pH, pOH and the ionisation of water

The self-ionisation of water and Kw, how Kw and neutral pH change with temperature, calculating pH, pOH, [H₃O⁺] and [OH⁻] for strong acids and bases, significant figures in pH, and the pH of solutions after dilution or mixing, with a practical to measure the pH of a range of acids and bases.

Pure water conducts electricity, very slightly. Even the purest water ever made, distilled again and again in quartz vessels, has a conductivity that never drops below a tiny floor. The reason is that water molecules are constantly passing protons to each other. At 25 °C, about two water molecules in every billion are ionised at any moment. That tiny fraction sets the whole pH scale.

The self-ionisation of water

Water is amphiprotic, so one molecule can donate a proton to another:

$$\mathrm{H_2O(l) + H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)}$$

The equilibrium constant for this reaction (water itself is left out, as a pure liquid) is the ionisation constant of water:

$$K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14} \text{ at } 25\ ^\circ\mathrm{C}$$

In pure water, [H₃O⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol/L. The key point is that $K_w$ holds in every aqueous solution, not just pure water. Add acid and [H₃O⁺] rises, so the equilibrium shifts left and [OH⁻] falls until their product is again 1.0 × 10⁻¹⁴. So even a strong acid solution contains a little OH⁻, and a strong base contains a little H₃O⁺.

A solution is neutral when [H₃O⁺] = [OH⁻], acidic when [H₃O⁺] > [OH⁻], and basic when [OH⁻] > [H₃O⁺].

Kw depends on temperature

Self-ionisation is endothermic (it breaks O–H bonds). By Le Chatelier's principle, raising the temperature shifts the equilibrium right and $K_w$ increases.

As temperature rises, Kw increases and the pH of pure water falls. Pure water is neutral at every temperature.

Temperature (°C)	0	25	50	60
$K_w$	1.14 × 10⁻¹⁵	1.0 × 10⁻¹⁴	5.5 × 10⁻¹⁴	9.6 × 10⁻¹⁴
pH of pure water	7.47	7.00	6.63	6.51

So pure water at 60 °C has pH 6.51 but is not acidic: [H₃O⁺] still equals [OH⁻]. "pH 7 is neutral" is only true at 25 °C. HSC questions use 25 °C unless they say otherwise.

Worked example: neutral but not pH 7

At 50 °C, $K_w$ = 5.5 × 10⁻¹⁴. Calculate the pH of pure water at this temperature.

In pure water [H₃O⁺] = [OH⁻] = $x$, so $x^2 = 5.5 \times 10^{-14}$.

set up

$x = \sqrt{5.5 \times 10^{-14}} = 2.3 \times 10^{-7}$ mol/L.

solve

pH = −log₁₀(2.3 × 10⁻⁷) = 6.63. It is still neutral.

pH and pOH

Concentrations of H₃O⁺ range over more than 14 powers of ten, so we use a logarithmic scale:

$$\mathrm{pH} = -\log_{10}[\mathrm{H_3O^+}] \qquad [\mathrm{H_3O^+}] = 10^{-\mathrm{pH}}$$

$$\mathrm{pOH} = -\log_{10}[\mathrm{OH^-}] \qquad [\mathrm{OH^-}] = 10^{-\mathrm{pOH}}$$

Taking −log₁₀ of both sides of $K_w$ = [H₃O⁺][OH⁻] gives, at 25 °C:

$$\mathrm{pH} + \mathrm{pOH} = 14.00$$

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Because the scale is logarithmic, a change of 1 pH unit is a tenfold change in [H₃O⁺]. Lemon juice at pH 2.3 has 1000 times the [H₃O⁺] of black coffee at pH 5.3, not "about half as much again". Count the units of difference and raise 10 to that power.

Significant figures: in a pH value, only the digits after the decimal point count. [H₃O⁺] = 2.5 × 10⁻³ mol/L (2 significant figures) gives pH = 2.60 (2 decimal places).

