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

Key Concepts

• Sulfuric acid, H₂SO₄ is a strong diprotic acid.
• Sulfuric acid dissociates (ionises) in two stages:

  Stage 1:
  H2SO4	→	H+(aq)	+	HSO4-(aq)		Ka1 very large
  Stage 2:
  HSO4-(aq)		H+(aq)	+	SO42-(aq)		Ka2 = 1.2 x 10-2 (at 25oC)1

• Sulfuric acid can be neutralised by a strong base such as sodium hydroxide, NaOH(aq).
• In the first stage of the neutralisation reaction, OH^-(aq) reacts with the H⁺(aq) released by the complete dissociation of the sulfuric acid:

  H⁺(aq) + OH^-(aq) → H₂O(l)

• In the second stage of the neutralisation reaction, OH^-(aq) reacts with the weaker HSO₄^-(aq) acid:

  HSO₄^-(aq) + OH^-(aq) → SO₄^2-(aq) + H₂O(l)

• At the equivalence point the moles of OH^- exactly equal the moles of H⁺ available from both stages of the dissociation of sulfuric acid so that we can write the overall reaction as:

  H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)
  moles OH^-(aq) = 2 x moles H₂SO₄
  moles H⁺(aq) = moles OH^-(aq)

• Adding more OH^-(aq) past the equivalence point simply increases the concentration of OH^-(aq) in solution.

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Concepts

• First stage: moles of OH^- less than moles of H₂SO₄
  (a) dissociation of sulfuric acid goes to completion: H₂SO₄ → H⁺(aq) + HSO₄^2-(aq)
  (b) hydroxide ions react with protons in solution: OH^-(aq) + H⁺(aq) → H₂O(l)
  (c) moles OH^- < moles H⁺ from dissociation of H₂SO₄ (H⁺ from dissociation of H₂SO₄ in excess)
  (d) species in solution (ignoring water) are:
      HSO₄^-(aq) from dissociation of H₂SO₄
      SO₄^2-(aq) from dissociation of HSO₄^-(aq)
      unreacted H⁺(aq) from dissociation of H₂SO₄ AND from dissociation of HSO₄^-(aq)
  (e) pH = -log₁₀[H⁺(aq)]
• Second stage: moles of OH^- less than 2 times the moles of H₂SO₄ but greater than moles of H₂SO₄
  (a) all the H⁺(aq) produced by the dissociation of H₂SO₄ has been consumed by reaction with OH^-(aq).
  (b) OH^- is now reacting with HSO₄^-: OH^-(aq) + HSO₄^-(aq) → H₂O(l) + SO₄^2-(aq)
  (c) species in solution (ignoring water) are:
      HSO₄^-(aq) from dissociation of H₂SO₄
      SO₄^2-(aq) from dissociation of HSO₄^-(aq)
      unreacted H⁺(aq) from dissociation of HSO₄^-(aq)
  (d) pH = -log₁₀[H⁺(aq)]
• Third stage: moles OH^- = 2 x moles of H₂SO₄ (equivalence point)
  overall reaction: H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)
  pH of solution is due to hydrolysis of SO₄^2-(aq)
• Fourth stage: moles OH^- greater than 2 x moles H₂SO₄
  (a) OH^- is in excess
  (b) pOH = -log₁₀[OH^-_(excess)]
  (c) pH = 14 - pOH (at 25^oC)²

Using these equations, and the known second dissociation constant for sulfuric acid, allows us to calculate the pH of the solution in the conical (erlenmeyer) flask at any stage as the hydroxide ions are being added.
The next section goes through these calculations in detail.

Calculating a Titration Curve

In this titration experiment, 0.10 mol L-1 NaOH(aq) is added in 1.00 mL increments from a burette to a conical (erlenmeyer) flask containing 5.00 mL of 0.10 mol L-1 H2SO4(aq) solution at 25oC. We will calculate the resulting pH of the solution in the conical (erlenmeyer) flask after each addition of NaOH(aq). The relevant chemical equations are: H2SO4 → H+(aq) + HSO4-(aq)   Ka1 is very large HSO4-(aq) H+(aq) + SO42-(aq)   Ka2 = 1.2 x 10-2

← NaOH (aq)   ← H2SO4(aq)

0 mL NaOH(aq) added to 5.00 mL of 0.10 mol L^-1 H₂SO₄ at 25^oC

Calculate the initial pH of the sulfuric acid in the conical flask.

