acid A + base B → conjugate acid of base B + conjugate base of acid A
(Lowry-Brönsted theory)

H^{+} + OH^{-} → H_{2}O is the most general neutralisation reaction

Equivalence point is the point at which the moles of H^{+} is equal to the moles of OH^{-} An indicator is used to show the equivalence point during a titration

A titration involves the progressive addition of one reactant from a burette
(usually the acid), to a known volume of the other reactant in a conical flask
(usually the base)

Calculations

Write the balanced chemical equation for the reaction

Extract all the relevant information from the question

Check that data for consistency, for example, concentrations are usually given in M or mol L^{-1} but volumes are often given in mL. You will need to convert the mL to L for consistency. You need to divide the volume in mL by 1000, and the easiest way to do this on your calculator is to multiply the volume in mL x 10^{-3} (enter the volume in mL into your calculator, then click "EXP" then -3, and you've done the coversion to L !)

Calculate the moles of reactant (n) for which you have both the volume(V) and concentration(c) : n = c x V

From the balanced chemical equation find the mole ratio known reactant : unknown reactant

Use the mole ratio to calculate the moles of the unknown reactant

From the volume(V) of unknown reactant and its previously calculated moles(n), calculate its concentration(c): c = n ÷ V

Examples

1. 30 mL of 0.10 mol L^{-1} NaOH neutralised 25.0 mL of hydrochloric acid.
Determine the concentration of the acid

Write the balanced chemical equation for the reaction NaOH(aq) + HCl(aq) → NaCl(aq) + H_{2}O(l)

Extract the relevant information from the question: NaOH V = 30 mL , M = 0.10 mol L^{-1} HCl V = 25.0 mL, c = mol L^{-1} ?

Check the data for consistency NaOH V = 30 x 10^{-3} L , c = 0.10 mol L^{-1} HCl V = 25.0 x 10^{-3} L, c = mol L^{-1} ?

Calculate moles NaOH n(NaOH) = c x V = 0.10 x 30 x 10^{-3} = 3 x 10^{-3} moles

From the balanced chemical equation find the mole ratio NaOH:HCl 1:1

Find moles HCl NaOH: HCl is 1:1 So n(NaOH) = n(HCl) = 3 x 10^{-3} moles at the equivalence point

Calculate concentration of HCl: c = n ÷ V n = 3 x 10^{-3} mol, V = 25.0 x 10^{-3} L c(HCl) = 3 x 10^{-3} ÷ 25.0 x 10^{-3} = 0.12 M or 0.12 mol L^{-1}

2. 50mL of 0.2 mol L^{-1} NaOH neutralised 20 mL of sulfuric acid.
Determine the concentration of the acid.

Write the balanced chemical equation for the reaction 2NaOH(aq) + H_{2}SO_{4}(aq) → Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Extract the relevant information from the question: NaOH V = 50 mL, c = 0.2 mol L^{-1} H_{2}SO_{4} V = 20 mL, c = mol L^{-1}?

Check the data for consistency NaOH V = 50 x 10^{-3} L, c = 0.2 mol L^{-1} H_{2}SO_{4} V = 20 x 10^{-3} L, c = mol L^{-1}?

Calculate moles NaOH n(NaOH) = c x V = 0.2 x 50 x 10^{-3} = 0.01 mol

From the balanced chemical equation find the mole ratio NaOH:H_{2}SO_{4} 2:1

Find moles H_{2}SO_{4} NaOH: H_{2}SO_{4} is 2:1 So n(H_{2}SO_{4}) = ½ x n(NaOH) = ½ x 0.01 = 5 x 10^{-3} moles H_{2}SO_{4} at the equivalence point

Calculate concentration of H_{2}SO_{4}: c = n ÷ V n = 5 x 10^{-3} mol, V = 20 x 10^{-3}L c(H_{2}SO_{4}) = 5 x 10^{-3} ÷ 20 x 10^{-3} = 0.25 M or 0.25 mol L^{-1}

3. 25.0 mL of 0.05 mol L^{-1} Ba(OH)_{2} neutralised 40.0 mL of nitric acid.
Determine the concentration of the acid.

Write the balanced chemical equation for the reaction Ba(OH)_{2}(aq) + 2HNO_{3}(aq) → Ba(NO_{3})_{2}(aq) + 2H_{2}O(l)

Extract the relevant information from the question: Ba(OH)_{2} V = 25.0 mL, M = 0.05 mol L^{-1} HNO_{3} V = 40.0 mL, c = mol L^{-1}?

Check the data for consistency Ba(OH)_{2} V = 25.0 x 10^{-3} L, c = 0.05 mol L^{-1} HNO_{3} V = 40.0 x 10^{-3} L, c = mol L^{-1}?

Calculate moles Ba(OH)_{2} n = c x V = 0.05 x 25.0 x 10^{-3} = 1.25 x 10^{-3} mol

From the balanced chemical equation find the mole ratio Ba(OH)_{2} : HNO_{3} 1 : 2

Find moles HNO_{3} Ba(OH)_{2} : HNO_{3} is 1 : 2 So n(HNO_{3}) = 2 x n(Ba(OH)_{2}) = 2 x 1.25 x 10^{-3} = 2.5 x 10^{-3} mol

Calculate concentration of HNO_{3}: c = n ÷ V n = 2.5 x 10^{-3}moles, V = 40.0 x 10^{-3} L c = 2.5 x 10^{-3} ÷ 40.0 x 10^{-3} = 0.0625 M or 0.0625 mol L^{-1}