Skip to main content
ExamExplained

Acids, bases and pH: TCE Chemistry (Tasmania)

Syllabus dot point

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

9 min read

Quick answer

An acid is a proton (H+\text{H}^+) donor and a base is a proton acceptor (the Bronsted-Lowry model). Strength describes how completely an acid or base ionises in water, which is different from concentration. pH measures hydrogen ion concentration as pH=−log⁡10[H+]\text{pH} = -\log_{10}[\text{H}^+], and at 25 ∘C25\ ^\circ\text{C} the ionic product of water gives pH+pOH=14\text{pH} + \text{pOH} = 14.

What this dot point is asking

TASC expects you to apply the Bronsted-Lowry model, identify conjugate pairs, separate strength from concentration, and carry out pH, pOH and KwK_w calculations for strong acids and bases.

The Bronsted-Lowry model

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

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

Conjugate acid-base pair

Two species related by the gain or loss of a single proton. The stronger an acid, the weaker its conjugate base, and vice versa.

Strong versus weak (not the same as concentration)

Strength is about the degree of ionisation:

  • A strong acid (HCl\text{HCl}, HNO3\text{HNO}_3, H2SO4\text{H}_2\text{SO}_4) ionises essentially completely, so we write a single arrow.
  • A weak acid (ethanoic acid CH3COOH\text{CH}_3\text{COOH}) ionises only partly and sits at equilibrium, so we write ⇌\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; this distinction is a perennial exam discriminator.

Self-ionisation of water and Kw

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

Ionic product of water

Kw=[H+][OH−]=1.0×10−14 at 25 ∘CK_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}\ \text{at } 25\ ^\circ\text{C}

In any aqueous solution at 25 ∘C25\ ^\circ\text{C} this product holds, so knowing one ion concentration gives the other.

Because self-ionisation is endothermic, raising the temperature increases KwK_w, so the pH of neutral water falls below 77 at higher temperatures (it is still neutral because [H+]=[OH−][\text{H}^+] = [\text{OH}^-]).

pH and pOH

pH=−log⁡10[H+]pOH=−log⁡10[OH−]pH+pOH=14 at 25 ∘C\text{pH} = -\log_{10}[\text{H}^+] \qquad \text{pOH} = -\log_{10}[\text{OH}^-] \qquad \text{pH} + \text{pOH} = 14 \text{ at } 25\ ^\circ\text{C}

A neutral solution at 25 ∘C25\ ^\circ\text{C} has pH=7\text{pH} = 7; acidic solutions are below 77 and basic above 77. To reverse a log, [H+]=10−pH[\text{H}^+] = 10^{-\text{pH}}.

pH of a strong base

Find the pH of 0.0050 mol L−10.0050\ \text{mol L}^{-1} sodium hydroxide solution at 25 ∘C25\ ^\circ\text{C}.

Step 1: Identify the ion concentration from complete ionisation

Sodium hydroxide is a strong base, so it dissociates completely in water. Every formula unit produces one hydroxide ion, meaning the hydroxide concentration equals the stated concentration with no further calculation needed:

[OH−]=0.0050 mol L−1.[\text{OH}^-] = 0.0050\ \text{mol L}^{-1}.

Step 2: Calculate pOH using the definition

The pOH is the negative base-ten logarithm of the hydroxide concentration, just as pH is the negative log of the hydrogen ion concentration. Applying that definition:

pOH=−log⁡10(0.0050)=−log⁡10(5.0×10−3)=2.30.\text{pOH} = -\log_{10}(0.0050) = -\log_{10}(5.0 \times 10^{-3}) = 2.30.

Step 3: Convert pOH to pH using the water relationship

At 25 ∘C25\ ^\circ\text{C} the ionic product of water gives pH+pOH=14.00\text{pH} + \text{pOH} = 14.00. Rearranging to find pH:

pH=14.00−pOH=14.00−2.30=11.70.\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.30 = 11.70.

Step 4: Verify using KwK_w directly

As a check, calculate [H+][\text{H}^+] from Kw=[H+][OH−]K_w = [\text{H}^+][\text{OH}^-] and take its negative log:

[H+]=Kw[OH−]=1.0×10−140.0050=2.0×10−12 mol L−1,[\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{0.0050} = 2.0 \times 10^{-12}\ \text{mol L}^{-1},

pH=−log⁡10(2.0×10−12)=11.70.\text{pH} = -\log_{10}(2.0 \times 10^{-12}) = 11.70.

Final answer: pH = 11.70

Polyprotic acids and diprotic bases

Some acids donate more than one proton. Sulfuric acid, H2SO4\text{H}_2\text{SO}_4, is diprotic: the first ionisation is essentially complete (strong), while the second, HSO4−⇌H++SO42−\text{HSO}_4^- \rightleftharpoons \text{H}^+ + \text{SO}_4^{2-}, is only moderately strong. For a dilute strong diprotic acid you can often treat both protons as fully released, so [H+]≈2c[\text{H}^+] \approx 2c. Phosphoric acid, H3PO4\text{H}_3\text{PO}_4, releases three protons in successive, progressively weaker steps. When a question gives a diprotic species, check whether each ionisation is strong or weak before assuming complete release.

