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expressing the concentration of solutions (mol L^-1, g L^-1, %m/v, %m/m, %v/v and ppm) including dilution calculations, and the Brønsted-Lowry model of acids and bases including conjugate acid-base pairs, the distinction between strong and weak (and concentrated and dilute) acids and bases, and the calculation of pH from [H+]
A focused VCE Chemistry Unit 2 answer on concentration and acid-base chemistry. Covers concentration units (mol L^-1, g L^-1, %m/v, %m/m, %v/v, ppm) and dilution calculations, the Brønsted-Lowry model with conjugate acid-base pairs, strong vs weak and concentrated vs dilute, and the calculation of pH from [H+].
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What this dot point is asking
VCAA wants you to express the concentration of a solution in any of the standard VCE units, perform dilution calculations, describe the Brønsted-Lowry model of acids and bases including conjugate acid-base pairs, distinguish strong vs weak and concentrated vs dilute, and calculate pH from a hydrogen-ion concentration.
The answer
Concentration units
Pick the unit that matches the question's data:
| Unit | Meaning | Formula |
|---|---|---|
| mol L^-1 (molar, M) | moles of solute per litre of solution | c = n / V |
| g L^-1 | grams of solute per litre of solution | c = mass / V |
| %m/v | grams of solute per 100 mL of solution | %m/v = (mass(g) / V(mL)) x 100 |
| %m/m | grams of solute per 100 g of solution | %m/m = (mass solute / mass solution) x 100 |
| %v/v | mL of solute per 100 mL of solution | %v/v = (V solute / V solution) x 100 |
| ppm | parts per million by mass (mg/kg or, for dilute aqueous, mg/L) | ppm = (mass solute / mass solution) x 10^6 |
Conversions you will need:
- mol L^-1 to g L^-1: multiply by molar mass M.
- ppm to mg/L for dilute aqueous solutions: 1 ppm = 1 mg/L.
- %m/v to g L^-1: 1 %m/v = 10 g L^-1.
Dilution
Adding more solvent decreases concentration but not the moles of solute. The dilution formula follows from n = cV being constant:
c1 V1 = c2 V2
Use it directly when a more concentrated stock is diluted to a less concentrated working solution. Units of V can be anything (mL or L) as long as you use the same units on both sides.
The Brønsted-Lowry model
A Brønsted-Lowry acid is a proton donor; a Brønsted-Lowry base is a proton acceptor. An acid-base reaction is a proton transfer:
HA + B <=> A^- + HB^+
The pair (HA, A^-) is a conjugate acid-base pair: HA is the acid, A^- is its conjugate base. Similarly (HB^+, B) is a pair: HB^+ is the conjugate acid of base B.
Common examples:
| Acid (donates H+) | Conjugate base |
|---|---|
| HCl | Cl^- |
| HNO3 | NO3^- |
| H2SO4 | HSO4^- (still acidic) |
| CH3COOH | CH3COO^- |
| H3O^+ | H2O |
| H2O | OH^- |
| NH4^+ | NH3 |
Water is amphiprotic: it can act as either an acid (donating H+ to give OH^-) or a base (accepting H+ to give H3O^+).
Strong vs weak; concentrated vs dilute
Strong = ionises (or dissociates) completely in water. All HCl molecules become H+ and Cl^-. The position of the ionisation equilibrium lies essentially fully to the right.
Weak = ionises partially. Only a small fraction of CH3COOH molecules give up a proton at any moment; the rest stay molecular.
Concentrated = a high number of moles of solute per litre of solution.
Dilute = a low number of moles of solute per litre.
These are independent axes. A weak acid can be concentrated (concentrated acetic acid, "glacial" acetic acid, is 17 mol L^-1 and weak); a strong acid can be dilute (0.0001 mol L^-1 HCl is dilute and strong).
Common strong acids you should recognise: HCl, HBr, HI, HNO3, H2SO4 (first proton). Common strong bases: NaOH, KOH, Ca(OH)2 (where dissolved), Ba(OH)2.
pH
pH is defined as:
pH = -log10([H+])
where [H+] is in mol L^-1. For pure water at 25 deg C, [H+] = 10^-7 mol L^-1 and pH = 7.
Useful shortcuts:
- A pH of 1 has [H+] = 0.1; pH of 2 has [H+] = 0.01; each pH unit is a factor of 10 in [H+].
- For a strong monoprotic acid, [H+] equals the acid concentration. 0.05 mol L^-1 HCl gives [H+] = 0.05, pH = 1.30.
- For a strong base, calculate [OH-] then use [H+][OH-] = 10^-14 at 25 deg C, then pH = -log[H+]. Or pOH = -log[OH-] and pH = 14 - pOH.
- For a weak acid or base, you cannot assume [H+] equals concentration. Weak acids have higher pH than strong acids of the same concentration; weak bases have lower pH than strong bases of the same concentration.
