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Arrhenius Acid-Base Direct Titration Calculations

Key Concepts

  • An acid-base titration is used to determine the concentration of an acid or a base.

  • An acid-base titration involves the progressive addition of one reactant from a burette (buret), often the acid, to a known volume of the other reactant in a conical (erlenmeyer) flask, often the base.

  • An Arrhenius acid is a species that dissociates in water to produce hydrogen ions or protons, H+(aq).
    An Arrhenius base is a species that dissociates in water to produce hydroxide ions, OH-(aq)

  • An Arrhenius acid reacts with an Arrhenius base to produce a salt and water in a neutralisation (neutralization) reaction:
    General word equation : Arrhenius acid + Arrhenius base salt + water
    H-X + M-OH M-X + H-OH
    General chemical equation : HX + MOH MX + H2O
    Net ionic equation : H+(aq) + OH-(aq)     H2O(l)

  • The equivalence point of the neutralisation titration is the point at which the moles of H+ is equal to the moles of OH-.
        An indicator is used to indicate the equivalence point during a titration by changing colour1.

  • The titration experiment is usually conducted several times carefully and the volume of solution used from the burette (buret) recorded (known as a titre).
    The volume of solution used in calculations is then the average of all these titres.

  • In order to calculate the concentration of an acid, c(acid), we need to know accurately:
          - the concentration of the base used, c(base), usually in mol L-1
          - the volume of the base used, V(base), usually in mL which we convert to L
          - the volume of the acid used to neutralise the base, V(acid), usually in mL which we convert to L

  • In order to calculate the concentration of a base, c(base), we need to know accurately:
          - the concentration of the acid used, c(acid), usually in mol L-1
          - the volume of the acid used, V(acid), usually in mL which we convert to L
          - the volume of the base used to neutralise the acid, V(base), usually in mL which we convert to L

Calculations at the Equivalence Point of an Acid-Base Titration2

  1. Write the balanced chemical equation for the reaction

  2. Extract all the relevant data from the question

  3. 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 (which is the same as dividing the volume in mL by 1000)

  4. Calculate the moles of reactant (n) for which you have both the volume (V) and concentration in mol L-1 (c)
    moles = concentration (mol L-1) x volume (L)
    n = c x V

  5. Use the balanced chemical equation to determine the stoichiometric (mole) ratio of acid to base:
    n(acid):n(base)

  6. Use the stoichiometric (mole) ratio to calculate the moles of the unknown reactant

  7. From the volume (V) of unknown reactant and its previously calculated moles (n), calculate its concentration (c) in mol L-1 :
    concentration (mol L-1) = moles ÷ volume (L)
    c = n ÷ V

Examples of Calculations at the Equivalence Point

1. 25.0 mL of aqueous sodium hydroxide solution of unknown concentration was placed in the conical (erlenmeyer) flask.
The burette (buret) was filled to the 50 mL mark with 0.10 mol L-1 aqueous hydrochloric acid solution.
The sodium hydroxide solution was neutralised when 20.0 mL of hydrochloric acid had been added.
Determine the concentration of the sodium hydroxide solution.

  1. Write the balanced chemical equation for the reaction:
    word equation : acid + basesalt + water
    chemical equation : HCl(aq) + NaOH(aq)NaCl(aq) + H2O(l)

  2. Extract the relevant data from the question:
    volume of HCl(aq) =V(HCl) = 20.0 mL
    concentration of HCl(aq) = c(HCl) = 0.10 mol L-1
    volume of NaOH(aq) = V(NaOH) = 25.0 mL
    concentration of NaOH(aq) = c(NaOH) = ? mol L-1

  3. Check the data for consistency: volumes are given in mL but concentrations are given in moles per L
    Convert volumes in mL to volumes in L
    volume of HCl(aq) =V(HCl) = 20.0 mL = 20.0 x 10-3 L
            concentration of HCl(aq) = c(HCl) = 0.10 mol L-1
    volume of NaOH(aq) = V(NaOH) = 25.0 mL = 25.0 x 10-3 L
            concentration of NaOH(aq) = c(NaOH) = ? mol L-1

