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

Key Concepts

A strong acid will neutralise a strong base in a neutralisation reaction:
H⁺_(aq) + OH^-_(aq) → H₂O_(l)
or
H₃O⁺_(aq) + OH^-_(aq) → 2H₂O_(l)

In an acid-base titration, a known volume of one reactant is placed in the conical (erlenmeyer) flask.

The other reactant is placed in the burette.

The reactant from the burette is slowly added to the reactant in the conical flask.

Before the equivalence point⁽¹⁾, the reactant in the conical (erlenmeyer) flask is in excess.
that is, either [OH^-] > [H⁺]
or, [H⁺] > [OH^-]

At the equivalence point, neither reactant is in excess.
that is, in solution [H⁺] = [OH^-]

After the equivalence point, the reactant added from the burette is in excess.
that is, either [OH^-] > [H⁺]
or, [H⁺] > [OH^-]

For a neutralisation reaction in which the acidic reagent is in excess,
pH = -log₁₀[H⁺_{(in excess)}]

For a neutralisation reaction in which the basic reagent is in excess,
pOH = -log₁₀[OH^-_{(in excess)}]

For aqueous solutions at 25°C, if the basic reagent is in excess, then
pH = 14 - pOH ⁽²⁾

You will need to be comfortable with calculating the pH of the resultant solution after mixing strong acids and strong bases together.

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Concepts

Adding Aqueous Strong Monoprotic Acid to Aqueous Strong Base at 25°C

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^-) × volume of solution (litres)

n(OH^-) = c(OH^-) × 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

At the equivalence point, just enough acid has been added so that all the acid has neutralised all the base and neither OH^- nor H⁺ is in excess.
n(H⁺_(added)) = n(OH^-_(initial))

[H⁺_{(in solution)}] = [OH^-_{(in solution)}] = concentration of H⁺ and OH^- as a result of the dissociation of H₂O

For neutral aqueous solutions at 25°C, [H⁺] = [OH^-] = 10^-7 mol L^-1

For neutral aqueous solutions at 25°C, pH = pOH = 7

After the equivalence point, each added volume of acid will:
(a) increase the total volume of the solution:

total volume of solution = initial volume of base in flask + volume of acid that has been added

(b) increase the moles of H⁺(aq) in solution (since all the OH^- from the base has already been neutralised)

n(H⁺_{(in excess)}) = n(H⁺_{(added from burette)}) - n(H⁺_{(reacted with OH^-)})

After the equivalence point:

(i) concentration of H⁺ = [H⁺] = n(H⁺_{(in excess)}) ÷ total volume of solution in litres

(ii) pH = -log₁₀[H⁺_{(in excess)}]

Adding Aqueous Strong Base to Aqueous Strong Monoprotic Acid at 25°C

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⁺) × volume of solution in L

n(H⁺) = c(H⁺) × 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

At the equivalence point, just enough base has been added so that all the base has neutralised all the acid and neither OH^- nor H⁺ is in excess.
n(OH^-_(added)) = n(H⁺_(initial))

[H⁺_{(in solution)}] = [OH^-_{(in solution)}] = concentration of H⁺ and OH^- as a result of the dissociation of H₂O

For neutral aqueous solutions at 25°C, [H⁺] = [OH^-] = 10^-7 mol L^-1

For neutral aqueous solutions at 25°C, pH = pOH = 7

After the equivalence point, each added volume of base will:
(a) increase the total volume of the solution:

total volume of solution = initial volume of acid in flask + volume of base that has been added

(b) increase the moles of OH^-(aq) in solution (since all the H⁺ from the acid has already been neutralised)

n(OH^-_{(in excess)}) = n(OH^-_{(added from burette)}) - n(OH^-_{(reacted with H⁺)})

After the equivalence point:

(i) concentration of OH^- = [OH^-] = n(OH^-_{(in excess)}) ÷ total volume of solution in litres

(ii) pOH = -log₁₀[OH^-_{(in excess)}]

(iii) pH = 14 - pOH

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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:

Extract the data from the question.

Write the balanced chemical equation for the neutralisation reaction.

Calculate the moles of acid present before reaction.

Calculate the moles of base present before reaction.

Use the stoichiometric ratio (mole ratio) to decide which reactant, acid or base, is in excess after the reaction occurs.

Calculate the excess moles of this reactant.

