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How are acid and base strength, pH and buffering described quantitatively?

Apply the Bronsted-Lowry theory, distinguish strong and weak acids and bases using Ka, Kb and pKa, calculate pH of strong and weak acids and of buffers, explain buffer action, and interpret titration curves and indicator choice

A focused answer to the H2 Chemistry learning outcome on ionic equilibria. Bronsted-Lowry acids and bases, Ka and pKa for weak acids, calculating pH of strong and weak acids, buffer action and the Henderson-Hasselbalch relationship, titration curves and indicator selection.

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What this dot point is asking

SEAB wants you to use the Bronsted-Lowry definition, distinguish strong from weak acids and bases through KaK_a, KbK_b and pKapK_a, calculate the pH of strong acids, weak acids and buffers, explain buffer action with equations, and interpret titration curves and choose indicators. The weak-acid pH and buffer calculations are reliable Paper 2 questions, and titration-curve interpretation appears in both written and practical papers.

The answer

Bronsted-Lowry theory

A Bronsted-Lowry acid is a proton (H+\text{H}^+) donor; a base is a proton acceptor. Each acid has a conjugate base (what remains after losing H+\text{H}^+), and each base has a conjugate acid:

CH3COOH+H2Oβ‡ŒCH3COOβˆ’+H3O+\text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+

Strong versus weak

A strong acid fully dissociates (HClβ†’H++Clβˆ’\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-); a weak acid only partially dissociates, described by an acid dissociation constant:

Ka=[H+][Aβˆ’][HA],pKa=βˆ’log⁑KaK_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}, \qquad pK_a = -\log K_a

A larger KaK_a (smaller pKapK_a) means a stronger acid. Strength (degree of dissociation) is distinct from concentration.

pH calculations

pH=βˆ’log⁑[H+],Kw=[H+][OHβˆ’]=1.0Γ—10βˆ’14Β mol2 dmβˆ’6Β (atΒ 298 K)\text{pH} = -\log[\text{H}^+], \qquad K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}\ \text{mol}^2\,\text{dm}^{-6}\ (\text{at } 298\,\text{K})

  • Strong acid: [H+][\text{H}^+] equals the acid concentration (full dissociation). For 0.100.10 mol dmβˆ’3^{-3} HCl, pH =1.0= 1.0.
  • Weak acid: [H+]=KaΓ—[HA][\text{H}^+] = \sqrt{K_a \times [\text{HA}]} using the weak-acid approximation.
  • Strong base: find [OHβˆ’][\text{OH}^-], then [H+]=Kw/[OHβˆ’][\text{H}^+] = K_w/[\text{OH}^-], then pH.

Buffers

A buffer resists pH change on adding small amounts of acid or base. It is made from a weak acid and its conjugate base (an acidic buffer) or a weak base and its conjugate acid (a basic buffer). Its pH is given by:

pH=pKa+log⁑[salt][acid]\text{pH} = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]}

How it works. The buffer holds large reservoirs of both HA and Aβˆ’\text{A}^-:

  • Added H+\text{H}^+ is removed by the conjugate base: Aβˆ’+H+β†’HA\text{A}^- + \text{H}^+ \rightarrow \text{HA}.
  • Added OHβˆ’\text{OH}^- is removed by the weak acid: HA+OHβˆ’β†’Aβˆ’+H2O\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}.

So the ratio [salt]/[acid][\text{salt}]/[\text{acid}] changes only slightly and pH is nearly constant.

Titration curves and indicators

A titration curve plots pH against volume of titrant. The equivalence point is the steep vertical section. Choose an indicator whose colour-change range falls within that vertical section:

Titration Equivalence pH Suitable indicator
strong acid + strong base 77 most indicators (e.g. bromothymol blue)
weak acid + strong base >7> 7 phenolphthalein
strong acid + weak base <7< 7 methyl orange
weak acid + weak base gradual, no sharp jump no suitable indicator

The half-equivalence point of a weak acid is where [HA]=[Aβˆ’][\text{HA}] = [\text{A}^-], so pH =pKa= pK_a, a quick way to read pKapK_a off a curve.

Examples in context

Example 1. Blood as a buffer. The hydrogencarbonate buffer keeps blood pH near 7.47.4. Added acid from metabolism reacts with HCO3βˆ’\text{HCO}_3^- to form carbonic acid, while added base reacts with the carbonic acid, so the pH stays within the narrow range cells tolerate. SEAB uses this physiological example to test the equations of buffer action in an applied context.

