Solution concentration and pH calculations

In this lesson you will learn

Calculate mass fractions and molar concentrations and convert between them using the density.
Solve dilution and mixing problems and explain how errors in preparing a solution change its concentration.
Calculate the pH of strong acids and bases, of diluted solutions and of mixtures, using Kw = 1.0 × 10⁻¹⁴ at 25 °C.

A hospital drip is labelled “0.9% NaCl”, a bottle of concentrated hydrochloric acid “36–38% HCl”, and a lab recipe asks for “0.100 mol/L NaOH”. All three answer the same question — how much solute is there? — in different units. CSCA expects you to switch between them quickly and then to take one more step, to pH, the scale of acidity. Everything rests on the mole from “Amount of substance: the mole”; why weak acids and salt solutions have the pH they have is explained in “Theories of electrolyte solutions”.

Mass fraction

Definition

Solution, solute, solvent (溶液 · 溶质 · 溶剂)

A solution is a homogeneous mixture of a solute (the dissolved substance) and a solvent (usually water). Its mass is the sum of both masses; its volume, in general, is not the sum of the volumes.

w = m(solute)/m(solution) · 100% m(solution) = m(solute) + m(solvent) = ρVw = m(solute)/m(solution) · 100% m(solution) = m(solute) + m(solvent) = ρV

where:

wmass fraction of the solute (质量分数)
ρ, Vdensity (g/mL = g/cm³) and volume (mL) of the solution

For a saturated solution with solubility S (grams of solute per 100 g of water): w = S/(100 + S). NaCl at 20 °C has S ≈ 36 g, so w ≈ 36/136 ≈ 26.5% — the most concentrated NaCl solution possible at that temperature.

Worked examples: mass fraction

1) 25 g of CuSO₄·5H₂O is dissolved in 75 g of water. Find the mass fraction of CuSO₄.
2) How much water must be added to 200 g of a 30% solution to make it 12%?
3) 100 g of a 10% solution is mixed with 300 g of a 30% solution of the same salt. Find the new mass fraction.

Show solution

1) The solute is CuSO₄ only: m = 25 · 160/250 = 16 g; the solution weighs 25 + 75 = 100 g → w = 16% (not 25%).
2) The solute stays 200 · 0.30 = 60 g; the new solution weighs 60/0.12 = 500 g → add 300 g of water.
3) Solute: 10 + 90 = 100 g in 100 + 300 = 400 g → w = 25% (not the average of 10% and 30%).

Molar concentration

Definition

Molar concentration (物质的量浓度)

c(B) — the amount of solute B in one litre of solution (not of solvent), in mol/L (Chinese books write mol·L⁻¹). A 1 mol/L NaCl solution is made by dissolving 58.5 g of NaCl and adding water up to 1 L, not by adding 1 L of water.

c = n/V n = c · Vc = n/V n = c · V

where:

cmolar concentration, mol/L
namount of solute, mol
Vvolume of the solution, L (mL ÷ 1000)

Ions follow the formula: in 0.1 mol/L Fe₂(SO₄)₃, c(Fe³⁺) = 0.2 mol/L and c(SO₄²⁻) = 0.3 mol/L. Pouring out part of a solution changes n and V, but not c.

Worked examples: molar concentration

1) 5.85 g of NaCl is dissolved and made up to 500 mL. Find c(NaCl).
2) Which has the larger c(Cl⁻): 0.2 mol/L CaCl₂ or 0.3 mol/L NaCl?
3) 50 mL is poured out of 1 L of 0.5 mol/L glucose solution. Find c and n in the portion.

Show solution

1) n = 5.85/58.5 = 0.1 mol; V = 0.5 L; c = 0.2 mol/L.
2) CaCl₂ → Ca²⁺ + 2Cl⁻: c(Cl⁻) = 2 · 0.2 = 0.4 mol/L, more than 0.3 mol/L in the NaCl solution. The volumes do not matter for a concentration.
3) c stays 0.5 mol/L; n = 0.5 · 0.050 = 0.025 mol.

c = 1000ρw/M w = cM/(1000ρ)c = 1000ρw/M w = cM/(1000ρ)

where:

ρdensity of the solution, g/cm³
wmass fraction as a decimal (36.5% → 0.365)
Mmolar mass of the solute, g/mol
10001 L = 1000 mL (cm³)

Where it comes from: take 1 L of solution. Its mass is 1000ρ g, the solute in it is 1000ρw g, which is 1000ρw/M mol — in one litre.

