Conduct investigations and perform calculations to determine the pH and pOH of strong and weak acids and bases, applying the formulae pH equals negative log of hydrogen ion concentration, and pH plus pOH equals 14

What this dot point is asking

NESA wants you to calculate the pH and pOH of strong and weak acid and base solutions, use the auto-ionisation constant of water ($K_w$), and apply the relationships $pH = -\log_{10}[H^+]$ and $pH + pOH = 14$. The chemistry of proton donation and acceptance is set up in the Brønsted-Lowry dot point. Expect a calculation question every year, with weak-acid problems carrying the highest marks.

The answer

The pH scale

A lower pH means a higher $[H^+]$ and a more acidic solution. Each unit of pH corresponds to a tenfold change in $[H^+]$.

The auto-ionisation of water

Water self-ionises:

At 25°C:

Taking $-\log_{10}$ of both sides gives:

For pure water, $[H^+] = [OH^-] = 10^{-7}$, so pH = pOH = 7. Water is neutral.

Strong acids and bases

Strong acids (HCl, $HNO_3$, $H_2SO_4$, $HClO_4$) dissociate completely. $[H^+]$ equals the acid concentration (for monoprotic acids).

Strong bases (NaOH, KOH, $Ca(OH)_2$, $Ba(OH)_2$) dissociate completely. Be careful with diprotic bases: $[OH^-] = 2 \times$ concentration of $Ca(OH)_2$.

Calculation is one step:

Weak acids and bases

Weak acids dissociate only partially. Use the dissociation constant Ka and an ICE table.

For a weak acid $HA \rightleftharpoons H^+ + A^-$:

If the initial concentration is $C_0$ and the extent of dissociation is $x$:

When $K_a$ is small and $C_0$ is reasonable (the 5% rule), approximate $C_0 - x \approx C_0$:

Significant figures for logs

Only the digits after the decimal point in a log are significant. A $[H^+]$ of $1.64 \times 10^{-3}$ (3 sig fig) gives pH = 2.79 (2 decimal places).

Worked example

Calculate the pH of three solutions at 25°C.

(a) 0.025 mol/L HCl (strong acid).

$[H^+] = 0.025$ mol/L.

(b) 0.025 mol/L NaOH (strong base).

$[OH^-] = 0.025$ mol/L.

**(c) 0.025 mol/L $CH_3COOH$** (weak acid, $K_a = 1.8 \times 10^{-5}$).

Check the 5% rule: $6.71 \times 10^{-4} / 0.025 = 2.7\%$. Valid.

Notice: the strong and weak acid had the same concentration but very different pH values (1.60 vs 3.17), because the weak acid is only partially dissociated.

Worked example 2: when the 5% rule fails

Calculate the pH and percent dissociation of 0.0010 mol/L formic acid ($HCOOH$, $K_a = 1.8 \times 10^{-4}$).

Step 1: Try the approximation.

Step 2: Check the 5% rule. $4.24 \times 10^{-4} / 0.0010 = 42\%$, well above 5%. The approximation fails. Solve the quadratic.

Step 3: Set up and solve.

So $[H^+] = 3.44 \times 10^{-4}$ mol/L.

Step 4: pH and percent dissociation.

The approximate $[H^+]$ (4.24 × 10⁻⁴) overestimated the true value (3.44 × 10⁻⁴) by ~23%. As a weak acid becomes more dilute, percent dissociation rises and the 5% rule begins to fail. Always check.

Common traps

Forgetting to double for $Ca(OH)_2$ or $Ba(OH)_2$. Each formula unit produces two $OH^-$.

Mixing $H_2SO_4$ assumptions. The first dissociation is complete; the second has Ka ≈ $1.2 \times 10^{-2}$. HSC usually treats $H_2SO_4$ as fully dissociating both protons for dilute solutions, but a careful answer states the assumption.

Skipping the 5% check for weak acids. If $x$ exceeds 5% of $C_0$, you must solve the quadratic. The approximation fails for concentrations below ~$10^{-3}$ mol/L of typical weak acids.

Wrong sig figs. For pH, the number of decimal places equals the sig figs of the concentration. pH = 2.79 (3 sig fig), not 2.794 or 2.8.

Dilution errors. When diluting a strong acid 10-fold, pH increases by 1 (more dilute, less acidic). When diluting a weak acid 10-fold, pH increases by about 0.5 because dissociation increases as concentration falls.

In one sentence

To find pH, use $pH = -\log_{10}[H^+]$ directly for strong acids (and $pH = 14 - pOH$ for strong bases), and for weak acids and bases use Ka or Kb with an ICE table to find $[H^+]$ or $[OH^-]$, applying the 5% approximation only when justified.