How to calculate pH: strong acids, weak acids, buffers and the titration curve, worked in the order a test asks them

To calculate pH, first decide what kind of solution you have. A strong acid gives all its hydrogen ions, so pH = −log[H⁺] straight from the concentration. A weak acid gives only some, so you need its Ka and a small equilibrium calculation. A buffer, a weak acid with its conjugate base, is one logarithm with the Henderson–Hasselbalch equation. A titration curve is those three cases in sequence. Below, each is worked with every line shown, the wrong line named, and a sibling left for you.

The four tools, and which solution calls for which

Four definitions carry the whole unit, all at 25 °C:

• pH = −log[H⁺], and [H⁺] = 10^(−pH)
• pOH = −log[OH⁻]
• pH + pOH = 14.00, because Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴
• pKa = −log Ka, the acid's strength on the same scale as pH

The skill a test marks first is the decision, not the arithmetic. One row of a table per case:

What is in the solution	The tool
A strong acid or a strong base alone	The log, straight from the concentration
A weak acid or a weak base alone	Ka or Kb with an ICE table
A weak acid with its conjugate base (its salt)	Henderson–Hasselbalch
A titration part way through	Decide which point you are at, then one of the three above

Which acids count as strong, in this course: HCl, HBr, HI, HNO₃, HClO₄, and the first proton of H₂SO₄. Every other acid you meet is weak and has a Ka. The strong bases are the group 1 hydroxides and the heavier group 2 ones.

In SCH4U this is two ideas of strand E, Acid–base theories, pH and strong/weak acids and then Buffers and acid–base titration; the SCH4U course map places them.

Case 1: a strong acid or base

A strong acid ionizes completely, so [H⁺] is the acid's concentration and the pH is one logarithm.

The problem. Find the pH of 0.0050 mol/L nitric acid, HNO₃.

• HNO₃ is on the strong list, so [H⁺] = 0.0050 mol/L (complete ionization)
• pH = −log(0.0050) = 2.30 (taking the logarithm)

A strong base. Find the pH of 0.020 mol/L sodium hydroxide.

• NaOH is a strong base, so [OH⁻] = 0.020 mol/L
• pOH = −log(0.020) = 1.70
• pH = 14.00 − 1.70 = 12.30 (the two add to 14.00)

Significant figures for logarithms. The digits after the decimal point in a pH match the significant figures of the concentration. Two figures in 0.0050 give two decimal places, 2.30. The digit before the point only records the power of ten.

The common wrong line. pH = −log(0.020) = 1.70 for the base. That number is the pOH, not the pH. A 0.020 mol/L solution of a strong base cannot have an acidic pH; the sense check is to ask whether the answer is on the right side of 7 before writing it down.

Case 2: a weak acid, with Ka

A weak acid ionizes only partly, so [H⁺] is smaller than the acid's concentration and you have to compute it from the equilibrium HA ⇌ H⁺ + A⁻, whose constant is Ka. The tool is the ICE table of the equilibrium page with Ka in Kc's place.

The problem. Find the pH of 0.100 mol/L acetic acid, given Ka = 1.8 × 10⁻⁵ (the value the standard tables list).

	[HA]	[H⁺]	[A⁻]
I	0.100	0	0
C	−x	+x	+x
E	0.100 − x	x	x

• Ka = x² ÷ (0.100 − x) = 1.8 × 10⁻⁵ (substituting the E row)
• Ka is small, so x will be small next to 0.100; approximate 0.100 − x ≈ 0.100: x² = 1.8 × 10⁻⁵ × 0.100 = 1.8 × 10⁻⁶ (multiplying)
• x = [H⁺] = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ mol/L (taking the root)
• Check the approximation: 1.34 × 10⁻³ ÷ 0.100 = 1.3 %, under 5 %, so it stands
• pH = −log(1.34 × 10⁻³) = 2.87 (taking the logarithm; two decimal places from the two figures in Ka)

Solving the full quadratic instead gives x = 1.33 × 10⁻³ and pH 2.88, which is why the 5 % rule is allowed: the approximation changes the second decimal place by less than the data's own precision.

The common wrong line. Treating the weak acid as strong: pH = −log(0.100) = 1.00. That is nearly two pH units too acidic, an error of about 75 times in [H⁺], and it is a common mark lost on this unit. The second: pH = −log(1.8 × 10⁻⁵) = 4.74. That is the pKa, a property of acetic acid, not the pH of this solution.

Case 3: a buffer, with Henderson–Hasselbalch

A buffer is a weak acid and its conjugate base present together in comparable amounts. It holds its pH because added acid is taken up by the base and added base by the acid; neither addition changes [H⁺] much. Its pH comes from one rearrangement of the Ka expression:

pH = pKa + log([A⁻] ÷ [HA])

For acetic acid, pKa = −log(1.8 × 10⁻⁵) = 4.745, kept to three decimals until the end and reported as 4.74.

The problem. A buffer holds 0.100 mol/L acetic acid and 0.150 mol/L sodium acetate. Find its pH, and then its pH after 0.010 mol of HCl is added to 1.00 L of it.

