pH and Kw calculations

The ionic product of water

• Water itself ionises very slightly:

$$H_2O \rightleftharpoons H^+ + OH^-$$

• The product of the two ion concentrations is a constant, $K_w$:

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

• This holds in every aqueous solution, not just pure water. Knowing one ion concentration therefore always gives you the other.

	Relationship	pH
Acidic	$[H^+] > [OH^-]$	< 7
Neutral	$[H^+] = [OH^-]$	= 7
Basic	$[H^+] < [OH^-]$	> 7

The relationships you need

$pH = -\log[H^+] \qquad [H^+] = 10^{-pH}$ $pOH = -\log[OH^-] \qquad pH + pOH = 14$ $K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$

• These are given on the resource booklet. The marks are for choosing the right one and using it correctly.

Important scope limit

pH calculations in this standard are restricted to strong acids and bases.

• Because a strong acid ionises completely, its $[H^+]$ equals its concentration (for a monoprotic acid), so you can go straight to the pH.
• You will not be asked to calculate the pH of a weak acid from its concentration — that requires $K_a$, which is Level 3.

Strong acids

• Monoprotic (HCl, $HNO_3$) — donates one $H^+$ per molecule, so $[H^+] =$ the acid concentration.
• Diprotic ($H_2SO_4$) — donates two, so $[H^+] = 2 \times$ the acid concentration.

Worked through for $0.010\ mol\ L^{-1}$ HCl:

$[H^+] = 0.010 = 1.0 \times 10^{-2}\ mol\ L^{-1}$ $pH = -\log(1.0 \times 10^{-2}) = 2.00$

Strong bases

A base gives you $[OH^-]$, so there is an extra step. There are two routes — use whichever you find easier:

• Route 1, via $K_w$: find $[H^+] = \dfrac{K_w}{[OH^-]}$, then take $-\log$.
• Route 2, via pOH: find $pOH = -\log[OH^-]$, then $pH = 14 - pOH$.

Worked through for $0.010\ mol\ L^{-1}$ NaOH:

• $$[OH^-] = 1.0 \times 10^{-2}\ mol\ L^{-1}$$
• Route 1: $[H^+] = \dfrac{1.0 \times 10^{-14}}{1.0 \times 10^{-2}} = 1.0 \times 10^{-12}$, so $pH = 12.00$
• Route 2: $pOH = 2.00$, so $pH = 14 - 2.00 = 12.00$ ✓

Significant figures in pH

• pH is a logarithm, so the digits before the decimal point come from the power of ten and are not significant figures.
• The number of decimal places in a pH should match the number of significant figures in the concentration.
  • $[H^+] = 1.0 \times 10^{-2}$ (2 s.f.) → pH = 2.00 (2 d.p.)

The method

• Identify whether you have an acid or a base, and whether it is strong.
• Find $[H^+]$ — directly for an acid, or via $K_w$ / pOH for a base.
• Watch for diprotic acids — double the concentration.
• Apply $pH = -\log[H^+]$.
• Round to the right number of decimal places.