Which assumptions are made when calculating the pH of a weak acid?

[H⁺] = [A⁻] and [HA] at equilibrium ≈ initial [HA] (negligible dissociation).

Define the ionic product of water.

Kw = [H⁺][OH⁻], 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 25 °C.

How do you find the pH of a strong base?

[H⁺] = Kw/[OH⁻] then pH = −log₁₀[H⁺] (or pH = pKw − pOH).

What is the pH of 0.0150 mol dm⁻³ Ba(OH)₂ at 25 °C?

[OH⁻] = 0.0300, so pH = 12.48.

Why is water at 50 °C (pH 6.63) still neutral?

Because [H⁺] = [OH⁻]; Kw rises with temperature so the neutral pH falls.

Why does a weak acid's pH rise by less than 1 on tenfold dilution?

More of it dissociates as it is diluted, so [H⁺] falls by only about √10.

Why is ammonium chloride solution acidic?

NH₄⁺ is the conjugate acid of a weak base and donates protons to water.

How is Ka found from a mass and a pH?

Find [HA] from mass ÷ M ÷ volume, [H⁺] = 10⁻ᵖᴴ, then Ka = [H⁺]²/[HA].

Weak acids, Ka, Kw and strong basesEdexcel International A Level Chemistry: Flashcards

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Write the expression for Ka of a weak acid HA.

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Write the expression for Ka of a weak acid HA.: Ka = [H⁺][A⁻] / [HA], in mol dm⁻³.
Define pKa.: pKa = −log₁₀Ka.
Which assumptions are made when calculating the pH of a weak acid?: [H⁺] = [A⁻] and [HA] at equilibrium ≈ initial [HA] (negligible dissociation).
Give the formula for [H⁺] of a weak acid.: [H⁺] = √(Ka × [HA]).
Define the ionic product of water.: Kw = [H⁺][OH⁻], 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 25 °C.
What is pKw at 25 °C?: 14.00.
How do you find the pH of a strong base?: [H⁺] = Kw/[OH⁻] then pH = −log₁₀[H⁺] (or pH = pKw − pOH).
What is the pH of 0.0150 mol dm⁻³ Ba(OH)₂ at 25 °C?: [OH⁻] = 0.0300, so pH = 12.48.
Why is water at 50 °C (pH 6.63) still neutral?: Because [H⁺] = [OH⁻]; Kw rises with temperature so the neutral pH falls.
How does pH change when a strong acid is diluted tenfold?: It rises by 1 unit.
Why does a weak acid's pH rise by less than 1 on tenfold dilution?: More of it dissociates as it is diluted, so [H⁺] falls by only about √10.
Why is ammonium chloride solution acidic?: NH₄⁺ is the conjugate acid of a weak base and donates protons to water.
How is Ka found from a mass and a pH?: Find [HA] from mass ÷ M ÷ volume, [H⁺] = 10⁻ᵖᴴ, then Ka = [H⁺]²/[HA].

Exam questions on Weak acids, Ka, Kw and strong bases

Benzoic acid, C₆H₅COOH, is a weak monoprotic acid used as a food preservative. At 298 K its acid dissociation constant, Ka, is 6.3 × 10⁻⁵ mol dm⁻³.
Calculate the pH of a 0.0100 mol dm⁻³ solution of benzoic acid, stating the assumptions you make.2 marks
A technician works with aqueous solutions at 25 °C, at which the ionic product of water, Kw, is 1.00 × 10⁻¹⁴ mol² dm⁻⁶. At 50 °C the value of pKw is 13.26.
Calculate the pH of 0.0150 mol dm⁻³ barium hydroxide solution, Ba(OH)₂, at 25 °C.2 marks
In Core Practical 11, a student dissolves 0.610 g of benzoic acid, C₆H₅COOH (M = 122.0 g mol⁻¹), in water and makes the solution up to 100 cm³ in a volumetric flask. The pH of the solution at 25 °C is measured as 2.75.
Calculate the value of Ka for benzoic acid from these data, stating the units.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).