Acid-Base Direct Titration Calculations Tutorial

Key Concepts

• An acid-base titration is used to determine the concentration of an acid or of 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 ¹:

  General word equation :	Arrhenius acid	+	Arrhenius base	→	salt	+	water
  General chemical equation :	H-X	+	M-OH	→	M-X	+	H-OH
  	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 colour².

• 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:

  (i) the concentration of the base used, c(base), usually in mol L^-1
  (ii) the volume of the base used, V(base), usually in mL which we convert to L
  (iii) 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:

  (i) the concentration of the acid used, c(acid), usually in mol L^-1
  (ii) the volume of the acid used, V(acid), usually in mL which we convert to L
  (iii) the volume of the base used to neutralise the acid, V(base), usually in mL which we convert to L

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Calculations at the Equivalence Point of an Acid-Base Titration³

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

Worked Example 1: Determine Concentration of Sodium Hydroxide

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.

Worked Solution:

1. Write the balanced chemical equation for the reaction:

   word equation :	acid	+	base	→	salt	+	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 (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

Worked Example 2: Determine Concentration of Sulfuric Acid

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:

Experiment	volume of H2SO4(aq) (mL)
1	20.2
2	19.8
3	20.0

Determine the concentration of the sulfuric acid.

Worked Solution:

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 H₂SO_4(aq) = c(H₂SO_4(aq)) = ? mol L^-1
   volume of H₂SO_4(aq) = average of all the volumes from all the experiments that were done carefully
   volume of H₂SO_4(aq) = V(H₂SO_4(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 H₂SO_4(aq) = V(H₂SO_4(aq)) = 20.0 mL = 20.0 x 10^-3 L
       concentration of H₂SO_4(aq) = c(H₂SO_4(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

Worked Example 3: Determine Concentration of Barium Hydroxide

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.

Experiment	initial volume of HNO3(aq) (mL)	final volume of HNO3(aq) (mL)
1	49.9	9.8
2	48.5	8.5
3	50.0	10.1

Determine the concentration of the barium hydroxide solution.

Worked Solution:

1. Write the balanced chemical equation for the reaction:

   Ba(OH)_2(aq) + 2HNO_3(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 HNO_3(aq) = c(HNO_3(aq)) = 0.062 mol L^-1
   volume of HNO_3(aq) used from the burette (buret) is the average of all the titres done carefully:
   

   Experiment	initial volume of HNO3(aq) (mL)	final volume of HNO3(aq) (mL)	volume of HNO3(aq) used (titre in mL)
   1	49.9 -	9.8 =	40.1
   2	48.5 -	8.5 =	40.0
   3	50.0 -	10.1 =	39.9
   average titre	(40.1 + 40.0 + 39.9) ÷ 3 =		40.0

   
   volume of HNO_3(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 HNO_3(aq) = c(HNO_3(aq)) = 0.062 mol L^-1
   volume of HNO_3(aq) = V( HNO₃) = 40.0 mL = 40.0 x 10^-3 L

4. Calculate moles HNO_3(aq), n(HNO₃) :

   n(HNO_3(aq)) = c(HNO_3(aq)) x V(HNO_3(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