Calculate the total volume of the solution (in litres).

Calculate the concentration of the reactant that is in excess.

Calculate the pH of the resultant solution.

Sample Calculation: pH of solution after mixing strong acid and strong base

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.

Solution:

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

Write the balanced chemical equation for the neutralisation reaction:
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Calculate the moles of acid present before reaction.
moles(HCl) = concentration(HCl) × volume(HCl) in litres

n(HCl) = 0.25 mol ~~L^-1~~ × 0.00628 L = 1.57 × 10^-3 mol

Calculate the moles of base present before reaction.
moles(NaOH) = concentration(NaOH) × volume(NaOH) in litres

n(NaOH) = 0.14 mol ~~L^-1~~ × 0.02000 L = 2.80 × 10^-3 mol

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 × 10^-3 mol HCl reacts with 1.57 × 10^-3 mol NaOH

1.57 × 10^-3 mol NaOH is less than the 2.80 × 10^-3 mol NaOH that are present.

NaOH is in excess. (that is, HCl is the limiting reagent)

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 × 10^-3) − (1.57 × 10^-3) = 1.23 × 10^-3 mol

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

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 × 10^-3 mol) ÷ (0.02628 L) = 0.0468 mol L^-1

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 25°C in aqueous solution: pH = 14.00 - pOH

pH = 14.00 - 1.33 = 12.67

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Worked 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 25°C.

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.

[OH^-(aq)] = [NaOH(aq)] = 0.10 mol L^-1

pOH = -log₁₀[OH^-(aq)] = -log₁₀[0.10] = 1.0

pH = 14 - pOH = 14 - 1.0 = 13

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

NaOH is in excess, HCl is the limiting reagent

Calculate moles of HCl added: moles = concentration (mol L^-1) × volume (L)
n(HCl) = c(HCl) × V(HCl)

c(HCl) = 0.10 mol L^-1

V(HCl) = 1.00 mL = 1.00 × 10^-3 L

n(HCl) = 0.10 mol ~~L^-1~~ × (1.00 × 10^-3)L = 1.00 × 10^-4 mol

Calculate moles NaOH unreacted = initial moles NaOH - moles NaOH reacted
initial moles NaOH = concentration (mol L^-1) × volume (L)

initial moles NaOH = c(NaOH) × V(NaOH)

c(NaOH) = 0.10 mol L^-1

V(NaOH) = 10.00 mL = 10.00 × 10^-3 L

initial moles NaOH = 0.10 mol ~~L^-1~~ × (10.00 × 10^-3)L = 1.00 × 10^-3 mol

moles NaOH reacted = moles HCl added = 1.00 × 10^-4 mol

moles NaOH unreacted = (1.00 × 10^-3) − (1.00 × 10^-4) = 9.00 × 10^-4 mol

Calculate [OH^-] = moles(unreacted OH^-) ÷ total volume of solution
moles(unreacted OH^-) = moles(unreacted NaOH) = 9.00 × 10^-4 mol

total volume of solution = 10.00 mL + 1.00 mL = 11.00 mL = 11.00 × 10^-3 L

[OH^-] = (9.00 × 10^-4)mol ÷ (11.00 × 10^-3)L = 0.082 mol L^-1

Calculate pOH:
pOH = −log₁₀[OH^-] = −log₁₀[0.082] = 1.09

Calculate pH:
pH = 14 − pOH = 14 − 1.09 = 12.91

Step 4: Continue these calculations, adding 1.00 mL HCl(aq) to the new solution, until a volume of 9.00 mL of the 0.10 mol L^-1 HCl is added.

That is, continue these calculations as above because the NaOH(aq) is still in excess

Step 5: When a total of 10.00 mL of HCl(aq) has been added to the base, the moles of H⁺(aq) added will be equivalent to the moles of OH^-(aq) present due to the dissociation of the base.

At this point neither the NaOH nor the HCl is in excess.

