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How do we calculate the pH of acids, bases and buffers, and why do buffers resist change?

The Bronsted-Lowry model and conjugate pairs, pH and the ionic product of water Kw, the acid dissociation constant Ka and pKa for weak acids, buffer action and pH, titration curves, and indicator choice.

An OCR H432 module 5 answer on acids and bases: Bronsted-Lowry conjugate pairs, pH and Kw, Ka and pKa for weak acids, buffer action and pH calculations, titration curves, and choosing an indicator.

Generated by Claude Opus 4.813 min answer

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  1. What this topic is asking
  2. The Bronsted-Lowry model
  3. pH and Kw
  4. Weak acids: Ka and pKa
  5. Buffers
  6. Titration curves and indicators
  7. Examples in context
  8. Try this

What this topic is asking

OCR specification point 5.1.3 wants you to use the Bronsted-Lowry model and identify conjugate acid-base pairs, calculate pH from [H+][\text{H}^+] (and the reverse), use KwK_w for strong bases, use KaK_a and pKa\text{p}K_a for weak acids, calculate the pH of buffers and explain how they work, and sketch and interpret titration curves to choose an indicator. This is one of the highest-yield calculation areas in the course.

The Bronsted-Lowry model

pH and Kw

For a strong monoprotic acid, [H+][\text{H}^+] equals the acid concentration. For a strong base, find [OH][\text{OH}^-] from the concentration, then [H+]=Kw/[OH][\text{H}^+] = K_w/[\text{OH}^-], then take the pH.

Weak acids: Ka and pKa

A larger KaK_a (smaller pKa\text{p}K_a) means a stronger weak acid.

Buffers

Titration curves and indicators

Examples in context

Example 1. Blood pH control. The carbonic acid and hydrogencarbonate buffer holds blood pH near 7.47.4; carbon dioxide from respiration forms carbonic acid, and the hydrogencarbonate ion mops up excess acid, a direct application of buffer chemistry.

Example 2. Choosing the right indicator in a titration. Titrating a weak acid (ethanoic acid) with a strong base (sodium hydroxide) has its equivalence point above pH 7, so phenolphthalein is used; methyl orange would change colour too early.

Try this

Q1. Calculate the pH of 0.0200 mol dm30.0200\ \text{mol dm}^{-3} nitric acid (a strong acid). [1 mark]

  • Cue. pH=log10(0.0200)=1.70\text{pH} = -\log_{10}(0.0200) = 1.70.

Q2. State why a buffer can resist the addition of a small amount of acid. [2 marks]

  • Cue. The conjugate base (A\text{A}^-) reacts with the added H+\text{H}^+ to form the weak acid, removing most of the added protons so the pH barely changes.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20194 marksEthanoic acid is a weak acid with Ka=1.7×105 mol dm3K_a = 1.7 \times 10^{-5}\ \text{mol dm}^{-3}. Calculate the pH of a 0.100 mol dm30.100\ \text{mol dm}^{-3} solution of ethanoic acid.
Show worked answer →

For a weak acid, [H+]=Ka×[HA][\text{H}^+] = \sqrt{K_a \times [\text{HA}]} using the approximations [H+]=[A][\text{H}^+] = [\text{A}^-] and [HA]eq[HA]initial[\text{HA}]_{\text{eq}} \approx [\text{HA}]_{\text{initial}} (1).

[H+]=(1.7×105)(0.100)=1.7×106=1.30×103 mol dm3[\text{H}^+] = \sqrt{(1.7 \times 10^{-5})(0.100)} = \sqrt{1.7 \times 10^{-6}} = 1.30 \times 10^{-3}\ \text{mol dm}^{-3} (1)(1).

pH=log10(1.30×103)=2.89\text{pH} = -\log_{10}(1.30 \times 10^{-3}) = 2.89 (1).

Markers reward the weak-acid expression, the value of [H+][\text{H}^+], and the final pH.

OCR 20214 marksA buffer is made by mixing 0.200 mol0.200\ \text{mol} of ethanoic acid (Ka=1.7×105K_a = 1.7 \times 10^{-5}) with 0.100 mol0.100\ \text{mol} of sodium ethanoate in 1.0 dm31.0\ \text{dm}^3 of solution. (a) Calculate the pH. (b) Explain how the buffer responds to a small addition of alkali.
Show worked answer →

(a) [H+]=Ka×[HA][A]=1.7×105×0.2000.100=3.4×105 mol dm3[\text{H}^+] = K_a \times \dfrac{[\text{HA}]}{[\text{A}^-]} = 1.7 \times 10^{-5} \times \dfrac{0.200}{0.100} = 3.4 \times 10^{-5}\ \text{mol dm}^{-3} (1). pH=log10(3.4×105)=4.47\text{pH} = -\log_{10}(3.4 \times 10^{-5}) = 4.47 (1).

(b) Added OH\text{OH}^- reacts with the weak acid: CH3COOH+OHCH3COO+H2O\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O} (1), removing most of the added hydroxide so the pH barely changes (1).

Markers reward the buffer calculation, the pH, the equation for removing OH\text{OH}^-, and the explanation that the pH is held roughly constant.

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