Use Kw and the relationship pH = -log10[H3O+] to calculate the pH of strong acid and strong base solutions, and to relate [H3O+] and [OH-] in any aqueous solution

What this dot point is asking

QCAA wants you to use the ionic product of water (Kw) and the definition pH = -log10[H3O+] to relate hydrogen ion and hydroxide ion concentrations in any aqueous solution at 25 degrees C, and to perform pH calculations on strong acid and strong base solutions including dilution and mixing. This is a calculation-heavy dot point that appears in every IA1 in some form and in EA Paper 1 multiple choice.

The answer

Water is amphiprotic and undergoes self-ionisation in equilibrium. The product of [H3O+] and [OH-] is constant at a given temperature (Kw). pH is a logarithmic measure of [H3O+]. Together these define every aqueous acid-base calculation.

The self-ionisation of water

Two water molecules can transfer a proton:

This equilibrium has the equilibrium constant (with water excluded as solvent):

At 25 degrees C, Kw = 1.0 x 10^-14 mol^2 L^-2. In pure water, [H3O+] = [OH-] = sqrt(Kw) = 1.0 x 10^-7 mol/L.

The relationship holds for any aqueous solution at 25 degrees C: add HCl and [H3O+] rises, but [OH-] falls so that the product remains 1.0 x 10^-14.

The pH definition

Taking the log of Kw at 25 degrees C:

Useful corollaries:

• pH 7 means [H3O+] = 1.0 x 10^-7 = [OH-] (neutral at 25 degrees C).
• pH < 7 means [H3O+] > [OH-] (acidic).
• pH > 7 means [H3O+] < [OH-] (basic).
• Each pH unit corresponds to a factor of 10 in [H3O+]. pH 2 has 100x the [H3O+] of pH 4.

Strong acid pH calculations

For monoprotic strong acids (HCl, HBr, HI, HNO3, HClO4), [H3O+] = [acid] for any concentration above about 10^-6 mol/L.

Worked example. 0.025 mol/L HCl at 25 degrees C:

For diprotic H2SO4 (assume fully ionised in both steps in dilute solution): [H3O+] = 2 x [acid]. 0.025 mol/L H2SO4 gives pH = -log(0.050) = 1.30.

Practical limit: very dilute strong acids (below about 10^-6 mol/L) cannot be calculated by [H3O+] = [acid] alone, because the water's own contribution becomes significant. QCAA does not typically test this regime; sticking to [acid] > 10^-5 keeps you safe.

Strong base pH calculations

For sodium hydroxide and other group 1 hydroxides, [OH-] = [base]. Calculate pOH then convert to pH.

Worked example. 0.025 mol/L NaOH at 25 degrees C:

For Ca(OH)2 (effectively fully ionised but releases 2 OH- per formula unit): [OH-] = 2 x [base]. Note solubility is the real-world limiter, not ionisation.

Mixed strong acid + strong base solutions

Three steps:

1. Calculate moles of H3O+ and OH- from the original solutions.
2. Identify which is in excess; the smaller amount is consumed completely.
3. Divide the excess moles by total volume to get the new concentration; calculate pH.

If the moles are equal, the solution is neutral (pH 7 at 25 degrees C).

Worked example. 25.0 mL of 0.10 mol/L HCl mixed with 35.0 mL of 0.10 mol/L NaOH:

• Moles H3O+ = 0.0025 mol.
• Moles OH- = 0.0035 mol.
• Excess OH- = 0.0010 mol.
• Total volume = 60.0 mL = 0.060 L.
• [OH-] = 0.0010 / 0.060 = 0.0167 mol/L.
• pOH = 1.78; pH = 12.22.

Dilution calculations

Diluting a strong acid by a factor of f decreases [H3O+] by f, so pH rises by log10(f).

Worked example. Diluting 0.10 mol/L HCl by a factor of 100: [H3O+] = 0.0010 mol/L; pH 3.00 (up from pH 1.00 originally).

Practical limit again: dilution past about 10^-6 mol/L of acid cannot push pH above 7. The asymptotic pH for infinite dilution of any acid in pure water is 7 (or rather pKw / 2 at the temperature of interest).

Temperature dependence of Kw

Self-ionisation is endothermic. As T rises:

• Kw increases.
• Both [H3O+] and [OH-] rise in pure water.
• Pure water's pH falls below 7 (5.86 at body temperature, around 6.5 at 60 degrees C).