The pH of the solution will be due to the self-dissociation of water:

H₂O(l) ⇋ H⁺(aq) + OH^-(aq) K_w = 10^-14 (25°C)

[H⁺(aq)] = [OH^-(aq)] = x

K_w = x² = 10^-14

√x² = √10^-14

x = 10^-7 = [H⁺(aq)] = 10^-7 mol L^-1

pH = −log₁₀[H⁺(aq)] = −log₁₀[10^-7] = 7.00

Step 6: Calculate the pH of the solution after 11.00 mL HCl has been added.

moles HCl: n(HCl) = concentration (mol L^-1) × volume (L)
n(HCl) = c(HCl) × V(HCl)

c(HCl) = 0.10 mol L^-1

V(HCl) = 11.00 mL = 11.00 × 10^-3 L

n(HCl) = 0.10 mol ~~L^-1~~ × (11.00 × 10^-3)L = 1.10 × 10^-3 mol

Calculate moles HCl in excess
moles(HCl) unreacted = total moles(HCl) − moles(HCl) reacted

total moles(HCl) = 1.10 × 10^-3 mol

moles(HCl) reacted = moles(NaOH) = 1.00 × 10^-3 mol

moles(HCl) unreacted = (1.10 × 10^-3) − (1.00 × 10^-3) = 1.00 × 10^-4 mol

Calculate [H⁺]:
[H⁺] = moles(H⁺ unreacted) ÷ total volume of solution

n(H⁺) unreacted = n(HCl) unreacted = 1.00 × 10^-4 mol

total volume = 10.00 mL + 11.00 mL = 21.00 mL = 21.00 × 10^-3 L

[H⁺] = (1.00 × 10^-4)mol ÷ (21.00 × 10^-3)L = 4.76 × 10^-3 mol L^-1

Calculate pH of the solution
pH = −log₁₀[H⁺] = −log₁₀[4.76 × 10^-3] = 2.32

Step 7: Continue these calculations, adding 1.00 mL of the HCl(aq) to the new soluton, until all the 14.00 mL 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 × 10^-3	10 × 10^-3	0.10	1	13
1 × 10^-3	1 × 10^-4	9 × 10^-4	11 × 10^-3	0.082	1.09	12.91
2 × 10^-3	2 × 10^-4	8 × 10^-4	12 × 10^-3	0.067	1.18	12.82
3 × 10^-3	3 × 10^-4	7 × 10^-4	13 × 10^-3	0.054	1.27	12.73
4 × 10^-3	4 × 10^-4	6 × 10^-4	14 × 10^-3	0.043	1.37	12.63
5 × 10^-3	5 × 10^-4	5 × 10^-4	15 × 10^-3	0.033	1.48	12.52
6 × 10^-3	6 × 10^-4	4 × 10^-4	16 × 10^-3	0.025	1.60	12.40
7 × 10^-3	7 × 10^-4	3 × 10^-4	17 × 10^-3	0.018	1.75	12.25
8 × 10^-3	8 × 10^-4	2 × 10^-4	18 × 10^-3	0.011	1.95	12.05
9 × 10^-3	9 × 10^-4	1 × 10^-4	19 × 10^-3	0.0053	2.28	11.72
10 × 10^-3	1 × 10^-3	0 (equivalence point)	20 × 10^-3	0 (from NaOH) (10^-7 from water dissociation)	7.00 (from water dissociation)	7.00 (from water dissociation)
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 × 10^-3	1.1 × 10^-3	1 × 10^-4	21 × 10^-3	4.76 × 10^-3	2.32
12 × 10^-3	1.2 × 10^-3	2 × 10^-4	22 × 10^-3	9.09 × 10^-3	2.04
13 × 10^-3	1.3 × 10^-3	3 × 10^-4	23 × 10^-3	0.013	1.88
14 × 10^-3	1.4 × 10^-3	4 × 10^-4	24 × 10^-3	0.017	1.78

Step 8. Plot a graph of pH vs Volume of HCl(aq)

Plotting the points on a graph 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 7.00 due to the self-dissociation of water:

H₂O(l) ⇋ H⁺(aq) + OH^-(aq) K_w = 10^-14 (25°C)

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Footnotes:

(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 25^oC you will need to use the appropriate value for K_w.
If the temperature of the solution has not been given in the question, assume 25^oC.

(3) We will assume additivity of volumes.

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

Key Concepts

Concepts

Adding Aqueous Strong Monoprotic Acid to Aqueous Strong Base at 25°C

Adding Aqueous Strong Base to Aqueous Strong Monoprotic Acid at 25°C

Calculations

Sample Calculation: pH of solution after mixing strong acid and strong base

Worked Example: Calculating a Titration Curve

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