Weak acid pH, percent ionisation, and the buffer that follows
SCH4U Grade 12 Chemistry · Chemical Systems and Equilibrium
A food-preservation lab works with propanoic acid, Ka = 1.34 × 10⁻⁵ at 25 °C. A 0.150 mol/L solution is prepared, and a second flask is made into a buffer holding 0.150 mol/L propanoic acid against 0.100 mol/L sodium propanoate. Find the acid's pKa, the pH of the plain 0.150 mol/L solution, the equilibrium hydrogen ion concentration, the percent of the acid that has ionised, and the pH of the buffer.
Put the acid on the logarithmic scale first. pKa 4.87 is the number the buffer step will need, and it is also the pH at which this acid is exactly half ionised.
Carried onward at full precision, not this rounded figure.
The acid alone. This solver uses the x-is-small approximation, [H⁺] = √(Ka·C), which is honest here because 0.150 mol/L is more than ten thousand times Ka.
Carried onward at full precision, not this rounded figure.
Undo the logarithm to recover the actual hydrogen ion concentration, 1.42 × 10⁻³ mol/L. A pH is a reading; a concentration is what the next step can divide.
Carried onward at full precision, not this rounded figure.
Under 1% of the molecules have given up their proton — and that number is the licence for the approximation used two steps ago. Below about 5%, the shortcut holds.
Carried onward at full precision, not this rounded figure.
Now add the conjugate base. The pKa from step 1 carries in, and the pH jumps almost two units — the propanoate suppresses the ionisation the whole chain has been tracking.
Carried onward at full precision, not this rounded figure.
Open the Henderson–Hasselbalch Equation (Weak Acid Buffer) solver →
Why this order
Steps 2 and 5 look like the same question and are not, which is the point of running them in one chain. A solution of a weak acid alone has only one source of both HA and A⁻ — the dissociation itself — so its pH comes from √(Ka·C) and depends on concentration. A buffer has both species poured in from bottles, so its pH comes from their ratio and barely depends on concentration at all. Dilute the buffer tenfold and its pH scarcely moves; dilute the plain acid tenfold and the pH rises half a unit. Reaching for Henderson–Hasselbalch on a solution with no added conjugate base is the commonest way to get step 2 wrong.
Both equations here carry assumptions worth naming out loud. The weak-acid solver is the x-is-small approximation: it drops the dissociated x from the denominator of Ka = x²/(C − x), which is fine while the acid is under about 5% ionised — step 4 exists precisely to check that, and 0.95% passes comfortably. Push a stronger weak acid to low concentration and the approximation overshoots and the quadratic is mandatory. Henderson–Hasselbalch makes the parallel assumption that the equilibrium concentrations equal the ones you weighed out, which fails for very dilute buffers and for ratios far from one. And the percent ionisation moves the opposite way from intuition: it rises on dilution, because [H⁺] follows a square root while C falls linearly.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.