Crucially, "neutral" means [H3O+] = [OH-], not "pH = 7". At body temperature pure water has pH 6.86 and is still neutral; blood at pH 7.4 is therefore slightly basic relative to neutrality at that temperature (though the bicarbonate buffer addressed in the buffer dot point keeps that balance very stable).

QCAA stimulus may present Kw at a non-25-degrees-C temperature and ask you to calculate the neutral pH. The method is the same: [H3O+] = sqrt(Kw), then take the log.

Significant figures and pH

A pH value has digits after the decimal that correspond to the significant figures of the [H3O+] value. pH 1.6 (one digit after decimal) corresponds to [H3O+] known to one significant figure. pH 1.60 corresponds to [H3O+] known to two significant figures.

QCAA typically expects pH values reported to two decimal places when starting from a two-significant-figure concentration. Watch for this in IA1 marking.

Worked IA1 stimulus

Stimulus: a student adds 10.0 mL of 0.500 mol/L NaOH to 40.0 mL of 0.200 mol/L HCl. Determine the pH of the final mixture.

Method:

1. Moles NaOH = 0.500 x 0.0100 = 5.00 x 10^-3 mol.
2. Moles HCl = 0.200 x 0.0400 = 8.00 x 10^-3 mol.
3. H+ in excess by 8.00 - 5.00 = 3.00 x 10^-3 mol.
4. Total volume = 50.0 mL = 0.0500 L.
5. [H3O+] = 3.00 x 10^-3 / 0.0500 = 0.0600 mol/L.
6. pH = -log10(0.0600) = 1.22.

This is exactly the calculation chain examiners are looking for: moles, excess, dilution into the combined volume, then the log.

Examples in context

Example 1. Coral reef calcification chemistry on Heron Island. Heron Island Research Station measures seawater pH near $8.05$ and $[\text{H}^+] = 10^{-8.05} = 8.9 \times 10^{-9} \, \text{mol L}^{-1}$. From $K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}$ at $25^{\circ}\text{C}$, $[\text{OH}^-] = 1.0 \times 10^{-14}/8.9 \times 10^{-9} = 1.1 \times 10^{-6} \, \text{mol L}^{-1}$, a hundred-fold excess over $\text{H}^+$. As $\text{CO}_2$ absorbs into the ocean, $[\text{H}^+]$ rises and $[\text{OH}^-]$ falls correspondingly so the product stays at $K_w$. Researchers use this water equilibrium framework to translate atmospheric $\text{pCO}_2$ scenarios into reef pH and carbonate-ion saturation maps.

Example 2. Polyprotic acid in QCAA IA2 titration scenario. A diprotic acid like sulfurous $\text{H}_2 \text{SO}_3$ from Gladstone refinery flue-gas scrubbing has two dissociations: $\text{p}K_{a1} = 1.81$ and $\text{p}K_{a2} = 7.20$. Titrating $25.0 \, \text{mL}$ of $0.100 \, \text{mol L}^{-1}$ $\text{H}_2 \text{SO}_3$ against $0.100 \, \text{mol L}^{-1}$ $\text{NaOH}$ shows two endpoints at $25.0$ and $50.0 \, \text{mL}$ NaOH added. Between them the system is buffered around $\text{pH} = 7.20$. At all points $K_w$ holds and the curve can be modelled by combining $K_{a1}$, $K_{a2}$ and water equilibrium expressions.

Try this

Q1. Write the expression for the autoionisation constant of water and its value at $25^{\circ}\text{C}$. [2 marks]

• Cue. $\text{H}_2 \text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$; $K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}$.

Q2. A NaOH solution has $\text{pH} = 12.30$. Calculate $[\text{OH}^-]$ and the original concentration. [3 marks]

• Cue. $[\text{H}^+] = 10^{-12.30} = 5.0 \times 10^{-13}$. $[\text{OH}^-] = K_w/[\text{H}^+] = 2.0 \times 10^{-2} \, \text{mol L}^{-1}$, so $[\text{NaOH}] = 0.020 \, \text{mol L}^{-1}$.

Q3. $K_w = 1.47 \times 10^{-14}$ at $30^{\circ}\text{C}$. (a) Calculate $\text{pH}$ of pure water at $30^{\circ}\text{C}$. (b) State whether it is still neutral. (c) Explain the temperature dependence. [3+2+2 marks]

• Cue. (a) $[\text{H}^+] = \sqrt{K_w} = 1.21 \times 10^{-7}$; $\text{pH} = 6.92$. (b) Still neutral: $[\text{H}^+] = [\text{OH}^-]$. (c) Autoionisation endothermic, Le Chatelier shifts right. ISMG analysis.