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Acid-base Titration Calculations

Key Concepts

acid-base reactions involve a proton transfer
the acid donates a proton to the base
acid-base reactions are also known as neutralisation reactions
acid + base -----> salt + water
acid A + base B -----> conjugate acid of base B + conjugate base of acid A
(Lowry-Brönsted theory)
H⁺ + OH^- ------> H₂O 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. The easiest way to do this is to multiply the volume in mL x 10^-3
Calculate the moles of reactant (n) for which you have both the volume(V) and concentration(M) : n = M 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(M): M = n ÷ V

Examples

30 mL of 0.10M NaOH neutralised 25.0mL of hydrochloric acid.
Determine the concentration of the acid
1. Write the balanced chemical equation for the reaction
  NaOH(aq) + HCl(aq) -----> NaCl(aq) + H₂O(l)
2. Extract the relevant information from the question:
  NaOH V = 30mL , M = 0.10M HCl V = 25.0mL, M = ?
3. Check the data for consistency
  NaOH V = 30 x 10^-3L , M = 0.10M HCl V = 25.0 x 10^-3L, M = ?
4. Calculate moles NaOH
  n(NaOH) = M x V = 0.10 x 30 x 10^-3 = 3 x 10^-3 moles
5. From the balanced chemical equation find the mole ratio
  NaOH:HCl
  1:1
6. Find moles HCl
  NaOH: HCl is 1:1
  So n(NaOH) = n(HCl) = 3 x 10^-3 moles at the equivalence point
7. Calculate concentration of HCl: M = n ÷ V
  n = 3 x 10^-3 mol, V = 25.0 x 10^-3L
  M(HCl) = 3 x 10^-3 ÷ 25.0 x 10^-3 = 0.12M or 0.12 mol L^-1
50mL of 0.2mol L^-1 NaOH neutralised 20mL of sulfuric acid.
Determine the concentration of the acid
1. Write the balanced chemical equation for the reaction
  2NaOH(aq) + H₂SO₄(aq) -----> Na₂SO₄(aq) + 2H₂O(l)
2. Extract the relevant information from the question:
  NaOH V = 50mL, M = 0.2M H₂SO₄ V = 20mL, M = ?
3. Check the data for consistency
  NaOH V = 50 x 10^-3L, M = 0.2M H₂SO₄ V = 20 x 10^-3L, M = ?
4. Calculate moles NaOH
  n(NaOH) = M x V = 0.2 x 50 x 10^-3 = 0.01 mol
5. From the balanced chemical equation find the mole ratio
  NaOH:H₂SO₄
  2:1
6. Find moles H₂SO₄
  NaOH: H₂SO₄ is 2:1
  So n(H₂SO₄) = ½ x n(NaOH) = ½ x 0.01 = 5 x 10^-3 moles H₂SO₄ at the equivalence point
7. Calculate concentration of H₂SO₄: M = n ÷ V
  n = 5 x 10^-3 mol, V = 20 x 10^-3L
  M(H₂SO₄) = 5 x 10^-3 ÷ 20 x 10^-3 = 0.25M or 0.25 mol L^-1
25.0mL of 0.05M Ba(OH)₂ neutralised 40.0mL of nitric acid.
Determine the concentration of the acid.
1. Write the balanced chemical equation for the reaction
  Ba(OH)₂(aq) + 2HNO₃(aq) -----> Ba(NO₃)₂(aq) + 2H₂O(l)
2. Extract the relevant information from the question:
  Ba(OH)₂ V = 25.0mL, M = 0.05M HNO₃ V = 40.0mL, M = ?
3. Check the data for consistency
  Ba(OH)₂ V = 25.0 x 10^-3L, M = 0.05M HNO₃ V = 40.0 x 10^-3L, M = ?
4. Calculate moles Ba(OH)₂
  n = M x V = 0.05 x 25.0 x 10^-3 = 1.25 x 10^-3 mol
5. From the balanced chemical equation find the mole ratio
  Ba(OH)₂ : HNO₃
  1 : 2
6. Find moles HNO₃
  Ba(OH)₂ : HNO₃ is 1 : 2
  So n(HNO₃) = 2 x n(Ba(OH)₂) = 2 x 1.25 x 10^-3 = 2.5 x 10^-3 mol
7. Calculate concentration of HNO₃: M = n ÷ V
  n = 2.5 x 10^-3moles, V = 40.0 x 10^-3L
  M = 2.5 x 10^-3 ÷ 40.0 x 10^-3 = 0.0625M or 0.0625 mol L^-1