  4. Calculate the moles HCl(aq)
          n(HCl) = c(HCl) x V(HCl) = 0.10 x 20.0 x 10-3 = 2.00 x 10-3 moles

  5. From the balanced chemical equation find the stoichiometric (mole) ratio of acid to base:
    n(HCl) : n(NaOH)
            1 : 1

  6. Find moles NaOH
    HCl : NaOH is 1:1
    So n(NaOH) = n(HCl) = 2.00 x 10-3 moles at the equivalence point

  7. Calculate concentration of NaOH:
    c(NaOH) = n(NaOH) ÷ V(NaOH)
        n(NaOH) = 2.00 x 10-3 mol
          V(NaOH) = 25.0 x 10-3 L
    c(NaOH) = 2.00 x 10-3 ÷ 25.0 x 10-3 = 0.080 M or 0.080 mol L-1

2.50.0 mL of 0.20 mol L-1 NaOH was placed in the conical (erlenmeyer) flask.
The burette (buret) was filled with sulfuric acid of unknown concentration.
After determining that the volume of acid required to neutralise the sodium hydroxide was about 22 mL, the experiment was repeated 3 times and the following results obtained:

Experimentvolume of H2SO4(aq) (mL)
120.2
219.8
320.0

Determine the concentration of the sulfuric acid.

  1. Write the balanced chemical equation for the reaction
    word equation :sulfuric acid + sodium hydroxide sodium sulfate + water
    balanced chemical equation : H2SO4(aq) + 2NaOH(aq)Na2SO4(aq) + 2H2O(l)

  2. Extract the relevant data from the question:
    volume of NaOH(aq) = V(NaOH) = 50.0 mL
    concentration of NaOH(aq) = c(NaOH) = 0.20 mol L-1
    concentration of H2SO4(aq) = c(H2SO4(aq)) = ? mol L-1
    volume of H2SO4(aq) = average of all the volumes from all the experiments that were done carefully
    volume of H2SO4(aq) = V(H2SO4(aq)) = (20.2 + 19.8 + 20.0) ÷ 3 = 20.0 mL

  3. Check the data for consistency: volumes are in mL but concentrations are in mol L-1
    Convert volumes in mL to volumes in L
    volume of NaOH(aq) = V(NaOH) = 50.0 mL = 50.0 x 10-3 L
            concentration of NaOH(aq) = c(NaOH) = 0.20 mol L-1
    volume of H2SO4(aq) = V(H2SO4(aq)) = 20.0 mL = 20.0 x 10-3 L
            concentration of H2SO4(aq) = c(H2SO4(aq)) = ? mol L-1

  4. Calculate moles NaOH, n(NaOH) :
    n(NaOH) = c(NaOH) x V(NaOH) = 0.20 x 50.0 x 10-3 = 0.010 mol

  5. From the balanced chemical equation find the stoichiometric (mole) ratio of acid to base
    H2SO4 : NaOH
          1 : 2

  6. Find moles H2SO4, n(H2SO4) :
    H2SO4 : NaOH is 1 : 2
    and H2SO4 : NaOH is ½ : 1
    So, 1 mole of NaOH neutralises ½ mole of H2SO4
    Therefore 0.01 moles NaOH will neutralise ½ x 0.01 moles of H2SO4
    n(H2SO4) = ½ x 0.01 = 5.00 x 10-3 mol

  7. Calculate concentration of H2SO4, c(H2SO4):
    c(H2SO4) = n(H2SO4) ÷ V(H2SO4)
          n(H2SO4) = 5.00 x 10-3 mol
          V(H2SO4) = 20.0 x 10-3 L
    c(H2SO4) = 5.00 x 10-3 ÷ 20.0 x 10-3 = 0.25 M or 0.25 mol L-1

3.25.0 mL of Ba(OH)2(aq) of unknown concentration was placed in the conical (erlenmeyer) flask.
The burette (buret) was filled with 0.062 mol L-1 nitric acid.
A titration was performed quickly to determine that about 45 mL of nitric acid was required to neutralise the barium hydroxide.
3 more titrations were performed very carefully to determine the volume of nitric acid required, the results of these experiments are shown below.