• pH = 4.745 + log(0.150 ÷ 0.100) = 4.745 + log(1.50) = 4.745 + 0.176 = 4.921 → 4.92 (base over acid, inside the log)
• The added strong acid converts acetate to acetic acid, one for one: 0.010 mol of A⁻ becomes HA
• New [HA] = 0.100 + 0.010 = 0.110 mol/L; new [A⁻] = 0.150 − 0.010 = 0.140 mol/L (both change)
• pH = 4.745 + log(0.140 ÷ 0.110) = 4.745 + log(1.273) = 4.745 + 0.105 = 4.850 → 4.85

The pH fell by 0.07. The same 0.010 mol of HCl in 1.00 L of pure water gives [H⁺] = 0.010 mol/L and pH 2.00, a fall of five units from 7. That comparison is the whole point of a buffer, and it is the sentence to write when a question asks what a buffer does.

The common wrong line. The ratio upside down, acid over base, which gives 4.745 + log(0.100 ÷ 0.150) = 4.745 − 0.176 = 4.57, a pH that falls when base is added. The second: changing only one concentration after the acid is added. The strong acid consumes base and makes acid at the same time, so both numbers move.

Case 4: the titration curve, read as three regions

A titration curve plots pH against the volume of base added. For a weak acid titrated with a strong base, the curve is the three cases above in sequence, and you read it region by region.

Before any base is added. The weak acid alone: Case 2. For 0.100 mol/L acetic acid, pH 2.87.

Between the start and the equivalence point. Some acid has been converted to its conjugate base, so the solution is a buffer: Case 3. At the half-equivalence point, exactly half the acid has been converted, [A⁻] = [HA], the log is zero, and pH = pKa, 4.74 here. That is how a titration measures a pKa: read the pH at half the equivalence volume.

At the equivalence point. All the acid has become its conjugate base, and a conjugate base of a weak acid is a weak base, with Kb = Kw ÷ Ka. The pH is above 7, how far above depends on Kb and the concentration, and the course does not usually ask for the number.

After the equivalence point. Excess strong base, and the pH is set by the leftover hydroxide: Case 1 on the excess.

In words, the curve starts low, rises slowly through the buffer region, jumps steeply near the equivalence point, and flattens at a high pH. The indicator has to change colour inside the steep jump, which for a weak acid against a strong base is above 7, so an indicator that changes colour at about 8 to 10 is the right choice and one that changes at about 4 would turn before the equivalence point.

The common wrong line. "The equivalence point is always pH 7." That is true only for a strong acid against a strong base, where the salt at equivalence is neutral. Here the salt is weakly basic.

pH on one screen

Case	The condition	The line
Strong acid	On the strong list	pH = −log[acid]
Strong base	Group 1 or heavy group 2 hydroxide	pOH = −log[OH⁻]; pH = 14.00 − pOH
Weak acid	Has a Ka	x = √(Ka × C), check x ÷ C < 5 %, pH = −log x
Buffer	Weak acid with its conjugate base	pH = pKa + log(base ÷ acid)
Half-equivalence	[A⁻] = [HA]	pH = pKa

Attempting the problems below before reading their answers is chapter 3 of the free playbook.

Try one cold

Cover the page above. Three short problems.

(a) Find the pH of 0.250 mol/L formic acid, HCOOH, given Ka = 1.8 × 10⁻⁴ (the tables' value).

(b) Find the pH of a buffer that holds 0.200 mol/L formic acid and 0.200 mol/L sodium formate.

(c) Find the pH of 0.0025 mol/L hydrochloric acid.

Check

• (a) x = √(1.8 × 10⁻⁴ × 0.250) = √(4.5 × 10⁻⁵) = 6.71 × 10⁻³ mol/L. Check: 6.71 × 10⁻³ ÷ 0.250 = 2.7 %, under 5 %. pH = −log(6.71 × 10⁻³) = 2.17.
• (b) Equal amounts of acid and base, so the log is zero and pH = pKa = −log(1.8 × 10⁻⁴) = 3.74.
• (c) HCl is strong: pH = −log(0.0025) = 2.60.

If (a) came out as 0.60, you treated formic acid as strong. If (b) came out as 3.74 only after a long calculation, you did it the slow way and got there; the equal-amounts shortcut is worth remembering. If (c) has you reaching for a Ka, look at the strong list again.

Where this sits in the course

These are ideas 19 and 20 of 24 in SCH4U, Acid–base theories, pH and strong/weak acids and Buffers and acid–base titration, the last two ideas of strand E. They follow Solubility equilibria and Ksp and rest on Dynamic equilibrium and the equilibrium constant, whose ICE table and 5 % rule are reused here unchanged, and they lean on strand C's Ionic and covalent bonding; Lewis structures; electronegativity, because bond polarity is part of why one acid gives up its proton more readily than another.

The idea after them, Oxidation numbers and balancing redox reactions, opens strand F, the electrochemistry strand, where balancing redox reactions in acid uses H⁺ and water as the balancing tools, the acid–base idea in a new place.

For US readers, the AP Chemistry exam covers the same four cases. Every idea of the three science courses has a page like this one.

How to practise it

One problem of each case a night, shuffled, so that the first decision, which case is this, is the skill you practise. Rohrer and Taylor (2007) found that students who practised problem types mixed together did worse during practice and much better on a test a week later than students who practised each type in its own block, and the reason is that the mixed set makes you choose. Sweller and Cooper (1985) found that a worked example followed by a matching problem teaches a procedure faster than problems alone; each case above is one example and the sibling at the end.

Keep the error log by case. Most strong-acid errors are the pOH written as the pH; most weak-acid errors are the acid treated as strong; most buffer errors are the ratio upside down. Mixed practice has the schedule, and how to study chemistry treats this strand as a decision followed by a short procedure.