الكيمياء (بالإنجليزية) الصف الثالث الثانوي، الأسبوع 9: Titration calculations: MaVa/na = MbVb/nb

Calculations in analysis: titration and mass analysis

الأسبوع 9 · الباب الثاني (٤): حسابات المعايرة بالعلاقة Ma Va / na = Mb Vb / nb ونسبة المادة في خليط من المعايرة والتحليل الكتلي بطريقتي التطاير والترسيب (ماء التبلور وكتلة أيون من راسب) · 3 دروس

Titration calculations: MaVa/na = MbVb/nb

Chemistry Sec 3 Term 1, Chapter 2: pp. 43–48

GoalCalculate an unknown concentration from a titration with MaVa/na = MbVb/nb, using the coefficients of the balanced equation.

New words

molarity (M) · moles of solute in one litre of solution, mol/Lna and nb · the coefficients of the acid and of the alkali in the balanced equationneutralisation · the reaction of an acid with an alkali to give a salt and water

Explanation

A titration gives a volume, and a volume with a concentration gives moles: n = V × M ÷ 1000, with V in mL. Because the reaction uses the two solutions in a fixed ratio, the balanced equation links them. For an acid a and an alkali b, MaVa ÷ na = MbVb ÷ nb, where na and nb are the coefficients in front of the acid and the alkali. Both volumes must be in the same unit, because the unit cancels. Method: write the balanced equation, read na and nb, put the known numbers in, solve for the unknown. The second way gives the same answer by moles: find the moles of the known solution, use the ratio to get the moles of the other, then divide by its volume in litres. Three slips are common: forgetting a coefficient (sulphuric acid and sodium hydroxide react 1 : 2), mixing mL and L, and using the acid’s coefficient for the alkali.

Step	What to do
1	balanced equation, read na and nb
2	put in M, V of both solutions
3	Ma × Va ÷ na = Mb × Vb ÷ nb
4	solve for the unknown; check units

Worked examples

	HCl	NaOH
M	0.10	?
V (mL)	16.0	20.0
n (coefficient)	1	1

One-to-one. 20.0 mL of NaOH solution needs 16.0 mL of 0.10 M HCl. HCl + NaOH → NaCl + H₂O, so na = nb = 1. Mb = 0.10 × 16.0 ÷ 20.0 = 0.080 mol/L. By moles: n(HCl) = 16.0 × 0.10 ÷ 1000 = 0.00160 mol = n(NaOH), and M = 0.00160 ÷ 0.0200 L = 0.080 mol/L.

	H₂SO₄	NaOH
M	0.050	?
V (mL)	25.0	20.0
n (coefficient)	1	2

One-to-two. 25.0 mL of 0.050 M H₂SO₄ is neutralised by 20.0 mL of NaOH solution. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, so na = 1 and nb = 2. 0.050 × 25.0 ÷ 1 = Mb × 20.0 ÷ 2, so Mb = 0.050 × 25.0 × 2 ÷ 20.0 = 0.125 mol/L.

Method	Result
with coefficients	0.125 mol/L
coefficient forgotten	0.0625 mol/L

Spot the slip. A student forgets the coefficient in the last example and writes Mb = 0.050 × 25.0 ÷ 20.0 = 0.0625 mol/L. That value is half the right answer, because each H₂SO₄ gives two H⁺ and needs two OH⁻. Check: one H₂SO₄ needs two NaOH, so the NaOH solution must be twice as concentrated as the raw ratio suggests.

Class activity

Strong and weak vinegar, in the kitchen with a grown-up. Make cabbage juice as in last week. In a cup of baking-soda solution (one spoon in water), count the drops of vinegar needed to turn the colour back to purple. Then dilute the vinegar with an equal amount of water and count again. The diluted vinegar should need about twice as many drops. This is the idea behind MaVa: when the concentration halves, the volume needed doubles. If you do not do it, say that any numbers are invented. No tasting, wash the cups.

Practice

25.0 mL of NaOH solution needs 18.0 mL of 0.100 M HCl. Find the concentration of the NaOH.

20.0 mL of Ca(OH)₂ solution is neutralised by 15.0 mL of 0.0800 M HCl. The equation is Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O. Find the concentration of Ca(OH)₂.

10.0 mL of 0.150 M H₂SO₄ is titrated with 0.200 M KOH. What volume of KOH is needed (H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O)?

For H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O, what are na and nb? What volume of 0.120 M NaOH neutralises 20.0 mL of 0.060 M H₃PO₄?