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How do we describe acid and base strength and calculate the pH of solutions?

Apply the Bronsted-Lowry model, distinguish strong and weak acids, and calculate pH, pOH and Kw relationships.

The Bronsted-Lowry model, conjugate acid-base pairs, strong versus weak acids and bases, the self-ionisation of water, and pH and pOH calculations.

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What this dot point is asking

You must use the Bronsted-Lowry model, identify conjugate pairs, distinguish strength from concentration, and carry out pH calculations.

The Bronsted-Lowry model

A Bronsted-Lowry acid donates a proton (H+\text{H}^+); a base accepts one. When an acid donates a proton it forms its conjugate base, and when a base accepts one it forms its conjugate acid. The two species differ by a single proton.

For example, in HCl+H2O→H3O++Cl−\text{HCl} + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+ + \text{Cl}^-, HCl is the acid and Cl−\text{Cl}^- is its conjugate base; water is the base and H3O+\text{H}_3\text{O}^+ is its conjugate acid. Substances like water that can act as either acid or base are amphiprotic.

Strong versus weak

Strength is about the degree of ionisation, not concentration.

  • A strong acid (such as HCl, HNO3\text{HNO}_3 or H2SO4\text{H}_2\text{SO}_4) ionises essentially completely, so we use a single arrow.
  • A weak acid (such as ethanoic acid, CH3COOH\text{CH}_3\text{COOH}) ionises only partly and sits at equilibrium, so we use the reversible arrow ⇌\rightleftharpoons.

Concentration is how many moles of acid are dissolved per litre. A solution can be dilute but strong, or concentrated but weak. Keep the two ideas separate.

Self-ionisation of water

Water ionises slightly: 2H2O⇌H3O++OH−2\text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-.

In any aqueous solution at 25 degrees Celsius this product holds, so if you know one ion concentration you can find the other.

pH and pOH

pH=−log⁡10[H+]pOH=−log⁡10[OH−]pH+pOH=14\text{pH} = -\log_{10}[\text{H}^+] \qquad \text{pOH} = -\log_{10}[\text{OH}^-] \qquad \text{pH} + \text{pOH} = 14

A neutral solution at 25 degrees has pH=7\text{pH} = 7; acidic solutions are below 7 and basic above 7.

In the exam, always check whether the acid or base is strong or weak before choosing your method, and state the temperature assumption when you use Kw=1.0×10−14K_w = 1.0 \times 10^{-14}.

Exam-style practice questions

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

2021 TASC3 marksStatement found online: the pH scale extends from 0 to 14. Hydrochloric acid solution, HCl(aq) (Mr = 36.46), with a concentration of 320 g L-1, is available at a hardware shop. By determining the pH of this solution, show that the statement is incorrect.
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Find the molar concentration: c = (320 g L-1) / (36.46 g mol-1) = 8.78 mol L-1.

HCl is a strong monoprotic acid, so it fully ionises: [H+] = 8.78 mol L-1.

pH = -log10[H+] = -log10(8.78) = -0.94.

Because the pH is negative (about -0.9, below 0), the claim that the pH scale only runs from 0 to 14 is incorrect: concentrated strong acids can have pH below 0. (3 marks: concentration, [H+], negative pH.)

2021 TASC3 marksWhen 1.08 x 10-2 mol of sodium is placed in 500 mL of water, the reaction 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g) occurs. Calculate the pH of the resulting solution at 25 degrees C, assuming negligible loss of water.
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From the 2:2 ratio, n(NaOH) = n(Na) = 1.08 x 10-2 mol.

Concentration of OH-: NaOH is a strong base, so [OH-] = 1.08 x 10-2 / 0.500 = 2.16 x 10-2 mol L-1.

pOH = -log10(2.16 x 10-2) = 1.67. At 25 degrees C, pH = 14.00 - pOH = 14.00 - 1.67 = 12.33.

(Alternatively, [H+] = Kw / [OH-] = 1.0 x 10-14 / 2.16 x 10-2 = 4.63 x 10-13, giving pH = 12.33.) (3 marks: moles of NaOH, [OH-], pH.)

2021 TASC2 marksFor the equilibrium NH3(g) + H2O(l) reversible NH4+(aq) + OH-(aq), which of the reactants, NH3(g) or H2O(l), is a Bronsted-Lowry acid? Justify your answer.
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A Bronsted-Lowry acid is a proton (H+) donor.

In the forward reaction water gives a proton to ammonia: H2O -> OH- + H+, and the NH3 accepts that proton to become NH4+. So water is the Bronsted-Lowry acid (proton donor) and ammonia is the Bronsted-Lowry base (proton acceptor).

This can be seen from the conjugate pairs: H2O/OH- and NH3/NH4+. (2 marks: identify water as the acid and justify by proton donation.)

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