Experimentinitial volume
of HNO3(aq) (mL)
final volume
of HNO3(aq) (mL)
149.99.8
248.58.5
350.010.1

    Determine the concentration of the barium hydroxide solution.

  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 data from the question:
    volume of Ba(OH)2(aq) = V(Ba(OH)2) = 25.0mL
    concentration of Ba(OH)2(aq) = c(Ba(OH)2) = ? mol L-1
    concentration of HNO3(aq) = c(HNO3(aq)) = 0.062 mol L-1
    volume of HNO3(aq) used from the burette (buret) is the average of all the titres done carefully:

    Experimentinitial volume
    of HNO3(aq) (mL)
    final volume
    of HNO3(aq) (mL)
    volume of HNO3(aq)
    used (titre in mL)
    149.9 - 9.8 = 40.1
    248.5 - 8.5 = 40.0
    350.0 - 10.1 = 39.9

    average titre(40.1 + 40.0 + 39.9) ÷ 3 = 40.0

    volume of HNO3(aq) = V( HNO3) = 40.0 mL

  3. Check the data for consistency: volumes are in mL but concentrations are in mol L-1
    Convert mL to L
    volume of Ba(OH)2(aq) = V(Ba(OH)2) = 25.0 mL = 25.0 x 10-3 L
            concentration of Ba(OH)2(aq) = c(Ba(OH)2) = ? mol L-1
            concentration of HNO3(aq) = c(HNO3(aq)) = 0.062 mol L-1
    volume of HNO3(aq) = V( HNO3) = 40.0 mL = 40.0 x 10-3 L

  4. Calculate moles HNO3(aq), n(HNO3) :
    n(HNO3(aq)) = c(HNO3(aq)) x V(HNO3(aq)) = 0.062 x 40.0 x 10-3 = 2.48 x 10-3 mol

  5. From the balanced chemical equation find the stoichiometric (mole) ratio
    Ba(OH)2 : HNO3
              1 : 2

  6. Find moles Ba(OH)2, n(Ba(OH)2) :
    2 mol HNO3 neutralises 1 mole Ba(OH)2
    So 1 mol HNO3 neutralises ½ mol Ba(OH)2
    Therefore 2.48 x 10-3 mol HNO3 will neutralise ½ x 2.48 x 10-3 mol of Ba(OH)2
    n(Ba(OH)2) = ½ x 2.48 x 10-3 = 1.24 x 10-3 mol

  7. Calculate concentration of Ba(OH)2(aq), c(Ba(OH)2) :
    c(Ba(OH)2) = n(Ba(OH)2) ÷ V(Ba(OH)2)
          n(Ba(OH)2) = 1.24 x 10-3 mol
          V(Ba(OH)2) = 25.0 x 10-3 L
    c(Ba(OH)2) = 1.24 x 10-3 ÷ 25.0 x 10-3 = 0.0496 M or 0.0496 mol L-1


What would you like to do now?

1The end point is NOT the same as the equivalence point. The equivalence point refers to the acid-base reaction, but the end point refers to the colour change of the indicator.

2If either the acid or the base is in excess, then the reaction is not at the equivalence point!
For discussions of how to calculate the concentrations of acid and base, and the pH, at different stages during an acid-base titration, refer to:
Calculating Titration Curve: Strong Monoprotic Acid-Strong Base
Calculating Titration Curve: Strong Diprotic Acid-Strong Base

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