Topics

pH, pOH and Kw

Covers AS91392

Water ionises itself

Even pure water contains a few ions, because water molecules transfer protons to one another:

$2\text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-$

The equilibrium constant for this is the ionic product of water, $K_w$ :

$K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1 \times 10^{-14} \text{ at 25 °C}$

Both ions are present in every aqueous solution, however acidic or basic. Their product is always $1 \times 10^{-14}$ at 25 °C.
So if you know one, you can always find the other.

The relationships

All of these are supplied in the resource booklet:

$\text{pH} = -\log[\text{H}_3\text{O}^+] \qquad [\text{H}_3\text{O}^+] = 10^{-\text{pH}}$

$K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1 \times 10^{-14}$

Two more follow directly, and are worth memorising:

$\text{pOH} = -\log[\text{OH}^-] \qquad \text{pH} + \text{pOH} = 14$

Acidic, neutral and basic

	Relationship	pH
acidic	$[\text{H}_3\text{O}^+] > [\text{OH}^-]$	< 7
neutral	$[\text{H}_3\text{O}^+] = [\text{OH}^-]$	= 7
basic	$[\text{H}_3\text{O}^+] < [\text{OH}^-]$	> 7

In pure water both concentrations are $1 \times 10^{-7}$ mol L⁻¹, which gives pH 7.

pH of a strong acid

A strong acid is completely dissociated, so $[\text{H}_3\text{O}^+]$ equals the concentration of the acid.
Take the negative log and you have the pH — one step, no equilibrium.

$\text{pH} = -\log(c)$

pH of a strong base

A strong base is completely dissociated, so $[\text{OH}^-]$ equals the concentration of the base.
But pH is defined from $[\text{H}_3\text{O}^+]$ , so there is an extra step. Either route works:
- Via $K_w$ : find $[\text{H}_3\text{O}^+] = K_w / [\text{OH}^-]$ , then take the negative log.
- Via pOH: find $\text{pOH} = -\log[\text{OH}^-]$ , then $\text{pH} = 14 - \text{pOH}$ .
This extra step is where marks are lost. Taking $-\log$ of a hydroxide concentration gives the pOH, not the pH.

The pH scale is logarithmic

Each whole pH unit is a tenfold change in $[\text{H}_3\text{O}^+]$ .
pH 3 is ten times more acidic than pH 4, and a hundred times more acidic than pH 5.
So diluting a strong acid tenfold raises its pH by exactly 1.

Significant figures for pH

Only the digits after the decimal point in a pH count as significant figures.
A concentration given to 3 significant figures gives a pH quoted to 2 decimal places — for example $[\text{H}_3\text{O}^+] = 1.32 \times 10^{-3}$ gives pH 2.88.
Going the other way, a pH given to 2 decimal places supports a concentration to 2 significant figures.

Tips and Tricks

For a strong base, do not stop at the log. $-\log[\text{OH}^-]$ gives you the pOH. You must then subtract from 14, or go through $K_w$ . This single omission is the most common error in the whole topic.

Quote pH to 2 decimal places. Writing "pH = 2.9" when the data supports 2.88 can cost a mark, and writing "pH = 2.879518..." looks careless.

Remember both ions are always present. A strongly acidic solution still contains hydroxide ions — just very few of them. A question asking for $[\text{OH}^-]$ in an acid is not a trick.

Use $K_w = 1 \times 10^{-14}$ only at 25 °C. The booklet states the temperature, and questions occasionally note it.

Check your answer is on the right side of 7. An acid must give pH < 7 and a base pH > 7. If your strong base comes out at pH 1.4, you have forgotten to subtract from 14.

Worked ExampleThree pH calculations

Calculate, at 25 °C:

(a) the pH of $0.0250$ mol L⁻¹ hydrochloric acid (b) the pH of $0.0400$ mol L⁻¹ sodium hydroxide (c) the concentration of hydroxide ions in a solution of pH $3.40$

Part (a) — a strong acid

Step 1 — Use the fact that it is strong. Hydrochloric acid is a strong acid, so it is completely dissociated:

$\text{HCl} + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+ + \text{Cl}^-$

Every mole of HCl gives one mole of $\text{H}_3\text{O}^+$ , so

$[\text{H}_3\text{O}^+] = 0.0250 \text{ mol L}^{-1}$

Step 2 — Take the negative log.

$\text{pH} = -\log(0.0250) = 1.602$

Part (b) — a strong base

Step 1 — Use the fact that it is strong. Sodium hydroxide is a strong base, completely dissociated:

$\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-$

$[\text{OH}^-] = 0.0400 \text{ mol L}^{-1}$

Step 2 — Find the pOH.

$\text{pOH} = -\log(0.0400) = 1.398$

Step 3 — Convert to pH. This is the step that must not be skipped.

$\text{pH} = 14 - 1.398 = 12.602$

Part (c) — working backwards from a pH

Step 1 — Find $[\text{H}_3\text{O}^+]$ from the pH.

$[\text{H}_3\text{O}^+] = 10^{-\text{pH}} = 10^{-3.40} = 3.98 \times 10^{-4} \text{ mol L}^{-1}$

Step 2 — Use $K_w$ to get the hydroxide concentration.

$[\text{OH}^-] = \frac{K_w}{[\text{H}_3\text{O}^+]} = \frac{1 \times 10^{-14}}{3.98 \times 10^{-4}}$