Worked examples: mass fraction ⇄ molar concentration

1) Concentrated hydrochloric acid: w = 36.5%, ρ = 1.2 g/cm³ (a rounded value). Find c(HCl).
2) A NaOH solution has c = 5 mol/L and ρ = 1.25 g/cm³. Find its mass fraction.

Show solution

1) c = 1000 · 1.2 · 0.365/36.5 = 12 mol/L.
2) w = 5 · 40/(1000 · 1.25) = 200/1250 = 16%.

CSCA-style item 1: using a reagent label

A bottle of concentrated nitric acid is labelled “HNO₃, w = 63%, ρ = 1.4 g/cm³”. What volume of it is needed to prepare 500 mL of 1.4 mol/L HNO₃? (HNO₃ 63 g/mol)
A) 50 mL B) 70 mL C) 5 mL D) 98 mL

Show solution

c = 1000 · 1.4 · 0.63/63 = 14 mol/L; V = 1.4 · 500/14 = 50 mL → A. Another route: n = 0.7 mol, that is 44.1 g of HNO₃, contained in 44.1/0.63 = 70 g of the acid, which takes up 70/1.4 = 50 mL. B takes these 70 g for 70 mL (it ignores the density), D multiplies by the density instead of dividing, and C is off by a factor of ten when converting litres to millilitres.

Dilution, mixing and preparing a solution

c₁V₁ = c₂V₂ m₁w₁ = m₂w₂

where:

c₁, V₁concentration and volume before dilution
c₂, V₂concentration and volume after dilution
m, wmass of the solution and its mass fraction

On dilution the amount (and the mass) of solute stays the same. V₁ and V₂ may both be in mL — the unit cancels.

c = (c₁V₁ + c₂V₂)/(V₁ + V₂)c = (c₁V₁ + c₂V₂)/(V₁ + V₂)

where:

cconcentration of the mixture

Mixing two solutions of the same solute: add the moles, then divide by the total volume. Volumes are added only when the problem says so (“assume the volumes are additive”); strictly, only masses add.

Worked examples: dilution and mixing

1) What volume of 12 mol/L hydrochloric acid is needed to prepare 250 mL of 0.6 mol/L acid?
2) 20 mL of 5 mol/L NaOH is diluted to 250 mL. Find the new concentration.
3) 100 mL of 0.1 mol/L NaCl is mixed with 300 mL of 0.3 mol/L NaCl (volumes additive). Find c.
4) 50 mL of 2 mol/L HCl is mixed with 150 mL of 0.8 mol/L HCl (volumes additive). Find c.

Show solution

1) V₁ = 0.6 · 250/12 = 12.5 mL (then water up to 250 mL).
2) c₂ = 5 · 20/250 = 0.4 mol/L.
3) n = 0.01 + 0.09 = 0.1 mol in 0.4 L → c = 0.25 mol/L (not the average, 0.2 mol/L).
4) n = 0.1 + 0.12 = 0.22 mol in 0.2 L → c = 1.1 mol/L.

A solution of exact molar concentration is prepared in a volumetric flask (容量瓶), which holds one fixed volume — 100, 250, 500 or 1000 mL — at 20 °C. If you need 450 mL, you prepare 500 mL and calculate for 500 mL. How to handle the flask and the other glassware is described in “Laboratory safety and apparatus”.

1
Calculate
For 500 mL of 0.2 mol/L NaOH: n = 0.2 · 0.5 = 0.1 mol, m = 0.1 · 40 = 4.0 g.
2
Weigh
Weigh 4.0 g of NaOH in a small beaker, not on paper: it absorbs moisture from the air and is corrosive.
3
Dissolve and cool
Dissolve it in a little water, stirring with a glass rod, and let the solution cool to room temperature.
4
Transfer
Pour the solution into the 500 mL flask along the glass rod.
5
Rinse
Rinse the beaker and the rod 2–3 times with water and add the rinsings to the flask.
6
Make up to the mark (定容)
Add water until the level is 1–2 cm below the mark, then use a dropper until the bottom of the meniscus touches the mark, with your eyes level with it.
7
Mix
Stopper the flask, turn it upside down several times, then pour the solution into a labelled bottle.