For strong acids and bases, ionisation is complete, so the concentration of H₃O⁺ or OH⁻ comes straight from the formula. Watch for acids and bases that release more than one: 0.010 mol/L Ba(OH)₂ gives [OH⁻] = 0.020 mol/L. For H₂SO₄ the HSC usually lets you assume both protons are released (the second step is actually only partial, so this slightly overestimates [H₃O⁺]).

Worked example: strong acid

Calculate the pH and pOH of 0.0040 mol/L HNO₃.

"]HNO₃ is a strong monoprotic acid: [H₃O⁺] = 0.0040 mol/L.

pH = −log₁₀(0.0040) = 2.40.

pOH = 14.00 − 2.40 = 11.60, so [OH⁻] = 10⁻¹¹·⁶⁰ = 2.5 × 10⁻¹² mol/L.

pOH

Worked example: strong base

Calculate the pH of a solution made by dissolving 0.37 g of Ca(OH)₂ (M = 74.10 g/mol) in water to make 500.0 mL of solution.

n(Ca(OH)₂) = 0.37 ÷ 74.10 = 4.99 × 10⁻³ mol.

moles

"]Each formula unit gives 2 OH⁻: [OH⁻] = 2 × 4.99 × 10⁻³ ÷ 0.5000 = 0.0200 mol/L.

pOH = −log₁₀(0.0200) = 1.70; pH = 14.00 − 1.70 = 12.30.

pOH and pH

Worked example: from pH back to concentration

Gastric juice has pH 1.80. Find [H₃O⁺] and [OH⁻].

"][H₃O⁺] = 10⁻¹·⁸⁰ = 1.6 × 10⁻² mol/L (2 sig. fig., from 2 decimal places).

"][OH⁻] = $K_w$ ÷ [H₃O⁺] = 1.0 × 10⁻¹⁴ ÷ 1.6 × 10⁻² = 6.3 × 10⁻¹³ mol/L.

Diluting acids and bases

Dilution doesn't change the number of moles of H₃O⁺ (or OH⁻), only the volume they're spread through. Use $c_1V_1 = c_2V_2$, then find the new pH. For a strong acid, a tenfold dilution raises the pH by exactly 1; for a strong base it lowers the pH by 1.

You can't dilute an acid until it becomes basic. As [H₃O⁺] from the acid falls towards 10⁻⁷ mol/L, the H₃O⁺ from water's own ionisation takes over, and the pH approaches 7 from below.

Worked example: dilution

25.0 mL of 0.200 mol/L HCl is diluted to 500.0 mL with distilled water. Find the new pH.

$c_2 = \dfrac{c_1V_1}{V_2} = \dfrac{0.200 \times 25.0}{500.0} = 0.0100$ mol/L

c₂

pH = −log₁₀(0.0100) = 2.000, up from 0.699 before dilution.

Mixing acids and bases

When you mix an acid and a base, they neutralise each other first, and the pH is set by whichever is left over. The method is always the same:

Calculate n(H₃O⁺) from the acid and n(OH⁻) from the base.
Subtract to find which is in excess and by how much.
Divide the excess by the total volume to get its concentration.
Calculate pH (or pOH, then pH). If neither is in excess, the pH is 7.00 for a strong acid and strong base.

Worked example: excess acid

Find the pH after 30.0 mL of 0.150 mol/L HCl is mixed with 20.0 mL of 0.100 mol/L NaOH.

n(H₃O⁺) = 0.150 × 0.0300 = 4.50 × 10⁻³ mol; n(OH⁻) = 0.100 × 0.0200 = 2.00 × 10⁻³ mol.

moles

Excess H₃O⁺ = 4.50 × 10⁻³ − 2.00 × 10⁻³ = 2.50 × 10⁻³ mol.

excess

Total volume = 50.0 mL: [H₃O⁺] = 2.50 × 10⁻³ ÷ 0.0500 = 0.0500 mol/L.

concentration

pH = −log₁₀(0.0500) = 1.30.

