First assumption when calculating the pH of a weak acid

[H^+] = [A^-] (water's ionisation negligible).

Second assumption when calculating the pH of a weak acid

The equilibrium [HA] is approximately its initial concentration.

Why is pH of a weak acid not −lg c?

Only part of the acid dissociates, so [H^+] is lower than c.

Find Ka from a mass of acid and its pH

[HA] = mass ÷ M ÷ volume; [H^+]=10^{-pH}; Ka=[H^+]^2/[HA].

Why is [HA]=[A^-] at half-neutralisation?

Half of the acid has reacted with alkali to form its salt.

Why is water omitted from the Ka expression?

Its concentration is huge and effectively constant.

Ka, pKa and weak acid pHEdexcel A-Level Chemistry: Flashcards

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Write the expression for $K_a$ of HA

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Write the expression for $K_a$ of HA: $K_a = \frac{[H^+][A^-]}{[HA]}$
Units of $K_a$ for a monobasic acid: mol dm⁻³
Define pKₐ: $pK_a = -\lg K_a$
How do $K_a$ and pKₐ show a stronger acid?: Larger $K_a$ and smaller pKₐ.
First assumption when calculating the pH of a weak acid: $[H^+] = [A^-]$ (water's ionisation negligible).
Second assumption when calculating the pH of a weak acid: The equilibrium [HA] is approximately its initial concentration.
Formula for $[H^+]$ of a weak acid: $[H^+] = \sqrt{K_a \times c}$
pH of 0.100 mol dm⁻³ ethanoic acid ( $K_a$ = 1.74 × 10⁻⁵): 2.88
Why is pH of a weak acid not −lg $c$ ?: Only part of the acid dissociates, so $[H^+]$ is lower than $c$ .
Find $K_a$ from a mass of acid and its pH: $[HA]$ = mass ÷ M ÷ volume; $[H^+]=10^{-pH}$ ; $K_a=[H^+]^2/[HA]$ .
Why is $[HA]=[A^-]$ at half-neutralisation?: Half of the acid has reacted with alkali to form its salt.
What is pKₐ at half-neutralisation?: pKₐ = pH, so $K_a=[H^+]$ .
Why is water omitted from the $K_a$ expression?: Its concentration is huge and effectively constant.

Exam questions on Ka, pKa and weak acid pH

A food technologist analyses a 0.100 mol dm⁻³ solution of ethanoic acid, CH₃COOH, a weak monobasic acid, at 298 K. For ethanoic acid, $K_a$ = 1.74 × 10⁻⁵ mol dm⁻³.
Calculate the pH of the 0.100 mol dm⁻³ ethanoic acid solution. State the assumptions you make.2 marks
A chemist prepares a 0.0500 mol dm⁻³ solution of methanoic acid, HCOOH, for use as a descaling agent. For methanoic acid, $K_a$ = 1.78 × 10⁻⁴ mol dm⁻³ at 298 K. She compares it with propanoic acid, which has pKₐ = 4.87.
The chemist states that the pH of the methanoic acid solution is greater than that of a 0.0500 mol dm⁻³ solution of a strong monobasic acid. Explain why.2 marks
A student dissolves 0.440 g of a pure solid weak monobasic acid, HA (molar mass 88.0 g mol⁻¹), in water and makes the solution up to 100 cm³. A calibrated pH meter reads 3.06 at 298 K.
Calculate $K_a$ for the acid HA. State the assumptions you make.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).