The logarithmic nature of pH

Because pH is a base-ten logarithm, a change of one pH unit corresponds to a tenfold change in [H+][\text{H}^+]. A solution of pH 33 is ten times more acidic than pH 44 and a hundred times more acidic than pH 55. This is why diluting a strong acid tenfold raises its pH by exactly one unit (until the concentration approaches that of water itself, where the self-ionisation of water must be included). Keep significant figures in mind: only the digits after the decimal point in a pH value are significant, because the integer part fixes only the power of ten.

Treating a weak acid like a strong one

For a weak acid, [H+][\text{H}^+] is not equal to the acid concentration, because the acid only partly ionises; you cannot simply take the negative log of the formal concentration (use KaK_a instead). A 0.1 mol L−10.1\ \text{mol L}^{-1} weak acid has a higher pH (less acidic) than a 0.1 mol L−10.1\ \text{mol L}^{-1} strong acid. Other common slips: forgetting that pH+pOH=14\text{pH} + \text{pOH} = 14 only at 25 ∘C25\ ^\circ\text{C}, and confusing strength with concentration.

In the exam, always decide first whether the acid or base is strong or weak, then choose the method. State the temperature assumption when you use Kw=1.0×10−14K_w = 1.0 \times 10^{-14}, and show the link between [H+][\text{H}^+], [OH−][\text{OH}^-] and KwK_w when a question gives you one and asks for the other.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice3 marks
A statement found online claims the pH scale extends only from 00 to 1414. Hydrochloric acid, HCl(aq)\text{HCl}(aq) (M=36.46 g mol−1M = 36.46\ \text{g mol}^{-1}), with a concentration of 320 g L−1320\ \text{g L}^{-1}, is sold at a hardware shop. By determining the pH of this solution, show that the statement is incorrect.
Show worked answer →

Convert to molar concentration: c=320/36.46=8.78 mol L−1c = 320 / 36.46 = 8.78\ \text{mol L}^{-1}. (1 mark)

HCl\text{HCl} is a strong monoprotic acid, so it fully ionises: [H+]=8.78 mol L−1[\text{H}^+] = 8.78\ \text{mol L}^{-1}. (1 mark)

pH=−log⁡10[H+]=−log⁡10(8.78)=−0.94.\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(8.78) = -0.94.

Because the pH is below 00 (about −0.9-0.9), the claim that the scale runs only from 00 to 1414 is incorrect: concentrated strong acids can have a negative pH. (1 mark)

Practice3 marks
When 1.08×10−2 mol1.08 \times 10^{-2}\ \text{mol} of sodium is placed in 500 mL500\ \text{mL} of water, the reaction 2Na(s)+2H2O(l)→2NaOH(aq)+H2(g)2\text{Na}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq) + \text{H}_2(g) occurs. Calculate the pH of the resulting solution at 25 ∘C25\ ^\circ\text{C}, assuming negligible change in water volume. (Kw=1.0×10−14K_w = 1.0 \times 10^{-14}.)
Show worked answer →

From the 2:22:2 ratio, n(NaOH)=n(Na)=1.08×10−2 moln(\text{NaOH}) = n(\text{Na}) = 1.08 \times 10^{-2}\ \text{mol}. (1 mark)

NaOH\text{NaOH} is a strong base, so [OH−]=1.08×10−2/0.500=2.16×10−2 mol L−1[\text{OH}^-] = 1.08 \times 10^{-2} / 0.500 = 2.16 \times 10^{-2}\ \text{mol L}^{-1}. (1 mark)

pOH=−log⁡10(2.16×10−2)=1.67\text{pOH} = -\log_{10}(2.16 \times 10^{-2}) = 1.67, so at 25 ∘C25\ ^\circ\text{C}, pH=14.00−1.67=12.33\text{pH} = 14.00 - 1.67 = 12.33. (1 mark)

Practice2 marks
For the equilibrium NH3(g)+H2O(l)⇌NH4+(aq)+OH−(aq)\text{NH}_3(g) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq) + \text{OH}^-(aq), which reactant is the Bronsted-Lowry acid? Justify your answer using conjugate pairs.
Show worked answer →

A Bronsted-Lowry acid is a proton (H+\text{H}^+) donor. In the forward reaction water gives a proton to ammonia: H2O\text{H}_2\text{O} loses H+\text{H}^+ to become OH−\text{OH}^-, while NH3\text{NH}_3 accepts that proton to become NH4+\text{NH}_4^+.

So water is the Bronsted-Lowry acid (proton donor) and ammonia is the base (proton acceptor). The conjugate pairs are H2O/OH−\text{H}_2\text{O}/\text{OH}^- and NH4+/NH3\text{NH}_4^+/\text{NH}_3. (2 marks: identify water as the acid and justify by proton donation.)

Practise this

Sources & how we know this

Based on the official TASC course document. Not yet individually reviewed by a person; always check TASC and your school's materials for current rules.

ExamExplained