pH & buffers: practice problems with step-by-step solutions
Free pH & buffers practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: strong acids and bases, weak acids, buffers.
Free practice problemEasy
Find the pH pH of a 0.05 mol/L solution of NaOH, a strong base, at 25 °C.
Self-check What is the pH of a 0.1 mol/L NaOH solution at 25 °C?
0.1
13
1
7
Answer: 13. [OH⁻] = 0.1 mol/L, so pOH = 1 and pH = 14 − 1 = 13. Giving 1 stops at the pOH: a base must have a pH above 7.
3Weak acids and buffers
A weak acid, such as acetic acid or HF, splits up only a little. It forms an equilibrium:
HA ⇌ H⁺ + A⁻ Ka = [H⁺][A⁻] / [HA]
Ka is the acid constant: the larger it is, the stronger the acid. Often you see pKa = −log Ka.
If x mol/L of an acid with concentration c splits up, then [H⁺] = [A⁻] = x and [HA] = c − x. So Ka = x² / (c − x).
When x is small compared with c, you may use c − x ≈ c:
[H⁺] ≈ √(Ka · c)
If x turns out to be more than about 5% of c, solve the quadratic x² + Kax − Kac = 0 instead.
A buffer is a mixture of a weak acid HA and its conjugate base A⁻, for example from its sodium salt. It keeps the pH almost constant when a little acid or base is added. Its pH follows the Henderson–Hasselbalch equation:
pH = pKa + log([A⁻] / [HA])
Adding a strong base turns some HA into A⁻: HA goes down and A⁻ goes up by the moles of base added.
Self-check A buffer contains equal concentrations of a weak acid and its conjugate base. The pKa is 4.75. What is the pH?
4.75
5.75
7
9.25
Answer: 4.75. With [A⁻] = [HA] the ratio is 1, and log 1 = 0, so pH = pKa = 4.75. A buffer is not automatically neutral.
4The method step by step
Decide what you have: a strong acid, a strong base, a weak acid, or a buffer (a weak acid together with its salt).
Strong acid: [H⁺] = c. Strong base: [OH⁻] = c × number of OH⁻, then pOH and pH = 14 − pOH.
Weak acid: [H⁺] ≈ √(Ka · c). Check that x is less than 5% of c; otherwise solve the quadratic.
Buffer: find pKa, then pH = pKa + log([A⁻]/[HA]). If acid or base is added, first update the moles of HA and A⁻.
Take pH = −log[H⁺] and check the answer: acids below 7, bases above 7, a buffer near its pKa.
5A worked example
Follow every step. Try to predict the next one before you open it.
Worked example
Find the pH of a 0.2 mol/L solution of benzoic acid (Ka = 6.3 × 10⁻⁵).
1
Recognise the acid
Ka is small, so benzoic acid is a weak acid. Only a small part of it splits into H⁺ and A⁻.
0.00355 / 0.2 = 1.8%, which is less than 5%. The approximation is fine. Solving the quadratic exactly gives 0.00352 mol/L, almost the same.
5
Find the pH
pH = −log(0.00352) = 2.45
The pH is below 7, as it must be for an acid, but higher than for 0.2 mol/L of a strong acid (pH 0.70).
Answer
pH ≈ 2.45
6Common mistakes
Giving the pOH when the pH is asked. For a base, always finish with pH = 14 − pOH.
Forgetting that Ba(OH)₂ and Ca(OH)₂ give two OH⁻ per formula unit.
Treating a weak acid as strong, so that [H⁺] = c.
Forgetting the minus sign in pH = −log[H⁺].
Turning the buffer ratio upside down: it is [A⁻]/[HA], base over acid.
Using the √(Ka·c) shortcut when x is more than 5% of c.
Self-check What is the pH of a 0.02 mol/L solution of Ba(OH)₂, a strong base, at 25 °C?
1.40
1.70
12.30
12.60
Answer: 12.60. Each Ba(OH)₂ gives two OH⁻, so [OH⁻] = 0.04 mol/L, pOH = 1.40 and pH = 12.60. 12.30 forgets the factor 2, and 1.40 is the pOH.