Worked example: excess base with a diprotic acid

15.0 mL of 0.100 mol/L H₂SO₄ is mixed with 40.0 mL of 0.100 mol/L KOH. Find the pH, assuming H₂SO₄ releases both protons.

n(H₃O⁺) = 2 × 0.100 × 0.0150 = 3.00 × 10⁻³ mol; n(OH⁻) = 0.100 × 0.0400 = 4.00 × 10⁻³ mol.

moles

Excess OH⁻ = 1.00 × 10⁻³ mol in 55.0 mL: [OH⁻] = 0.0182 mol/L.

excess

pOH = 1.74, so pH = 14.00 − 1.74 = 12.26.

InteractiveMix and dilute

Try this

The sim starts with 25 mL of 0.1 M HCl and 20 mL of 0.1 M NaOH. What is the pH, and how many moles of which ion are in excess?
Slide beaker B's volume to 25 mL, then to 30 mL. Record the pH each time and explain the big jump.
Dilution: set beaker A to 10 mL of 0.1 M HCl and beaker B to 90 mL of water. What is the pH? Now try 0.01 M HCl with the same volumes.
Mix 20 mL of 0.05 M H₂SO₄ with 20 mL of 0.1 M NaOH. Why is this exactly neutral even though the concentrations are different?
Set A to 10 mL of 0.05 M Ba(OH)₂ and B to 90 mL of water. Check the sim's pH with your own calculation.

Investigation: measuring the pH of a range of acids and bases

Aim: to measure the pH of laboratory and household acids and bases and compare methods.

Method: Calibrate a pH meter or probe with pH 4.00, 7.00 and 10.00 buffers. Place about 20 mL of each solution in a clean, labelled beaker: 0.1 mol/L HCl, 0.1 mol/L CH₃COOH, 0.1 mol/L NaOH, 0.1 mol/L NH₃, and household samples such as vinegar, lemon juice, soft drink, milk, bicarb solution, soapy water and diluted cloudy ammonia. Measure the pH of each with the meter, rinsing the probe with distilled water and blotting it dry between samples. Then test a small sample of each with universal indicator solution or paper and record the colour and estimated pH.

Typical results: 0.1 mol/L HCl about 1.0 and 0.1 mol/L NaOH about 13.0; 0.1 mol/L CH₃COOH about 2.9 and NH₃ about 11.1 (the weak acid and base are closer to 7 at the same concentration, which you'll explain in 6G); vinegar 2.4–3.0, lemon juice about 2.3, cola about 2.5, milk about 6.7, bicarb about 8.3, cloudy ammonia about 11.

Accuracy and reliability: the meter is more accurate and precise (to ±0.01) than universal indicator (about ±1). Recalibrate if readings drift, take three readings of each solution and average them, and keep all solutions at the same temperature. Coloured samples such as cola are hard to judge with indicator, which is a reason to prefer a meter.

Safety: wear safety glasses; 0.1 mol/L NaOH and ammonia are irritants. Never taste household samples in the lab.

Test yourself

What is the pH of 0.025 mol/L KOH at 25 °C?
Solution X has pH 3 and solution Y has pH 6. [H₃O⁺] in X is
At 40 °C, $K_w$ = 2.9 × 10⁻¹⁴. A sample of pure water at 40 °C
50.0 mL of 0.10 mol/L HCl is mixed with 50.0 mL of 0.30 mol/L NaOH. The pH of the mixture is
A 0.10 mol/L HCl solution is diluted by a factor of 100. The new pH is

Label the regions of the pH scale.

Order these solutions from lowest to highest pH.

At 25 °C, $K_w$ = . The self-ionisation of water is , so $K_w$ as temperature rises. pH + pOH = at 25 °C. A solution is neutral when [H₃O⁺] equals .

5 marks

A student mixes 20.0 mL of 0.120 mol/L H₂SO₄ with 30.0 mL of 0.150 mol/L NaOH and then dilutes the mixture to 250.0 mL with distilled water. Calculate the pH of the final solution. Assume H₂SO₄ is fully ionised.

Your answer