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Acid-base Titration Curve Calculations
(Strong Monoprotic Acid-Strong Base)

Key Concepts

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Concepts

Adding Aqueous Strong Monoprotic Acid to Aqueous Strong Base at 25oC

For the neutralisation reaction
H+(aq) + OH-(aq) → H2O

in which acid is added from a burette to a conical (erlenmeyer) flask containing base:

  • Initially, only base, OH-, is present in the conical flask. OH- is in excess.
    pOH = -log10[OH-]     and     pH = 14 - pOH
    moles(OH-) = concentration(OH-) x volume of solution (litres)
    n(OH-) = c(OH-) x V(solution) in L
  • Before the equivalence point, adding acid to the base will :
    (a) increase the volume of the solution in the conical flask:
    total volume 3 = initial volume of base + volume of acid added

    (b) consume some of the OH- since H+ + OH- → H2O
    n(OH-(in excess)] = n(OH-(initial)) - n(OH-(reacted with H+))

    Before the equivalence point:
    (i) concentration of OH- = [OH-(in excess)] = n(OH-(in excess)) ÷ total volume of solution (litres)
    (ii) pOH = -log10[OH-(in excess)]
    (iii) pH = 14 - pOH

 
← acid
 
← base
 

Adding Aqueous Strong Base to Aqueous Strong Monoprotic Acid at 25oC

For the neutralisation reaction H+(aq) + OH-(aq) → H2O
in which base is added from a burette to a conical (erlenmeyer) flask containing acid:

  • Initially, only acid, H+, is present in the conical flask. H+ is in excess.
    pH = -log10[H+]
    moles(H+) = concentration(H+) x volume of solution in L
    n(H+) = c(H+) x V(solution) in L
  • Before the equivalence point, adding base to the acid will :
    (a) increase the volume of the solution in the conical flask:
    total volume = initial volume of acid + volume of base added

    (b) consume some of the H+ since H+ + OH- → H2O
    n(H+(in excess)] = n(H+(initial)) - n(H+(reacted with OH-))

    Before the equivalence point:
    (i) concentration of H+ = [H+(in excess)] = n(H+(in excess)) ÷ total volume of solution in L
    (ii) pH = -log10[H+(in excess)]

 
← base
 
← acid
 

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Calculations

Below are the general steps you can use to the determine the pH of the resultant solution at any point during an strong acid-strong base titration:

  1. Extract the data from the question.
  2. Write the balanced chemical equation for the neutralisation reaction.

  3. Calculate the moles of acid present before reaction.
  4. Calculate the moles of base present before reaction.
  5. Use the stoichiometric (mole) ratio to decide which reactant, acid or base, is in excess after the reaction occurs.
  6. Calculate the excess moles of this reactant.
  7. Calculate the total volume of the solution.
  8. Calculate the concentration of the reactant that is in excess.
  9. Calculate the pH of the solution.

Sample Calculation

Question: 6.28 mL of 0.25 mol L-1 HCl(aq) has been added to 20.00 mL of 0.14 mol L-1 NaOH(aq).
Determine the pH of this solution.

  1. Extract the data from the question.
    V(HCl(aq)) = volume of HCl(aq) = 6.28 mL = 6.28/1000 = 0.00628 L
    [HCl(aq)] = c(HCl(aq)) = concentration of HCl(aq) = 0.25 mol L-1
    V(NaOH(aq)) = volume of NaOH(aq) = 20.00 mL = 20.00/1000 = 0.02000 L
    [NaOH(aq)] = c(NaOH(aq)) = concentration of NaOH(aq) = 0.14 mol L-1
  2. Write the balanced chemical equation for the neutralisation reaction:
    HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
  3. Calculate the moles of acid present before reaction.
    moles(HCl) = concentration(HCl) x volume(HCl) in litres
    n(HCl) = 0.25 x 0.00628 = 1.57 x 10-3 mol
  4. Calculate the moles of base present before reaction.
    moles(NaOH) = concentration(NaOH) x volume(NaOH) in litres
    n(NaOH) = 0.14 x 0.02000 = 2.80 x 10-3
  5. Use the stoichiometric (mole) ratio to decide which reactant, acid or base, is in excess after reaction occurs.
    stoichiometric (mole) ratio HCl:NaOH is 1:1
    1 mol HCl reacts with 1 mol NaOH
    1.57 x 10-3 mol HCl reacts with 1.57 x 10-3 mol NaOH
    1.57 x 10-3 mol NaOH is less than the 2.80 x 10-3 mol NaOH that are present.
    NaOH is in excess. (that is, HCl is the limiting reagent)
  6. Calculate the excess moles of this reactant.
    n(NaOH(in excess)) = n(NaOH(initial)) - n(NaOH(reacted with HCl))
    n(NaOH(in excess)) = 2.80 x 10-3 - 1.57 x 10-3 = 1.23 x 10-3 mol
  7. Calculate the total volume of the solution.
    total volume of solution = volume of NaOH(aq) + volume of HCl(aq) added
    total volume of solution = 0.0200 + 0.00628 = 0.02628 L
  8. Calculate the concentration of the reactant that is in excess.
    [NaOH(in excess)] = n(NaOH(in excess)) ÷ total volume of solution in litres
    [NaOH(in excess)] = 1.23 x 10-3 ÷ 0.02628 = 0.0468 mol L-1
  9. Calculate the pH of the solution.
    NaOH is a strong base, it dissociates completely in water:
    NaOH(aq) → Na+(aq) + OH-(aq)

    Therefore, [OH-(aq)(in excess)] = [NaOH(aq)(in excess)] = 0.0468 mol L-1
    pOH = -log10[OH-(aq)(in excess)] = -log10[0.0468] = 1.33
    At 25oC in aqueous solution: pH = 14.00 - pOH
    pH = 14.00 - 1.33 = 12.67

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Example: Calculating a Titration Curve

In an experiment, 14.00 mL of 0.10 mol L-1 HCl(aq) is added 1.00 mL at a time from a burette to a conical flask containing 10.00 mL 0.10 mol L-1 NaOH(aq) solution at 25oC.