Error	What changes	c
Beaker and rod not rinsed	n smaller	lower ↓
Some solution spilt during transfer	n smaller	lower ↓
Hot solution made up to the mark, then cooled	V smaller	higher ↑
Looking down at the mark when making up (俯视)	V smaller	higher ↑
Looking up at the mark when making up (仰视)	V larger	lower ↓
Water added past the mark, excess removed with a dropper	n smaller	lower ↓
Level below the mark after inverting, more water added	V larger	lower ↓
Flask still wet with distilled water before use	nothing	no effect

Every error is judged through c = n/V: what did it do to n, and what did it do to V?

CSCA-style item 2: error analysis

A student prepares 250 mL of 0.1 mol/L NaOH solution. Which error makes the concentration too LOW?
A) The solution was transferred and made up to the mark before it had cooled.
B) The volumetric flask contained a little distilled water before use.
C) After the flask was inverted, the level fell slightly below the mark, and more water was added up to it.
D) The student looked down at the mark while making up to volume.

Show solution

C: the extra water increases V (the level dropped only because some solution stayed on the neck and the stopper). A and D both make V too small, so c is too high; B has no effect, because water is added up to the mark anyway.

pH of strong acids, strong bases and their mixtures

Definition

pH (pH值)

pH = −lg c(H⁺), where lg is the common (base-10) logarithm. At 25 °C a solution is neutral at pH 7, acidic below 7 and basic above 7. One pH unit is a tenfold change in c(H⁺).

pH = −lg c(H⁺) pH = a ⇔ c(H⁺) = 10⁻ᵃ mol/LpH = −lg c(H⁺) pH = a ⇔ c(H⁺) = 10⁻ᵃ mol/L

where:

c(H⁺)molar concentration of hydrogen ions, mol/L

Strong acids (HCl, HNO₃, H₂SO₄) and strong bases (NaOH, KOH, Ba(OH)₂) ionise completely, so c(H⁺) or c(OH⁻) comes straight from the formula: 0.05 mol/L H₂SO₄ gives 0.1 mol/L of H⁺.

Kw = c(H⁺) · c(OH⁻) = 1.0 × 10⁻¹⁴ (25 °C) pH + pOH = 14

where:

Kwionic product of water (水的离子积)
pOH−lg c(OH⁻)

c(H⁺) · c(OH⁻) = Kw holds in every aqueous solution, acidic or basic. Kw grows with temperature: near 100 °C it is of the order of 10⁻¹² (problems give Kw = 1 × 10⁻¹²), so neutral water has pH ≈ 6 there — still neutral, because c(H⁺) = c(OH⁻). Where Kw comes from, the ionisation of water, is part of “Theories of electrolyte solutions”.

Worked examples: pH and Kw

1) Find the pH of 0.05 mol/L H₂SO₄.
2) 0.4 g of NaOH is dissolved in water and made up to 1 L. Find the pH.
3) A solution has pH = 3 at 25 °C. Find c(OH⁻).

Show solution

1) Two H⁺ per H₂SO₄: c(H⁺) = 0.1 = 10⁻¹ mol/L → pH = 1.
2) n = 0.4/40 = 0.01 mol, c(OH⁻) = 10⁻² mol/L → pOH = 2 → pH = 12. Or: c(H⁺) = 10⁻¹⁴/10⁻² = 10⁻¹² mol/L.
3) c(H⁺) = 10⁻³ mol/L → c(OH⁻) = 10⁻¹⁴/10⁻³ = 10⁻¹¹ mol/L.

strong acid diluted 10ⁿ times: pH → pH + n strong base: pH → pH − n (never across 7)

where:

nthe power of ten of the dilution (100 times → n = 2)

Dilution divides c(H⁺) of an acid (or c(OH⁻) of a base) by 10ⁿ. Very dilute solutions approach pH 7 from their own side: water itself supplies 10⁻⁷ mol/L of H⁺ and of OH⁻.

acid + base: c(H⁺ or OH⁻ left over) = |n(H⁺) − n(OH⁻)|/(V₁ + V₂)acid + base: c(H⁺ or OH⁻ left over) = |n(H⁺) − n(OH⁻)|/(V₁ + V₂)

where:

V₁ + V₂total volume of the mixture, L

First neutralise H⁺ with OH⁻ in moles, then divide what is left by the total volume and take the pH. Never average pH values.