Worked example
(1) Concentration. Dissolve 2.92 g of NaCl (M = 58.44 g mol^-1) in water and dilute to 250 mL. Find the concentration in mol L^-1 and in ppm.
n = 2.92 / 58.44 = 0.0500 mol.
c = 0.0500 / 0.250 = 0.200 mol L^-1.
In ppm (mg/L for dilute aqueous): 2.92 g = 2920 mg in 0.250 L gives 11,680 mg/L = 11,680 ppm (an unusually high value because the solution is not actually dilute).
(2) Dilution. How would you prepare 100 mL of 0.0500 mol L^-1 NaCl from this stock?
c1 V1 = c2 V2: V1 = (0.0500 x 100) / 0.200 = 25.0 mL. Pipette 25.0 mL of the stock into a 100 mL volumetric flask and fill to the mark.
(3) pH. Find the pH of 100 mL of 0.0500 mol L^-1 NaOH after diluting it to 500 mL.
After dilution: c2 = (0.0500 x 100) / 500 = 0.0100 mol L^-1 NaOH.
NaOH is a strong base: [OH-] = 0.0100 mol L^-1, pOH = 2.00, pH = 14 - 2.00 = 12.00.
Common traps
Confusing strong with concentrated. A strong acid can be dilute. A weak acid can be concentrated. Always say which axis you mean.
Treating weak acids as fully ionised when calculating pH. 0.10 mol L^-1 CH3COOH does not have pH = 1.00. Only a small fraction ionises, so pH is higher (~2.9).
Forgetting volumes are in litres for c = n / V. Or mixing mL and L in c1V1 = c2V2 (the units must match on both sides).
Confusing ppm with %m/m. 1 % = 10,000 ppm. So 0.001% is 10 ppm.
Writing H+ as the only acid species. Strictly, a proton in water exists as H3O^+ (hydronium). VCE accepts either notation; be consistent within a question.
Using log instead of -log. pH = negative log of [H+]. For [H+] = 1 x 10^-3, pH = 3, not -3.
In one sentence
Concentration is the amount of solute per amount of solution in any of mol L^-1, g L^-1, %m/v, %m/m, %v/v and ppm with c1V1 = c2V2 for dilutions; an acid in the Brønsted-Lowry model is a proton donor and the conjugate base is what is left after the proton goes; strong/weak describes the extent of ionisation, concentrated/dilute describes the amount, and pH is the negative base-10 log of [H+].
Past exam questions, worked
Real questions from past VCAA papers on this dot point, with our answer explainer.
2024 VCE3 marksA 250 mL stock solution of HCl has a concentration of 2.00 mol L^-1. Calculate (a) the volume of stock required to make 500 mL of 0.100 mol L^-1 HCl, and (b) the pH of the diluted solution.Show worked answer →
A 3-mark answer needs the dilution formula, the calculated volume, and the pH from -log [H+].
(a) Dilution formula c1V1 = c2V2:
V1 = (c2 V2) / c1 = (0.100 x 500) / 2.00 = 25.0 mL
Pipette 25.0 mL of stock into a 500 mL volumetric flask and fill to the line.
(b) HCl is a strong acid (fully ionised). [H+] = 0.100 mol L^-1.
pH = -log[H+] = -log(0.100) = 1.00
2025 VCE4 marksAmmonia (NH3) is a weak base in water. (a) Write the Brønsted-Lowry equation for ammonia reacting with water and identify both conjugate acid-base pairs. (b) Explain the difference between a weak base and a dilute base. (c) Predict whether a 0.10 mol L^-1 solution of NH3 has the same pH as a 0.10 mol L^-1 solution of NaOH.Show worked answer →
A 4-mark answer needs the equation, the pairs labelled, and a clear strong-vs-weak distinction.
(a) Equilibrium reaction:
NH3(aq) + H2O(l) <=> NH4^+(aq) + OH^-(aq)
Conjugate acid-base pairs
Pair 1: NH3 (base) / NH4^+ (its conjugate acid). NH3 gains a proton to become NH4^+.
Pair 2: H2O (acid) / OH^- (its conjugate base). H2O donates a proton to become OH^-.
(b) Weak vs dilute: a weak base only partially ionises in water (the equilibrium above lies to the left; only a small fraction of NH3 molecules are protonated). A dilute base has a low concentration of base in solution but says nothing about the extent of ionisation. NH3 is weak even at high concentration; NaOH is strong even when dilute.
(c) Same concentration, different pH: NaOH is a strong base, fully ionised, so [OH-] = 0.10 mol L^-1 and pH = 13.0. NH3 is a weak base, only ~1% ionised, so [OH-] is much lower (~10^-3 mol L^-1) and pH is about 11. The NH3 solution has a lower pH than NaOH at the same concentration.
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