Example 2. Reading pKapK_a from a titration curve. When a weak acid is titrated with strong base, the pH at the half-equivalence point equals the pKapK_a of the acid, because [HA]=[Aβˆ’][\text{HA}] = [\text{A}^-] there. Candidates are routinely asked to read this off a printed curve and identify the acid from a data-booklet-style list of pKapK_a values, linking the graph to the buffer equation.

Try this

Q1. Calculate the pH of 0.05000.0500 mol dmβˆ’3^{-3} nitric acid (a strong acid). [1 mark]

  • Cue. [H+]=0.0500[\text{H}^+] = 0.0500, pH =βˆ’log⁑(0.0500)=1.30= -\log(0.0500) = 1.30.

Q2. Calculate the pH of 0.01000.0100 mol dmβˆ’3^{-3} sodium hydroxide. [2 marks]

  • Cue. [OHβˆ’]=0.0100[\text{OH}^-] = 0.0100; [H+]=10βˆ’14/0.0100=10βˆ’12[\text{H}^+] = 10^{-14}/0.0100 = 10^{-12}; pH =12.0= 12.0.

Q3. Explain why a mixture of ammonia and ammonium chloride acts as a buffer, with equations. [3 marks]

  • Cue. NH3\text{NH}_3 (weak base) and NH4+\text{NH}_4^+ (conjugate acid). Added H+\text{H}^+: NH3+H+β†’NH4+\text{NH}_3 + \text{H}^+ \rightarrow \text{NH}_4^+. Added OHβˆ’\text{OH}^-: NH4++OHβˆ’β†’NH3+H2O\text{NH}_4^+ + \text{OH}^- \rightarrow \text{NH}_3 + \text{H}_2\text{O}.

Exam-style practice questions

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

Specimen (9729)4 marksEthanoic acid has Ka = 1.8 x 10^-5 mol per dm cubed. Calculate the pH of a 0.100 mol per dm cubed solution of ethanoic acid.
Show worked answer β†’

For a weak acid, set up the Ka expression and use the approximation that the acid is only slightly dissociated.

Ka=[H+]2[HA]K_a = \dfrac{[\text{H}^+]^2}{[\text{HA}]} assuming [H+] = [A-] and [HA] is approximately the initial concentration.

[H+]=KaΓ—[HA]=1.8Γ—10βˆ’5Γ—0.100[\text{H}^+] = \sqrt{K_a \times [\text{HA}]} = \sqrt{1.8 \times 10^{-5} \times 0.100}

[H+]=1.8Γ—10βˆ’6=1.34Γ—10βˆ’3[\text{H}^+] = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3} mol per dm cubed.

pH=βˆ’log⁑(1.34Γ—10βˆ’3)=2.87\text{pH} = -\log(1.34 \times 10^{-3}) = 2.87.

Markers reward the Ka expression, the weak-acid approximation, the value of [H+], and the pH to 2 decimal places.

2022 (style)5 marksA buffer is made by mixing 50.0 cm cubed of 0.200 mol per dm cubed ethanoic acid with 50.0 cm cubed of 0.200 mol per dm cubed sodium ethanoate. (a) Calculate its pH. (b) Explain how the buffer resists a change in pH when a small amount of acid is added. (Ka of ethanoic acid = 1.8 x 10^-5.)
Show worked answer β†’

(a) Use pH = pKa + log([salt]/[acid]). After mixing equal volumes, [acid] = [salt] = 0.100 mol per dm cubed.

pKa = -log(1.8 x 10^-5) = 4.74.

pH=4.74+log⁑(0.1000.100)=4.74+0=4.74\text{pH} = 4.74 + \log\left(\dfrac{0.100}{0.100}\right) = 4.74 + 0 = 4.74.

(b) The buffer contains a large reservoir of both the weak acid (CH3COOH) and its conjugate base (CH3COO-).

When acid (H+) is added, it reacts with the ethanoate ions: CH3COO- + H+ -> CH3COOH. The added H+ is removed, so pH changes very little.

Markers reward the equal concentrations after mixing, the correct pKa and pH, and the equation showing the conjugate base mopping up added H+.

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