Worked examples: pH after dilution and mixing

1) HCl with pH = 2 is diluted 100 times. Find the pH.
2) NaOH with pH = 12 is diluted 10 times, and HCl with pH = 6 is diluted 100 times. Find both pH values.
3) 99 mL of 0.1 mol/L HCl is mixed with 101 mL of 0.1 mol/L NaOH. Find the pH.

Show solution

1) c(H⁺): 10⁻² → 10⁻⁴ mol/L → pH = 4.
2) c(OH⁻): 10⁻² → 10⁻³ mol/L → pH = 11. For the HCl, pH = 8 is impossible — an acid stays acidic: the pH becomes slightly below 7.
3) n(H⁺) = 9.9 × 10⁻³ mol, n(OH⁻) = 10.1 × 10⁻³ mol; 2 × 10⁻⁴ mol of OH⁻ is left in 0.2 L → c(OH⁻) = 10⁻³ mol/L → pH = 11. An excess of only about 2% of base already gives pH 11.

Interactive

Loading simulation…

Move the value: each step of 1 is a tenfold change in c(H⁺). pH 1 and pH 3 differ a hundred times, not three times.

CSCA-style item 3: pH of an acid–base mixture

At 25 °C, 40 mL of 0.05 mol/L H₂SO₄ is mixed with 60 mL of 0.05 mol/L NaOH (volumes additive). What is the pH of the mixture?
A) 12 B) 7 C) 1 D) 2

Show solution

n(H⁺) = 0.040 · 0.05 · 2 = 0.004 mol; n(OH⁻) = 0.060 · 0.05 = 0.003 mol; 0.001 mol of H⁺ is left in 0.100 L → c(H⁺) = 0.01 mol/L → pH = 2 → D. A forgets that H₂SO₄ gives two H⁺ (then the base seems to be in excess); B assumes that equal concentrations always neutralise each other; C ignores the base altogether.

How CSCA asks about this

The test is taken in English or Chinese, so learn to recognise the key terms in both:

Term	中文	Pinyin
solution	溶液	róngyè
solute / solvent	溶质 / 溶剂	róngzhì / róngjì
mass fraction	质量分数	zhìliàng fēnshù
molar concentration	物质的量浓度	wùzhì de liàng nóngdù
density	密度	mìdù
solubility	溶解度	róngjiědù
saturated solution	饱和溶液	bǎohé róngyè
dilution	稀释	xīshì
volumetric flask	容量瓶	róngliàngpíng
making up to the mark	定容	dìngróng
ionic product of water	水的离子积	shuǐ de lízǐjī
neutralisation	中和	zhōnghé
acidic / basic / neutral	酸性 / 碱性 / 中性	suānxìng / jiǎnxìng / zhōngxìng

Concentration of an ion from a salt formula: c(ion) = c(salt) × its index; the volume does not matter.
Conversions: m → n → c, and w ⇄ c through c = 1000ρw/M.
Dilution and mixing: c₁V₁ = c₂V₂; add the moles, then divide by the total volume.
Preparing a solution: the flask size (450 mL → a 500 mL flask), the order of the steps, and errors judged through n and V.
pH: strong acids and bases, dilution by 10ⁿ, mixtures with an excess, Kw at another temperature.

Key points

w = m(solute)/m(solution); m(solution) = m(solute) + m(solvent) = ρV.
c = n/V with V of the solution in litres; c(ion) = c(salt) × index; c = 1000ρw/M.
Dilution: c₁V₁ = c₂V₂; mixing: add the moles and divide by the total volume.
Errors in preparing a solution are judged through c = n/V; looking down at the mark → higher, looking up → lower.
pH = −lg c(H⁺); Kw = c(H⁺) · c(OH⁻) = 1.0 × 10⁻¹⁴ at 25 °C; pH + pOH = 14.
A strong acid or base diluted 10ⁿ times changes its pH by n but never crosses 7; in acid–base mixtures neutralise first, then divide the excess by the total volume.

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