{ dm}^3$Vtotal​=0.0750 dm3 $[\text{H}^+] = \frac{2.50 \times 10^{-3}}{0.0750} = 3.33 \times 10^{-2} \text{ mol dm}^{-3}$[H+]=0.07502.50×10−3​=3.33×10−2 mol dm−3 $\text{pH} = -\log_{10}(3.33 \times 10^{-2}) = 1.48 \quad (2 \text{ dp})$pH=−log10​(3.33×10−2)=1.48(2 dp)

Worked example 9 — exact equivalence

20.0 cm³ of 0.100 mol dm⁻³ NaOH mixed with 20.0 cm³ of 0.100 mol dm⁻³ HCl.

$n(\text{OH}^-) = 2.00 \times 10^{-3} \text{ mol} = n(\text{H}^+)$n(OH−)=2.00×10−3 mol=n(H+)

Moles are exactly equal — full neutralisation. The remaining H⁺ and OH⁻ come only from water self-ionisation: $[\text{H}^+] = [\text{OH}^-] = 10^{-7}$[H+]=[OH−]=10−7 mol dm⁻³, pH = 7.00. This is the equivalence point of a strong-strong titration (lesson 9).

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Reverse calculation — pH to base concentration

Worked example 10 — KOH from pH

A KOH solution has pH = 11.60. Calculate [KOH].

$[\text{H}^+] = 10^{-11.60} = 2.51 \times 10^{-12} \text{ mol dm}^{-3}$[H+]=10−11.60=2.51×10−12 mol dm−3 $[\text{OH}^-] = \frac{K_w}{[\text{H}^+]} = \frac{1.00 \times 10^{-14}}{2.51 \times 10^{-12}} = 3.98 \times 10^{-3} \text{ mol dm}^{-3}$[OH−]=[H+]Kw​​=2.51×10−121.00×10−14​=3.98×10−3 mol dm−3 $[\text{KOH}] = [\text{OH}^-] = 3.98 \times 10^{-3} \text{ mol dm}^{-3}$[KOH]=[OH−]=3.98×10−3 mol dm−3

For Ba(OH)₂ you would divide $[\text{OH}^-]$[OH−] by 2 to get the base concentration.

Worked example 11 — solid base dissolved

0.400 g of NaOH ($M_r = 40.0$Mr​=40.0) is dissolved in water to make 500 cm³ of solution. Calculate the pH.

$n(\text{NaOH}) = \frac{0.400}{40.0} = 0.0100 \text{ mol}$n(NaOH)=40.00.400​=0.0100 mol $[\text{NaOH}] = \frac{0.0100}{0.500} = 0.0200 \text{ mol dm}^{-3}$[NaOH]=0.5000.0100​=0.0200 mol dm−3 $[\text{H}^+] = \frac{1.00 \times 10^{-14}}{0.0200} = 5.00 \times 10^{-13} \text{ mol dm}^{-3}$[H+]=0.02001.00×10−14​=5.00×10−13 mol dm−3 $\text{pH} = 12.30 \quad (2 \text{ dp})$pH=12.30(2 dp)

Limewater — Ca(OH)₂ pH from saturation

Saturated Ca(OH)₂ at 298 K has [Ca(OH)₂] ≈ 0.020 mol dm⁻³ (sparingly soluble). What dissolves is fully ionised, so:

$[\text{OH}^-] = 2 \times 0.020 = 0.040 \text{ mol dm}^{-3}$[OH−]=2×0.020=0.040 mol dm−3 $[\text{H}^+] = \frac{1.00 \times 10^{-14}}{0.040} = 2.50 \times 10^{-13} \text{ mol dm}^{-3}$[H+]=0.0401.00×10−14​=2.50×10−13 mol dm−3 $\text{pH} \approx 12.60$pH≈12.60

Limewater's modest alkalinity (pH ~12.6, not higher) is due to solubility limitation, not to weakness. Within the solubility limit Ca(OH)₂ is a strong base.

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Very dilute strong bases

At $[\text{OH}^-] < 10^{-6}$[OH−]<10−6 mol dm⁻³, the water self-ionisation contribution becomes significant and the simple formula breaks down. The pH of an extremely dilute base still approaches 7 from above and never crosses below 7. If you ever calculate pH < 7 for a base, you have made an arithmetic error.

Worked example 12 — staged dilution

A student takes 10.0 cm³ of 1.00 mol dm⁻³ NaOH and dilutes it first to 100 cm³, then takes 10.0 cm³ of that diluted solution and dilutes again to 100 cm³. Calculate the pH of the final solution.

Stage 1: $c_2 = (1.00 \times 10.0) / 100 = 0.100$c2​=(1.00×10.0)/100=0.100 mol dm⁻³. (pH = 13.00 if measured here.) Stage 2: $c_3 = (0.100 \times 10.0) / 100 = 0.0100$c3​=(0.100×10.0)/100=0.0100 mol dm⁻³.

$[\text{OH}^-] = 0.0100 \text{ mol dm}^{-3}$[OH−]=0.0100 mol dm−3 $[\text{H}^+] = \frac{1.00 \times 10^{-14}}{0.0100} = 1.00 \times 10^{-12} \text{ mol dm}^{-3}$[H+]=0.01001.00×10−14​=1.00×10−12 mol dm−3 $\text{pH} = 12.00$pH=12.00

The two 10× dilutions have dropped pH from 14.00 (original 1.00 mol dm⁻³) to 12.00 — exactly $\log_{10}(100) = 2.00$log10​(100)=2.00 units, as expected.

Worked example 13 — back-titrating Ba(OH)₂

A 25.0 cm³ aliquot of Ba(OH)₂ solution of unknown concentration is titrated against 0.100 mol dm⁻³ HCl. The mean titre is 18.40 cm³. Calculate the concentration of Ba(OH)₂.