7How exam questions look
Easy: strong acids and bases. The pH of a solution of HCl, NaOH or Ba(OH)₂.
Medium: weak acids. The pH of a weak acid such as HF or benzoic acid from its Ka and concentration.
Hard: buffers. The pH from Henderson–Hasselbalch, or the new pH after a strong base is added.
Example of a hard question: a 1.00 L buffer with 0.25 mol HF and 0.10 mol F⁻ (Ka = 0.00068) gets 0.05 mol NaOH. HF becomes 0.20 mol and F⁻ 0.15 mol, so pH = 3.17 + log(0.15/0.20) = 3.04.
Find the pH pH of a 0.5 mol/L solution of hydrochloric acid (HCl), a strong acid.
c = 0.5 mol/L
Find pH
Show hints
A strong acid ionises completely, so [H₃O⁺] equals the acid concentration.
pH = −log₁₀[H₃O⁺].
[H₃O⁺] = 0.5 mol/L.
Show the answer and the step-by-step solution
pH: pH = −log₁₀[H₃O⁺] and pOH = −log₁₀[OH⁻], with pH + pOH = 14 at 25 °C. Strong acids and bases ionise completely. A weak acid only partly: Ka = x²/(c − x), where x = [H₃O⁺]. A buffer of a weak acid and its conjugate base follows Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]).
pH from concentration
pH = −log(0.5) = 0.301
Answer
pH = 0.301
Example 2Medium
Find the pH pH of a 0.1 mol/L solution of hypochlorous acid (Ka = 3 × 10⁻⁸).
c = 0.1 mol/LKa = 3 × 10⁻⁸
Find pH
Show hints
Set up Ka = x²/(c − x), where x = [H₃O⁺].
Because Ka is small, x is much less than c, so x ≈ √(Ka·c) is a good approximation.
√(3 × 10⁻⁸·0.1) = 5.477 × 10⁻⁵.
Show the answer and the step-by-step solution
pH: pH = −log₁₀[H₃O⁺] and pOH = −log₁₀[OH⁻], with pH + pOH = 14 at 25 °C. Strong acids and bases ionise completely. A weak acid only partly: Ka = x²/(c − x), where x = [H₃O⁺]. A buffer of a weak acid and its conjugate base follows Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]).
Equilibrium
HA ⇌ H⁺ + A⁻: x mol/L of acid ionises.
Ka = x²/(0.1 − x) → x = 5.476 × 10⁻⁵ mol/L
pH
pH = −log(5.476 × 10⁻⁵) = 4.262
Answer
pH = 4.262
Example 3Hard
A 1.00 L buffer contains 0.3 mol hypochlorous acid and 0.2 mol of its conjugate base (Ka = 3 × 10⁻⁸). 0.05 mol of NaOH is added. Find the new pH pH.
The added hydroxide reacts with the weak acid: HA + OH⁻ → A⁻ + H₂O.
So HA goes down by the added amount and A⁻ goes up by the same amount.
n(HA) = 0.25, n(A⁻) = 0.25.
Show the answer and the step-by-step solution
pH: pH = −log₁₀[H₃O⁺] and pOH = −log₁₀[OH⁻], with pH + pOH = 14 at 25 °C. Strong acids and bases ionise completely. A weak acid only partly: Ka = x²/(c − x), where x = [H₃O⁺]. A buffer of a weak acid and its conjugate base follows Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]).
Neutralisation
n(HA) = 0.3 − 0.05 = 0.25 mol
n(A⁻) = 0.2 + 0.05 = 0.25 mol
Henderson–Hasselbalch
pH = 7.523 + log(0.25/0.25) = 7.523
Answer
pH = 7.523
Questions and answers
What is pH & buffers?
An acid gives off hydrogen ions, H⁺, in water. A base gives off hydroxide ions, OH⁻, or takes up H⁺. Lemon juice is acidic, soap is basic.
The concentration of H⁺ can range from about 1 mol/L to 10⁻¹⁴ mol/L. Such numbers are awkward, so chemists use the pH:
How do I solve pH & buffers problems step by step?
Decide whether the acid or base is strong (fully dissociated) or weak.
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.