Calculate the resulting pH of the solution in the conical (erlenmeyer) flask after each 1.00 mL addition of HCl(aq) and draw the resulting titration curve.

 
← HCl(aq)    
 
← NaOH(aq)
 

Step 1: Write the balanced chemical equation for the reaction

general word equation acid + base salt + water
word equation for reaction hydrochloric acid + sodium hydroxide sodium chloride + water
balanced chemical equation HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Since HCl(aq) is a strong monoprotic acid it fully dissociates in water:

HCl(aq) → H+(aq) + Cl-(aq)

SInce NaOH(aq) is a strong base it fully dissociates in water:
NaOH(aq) → Na+(aq) + OH-(aq)

So:
H+(aq) + OH-(aq) → H2O(l)

Step 2: Calculate the pH of the NaOH(aq) before any HCl is added.

Step 3: Calculate the pH of the solution after 1.00 mL 0.10 mol L-1 HCl has been added.

Step 4: Continue these calculations, adding 1.00 mL HCl(aq) to the new solution, until a volume of 11.00 mL of the 0.10 mol L-1 HCl is added.

At this point the NaOH is no longer in excess, rather it is now the HCl that is in excess.

Step 5: Calculate the pH of the solution after 11.00 mL HCl has been added.

Step 6: Continue these calculations, adding 1.00 mL of the HCl(aq) to the new soluton, until all the HCl has been added.
The results of the calculations you should have performed are shown in the table below:

volume HCl added in L moles (n)HCl added moles (n)NaOH present Total volume of solution [OH-] = n(NaOH) ÷ total volume pOH = -log10[OH-] pH = 14 - pOH
0 0 1 x 10-3 10 x 10-3 0.10 1 13
1 x 10-3 1 x 10-4 9 x 10-4 11 x 10-3 0.082 1.09 12.91
2 x 10-3 2 x 10-4 8 x 10-4 12 x 10-3 0.067 1.18 12.82
3 x 10-3 3 x 10-4 7 x 10-4 13 x 10-3 0.054 1.27 12.73
4 x 10-3 4 x 10-4 6 x 10-4 14 x 10-3 0.043 1.37 12.63
5 x 10-3 5 x 10-4 5 x 10-4 15 x 10-3 0.033 1.48 12.52
6 x 10-3 6 x 10-4 4 x 10-4 16 x 10-3 0.025 1.60 12.40
7 x 10-3 7 x 10-4 3 x 10-4 17 x 10-3 0.018 1.75 12.25
8 x 10-3 8 x 10-4 2 x 10-4 18 x 10-3 0.011 1.95 12.05
9 x 10-3 9 x 10-4 1 x 10-4 19 x 10-3 0.0053 2.28 11.72
10 x 10-3 1 x 10-3 0 20 x 10-3 0 undefined undefined
volume HCl added in L moles (n)HCl added moles (n)HCl unreacted Total volume of solution [H+] = n(HCl) unreacted ÷ total volume pH = -log10[H+]  
11 x 10-3 1.1 x 10-3 1 x 10-4 21 x 10-3 4.76 x 10-3 2.32
12 x 10-3 1.2 x 10-3 2 x 10-4 22 x 10-3 9.09 x 10-3 2.04
13 x 10-3 1.3 x 10-3 3 x 10-4 23 x 10-3 0.013 1.88
14 x 10-3 1.4 x 10-3 4 x 10-4 24 x 10-3 0.017 1.78

Step 7. Plot a graph of pH vs Volume of HCl(aq)

Plotting the points using the table above will result in a curve as shown below:

pH of solution 0.10 M NaOH - 0.10 M HCl Titration Curve

volume of HCl added (mL)
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

The equivalence point for the neutralisation reaction shown above has been marked on the curve.

At the equivalence point, the moles of H+ added will exactly equal the moles of OH- in the conical flask:

n(H+) = n(OH-)

At the equivalence point neither the HCl nor the NaOH is in excess.
At the equivalence point neither the HCl nor the NaOH is the limiting reagent.

For the addition of a strong monoprotic acid, like HCl(aq), to a strong base, like NaOH(aq), the pH at the equivalence point will be 7.

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1. 1 equivalent of an acid is the quantity of that acid which will donate 1 mole of H+.
1 equivalent of a base is the quantity which supplies 1 mole of OH-.
At the equivalence point, 1 equivalent of acid neutralises 1 equivalent of base.

2. If the temperature of the aqueous solution is not 25oC you will need to use the appropriate value for Kw.
If the temperature of the solution has not been given in the question, assume 25oC.

3. We will assume additivity of volumes.