Stage 1: 0.10 mol L^-1 H₂SO₄ fully dissociates to produce 0.10 mol L^-1 H⁺(aq) and 0.10 mol L^-1 HSO₄^-(aq)

Stage 2: some of the HSO₄^- dissociates to produce more H⁺(aq) and some SO₄^2-(aq)
Let X = change in concentration

	R.I.C.E. Table
Reaction	HSO4-(aq)		H+(aq)	+	SO42-(aq)
Initial concentration mol L-1	0.10		0.10		0
Change in concentration mol L-1	-X		+X		+X
Equilibrium concentration mol L-1	0.10 - X		0.10 + X		0 + X = X

Write the equilibrium expression for the second stage dissociation:

Substitute in the values:

Rearrange the expression:

and expand

and rearrange:

Write an expression to find X:

Substitute in the values to solve for X

Subsitute the value for X into the expression to find the final concentration of H⁺(aq):

[H⁺(aq)] = 0.10 + X = 0.10 + 9.76 x 10^-3 = 0.11 mol L^-1

Calculate the pH of the solution:

pH = -log₁₀[H⁺(aq)] = -log₁₀[0.11] = 0.96

1.00 mL 0.10 mol L^-1 NaOH(aq) added to 5.00 mL 0.010 H₂SO₄ at 25^oC

Calculate moles,n, of OH^- added:
n(OH^-) = n(NaOH) = c(NaOH) x V(NaOH in L) = 0.10 x 1.00/1000 = 1.00 x 10^-4 mol

Calculate moles of H⁺(aq) available to react:

Calculate moles of H⁺(aq) in excess = 5.00 x 10^-4 - 1.00 x 10^-4 = 4.00 x 10^-4 mol

Calculate concentrations of H⁺(aq) and HSO₄^- available for second stage of dissociation:

Calculate the change in concentrations as a result of the second stage dissociation:

	HSO4-(aq)		H+(aq)	+	SO42-(aq)
initial concentration / mol L-1	0.083		0.067		0
final concentration / mol L-1	0.083 - X		0.067 + X		0 + X = X

Calculate the value of X:

Calculate the concentration of H⁺(aq) in the final solution:
[H⁺(aq)] = 0.067 + 0.011 = 0.078 mol L^-1

Calculate the pH of the final solution:
pH = -log₁₀[H⁺(aq)] = -log₁₀[0.078] = 1.11

2.00 mL 0.10 mol L^-1 NaOH(aq) added to 5.00 mL 0.010 H₂SO₄ at 25^oC

Calculate moles,n, of OH^- added:
n(OH^-) = n(NaOH) = c(NaOH) x V(NaOH in L) = 0.10 x 2.00/1000 = 2.00 x 10^-4 mol

Calculate moles of H⁺(aq) available to react:

Calculate moles of H⁺(aq) that will react with OH^-(aq)

Calculate moles of H⁺(aq) in excess = 5.00 x 10^-4 - 2.00 x 10^-4 = 3.00 x 10^-4 mol

Calculate concentrations of H⁺(aq) and HSO₄^- available for second stage of dissociation:

Calculate the change in concentrations as a result of the second stage dissociation:

	HSO4-(aq)		H+(aq)	+	SO42-(aq)
initial concentration / mol L-1	0.071		0.043		0
final concentration / mol L-1	0.071 - X		0.043 + X		0 + X = X

Calculate the value of X:

Calculate the concentration of H⁺(aq) in the final solution:
[H⁺(aq)] = 0.043 + 0.013 = 0.056 mol L^-1

Calculate the pH of the final solution:
pH = -log₁₀[H⁺(aq)] = -log₁₀[0.056] = 1.25

3.00 mL 0.10 mol L^-1 NaOH(aq) added to 5.00 mL 0.010 H₂SO₄ at 25^oC

Calculate moles,n, of OH^- added:
n(OH^-) = n(NaOH) = c(NaOH) x V(NaOH in L) = 0.10 x 3.00/1000 = 3.00 x 10^-4 mol

Calculate moles of H⁺(aq) available to react:

Calculate moles of H⁺(aq) that will react with OH^-(aq)

Calculate moles of H⁺(aq) in excess = 5.00 x 10^-4 - 3.00 x 10^-4 = 2.00 x 10^-4 mol

Calculate concentrations of H⁺(aq) and HSO₄^- available for second stage of dissociation:

Calculate the change in concentrations as a result of the second stage dissociation:

	HSO4-(aq)		H+(aq)	+	SO42-(aq)
initial concentration / mol L-1	0.063		0.025		0
final concentration / mol L-1	0.063 - X		0.025 + X		0 + X = X

Calculate the value of X:

Calculate the concentration of H⁺(aq) in the final solution:
[H⁺(aq)] = 0.025 + 0.015 = 0.040 mol L^-1

Calculate the pH of the final solution:
pH = -log₁₀[H⁺(aq)] = -log₁₀[0.040] = 1.40

4.00 mL 0.10 mol L^-1 NaOH(aq) added to 5.00 mL 0.010 H₂SO₄ at 25^oC

Calculate moles,n, of OH^- added:
n(OH^-) = n(NaOH) = c(NaOH) x V(NaOH in L) = 0.10 x 4.00/1000 = 4.00 x 10^-4 mol

Calculate moles of H⁺(aq) available to react:

Calculate moles of H⁺(aq) that will react with OH^-(aq)

Calculate moles of H⁺(aq) in excess = 5.00 x 10^-4 - 4.00 x 10^-4 = 1.00 x 10^-4 mol

Calculate concentrations of H⁺(aq) and HSO₄^- available for second stage of dissociation:

Calculate the change in concentrations as a result of the second stage dissociation:

	HSO4-(aq)		H+(aq)	+	SO42-(aq)
initial concentration / mol L-1	0.056		0.011		0
final concentration / mol L-1	0.056 - X		0.011 + X		0 + X = X

Calculate the value of X:

Calculate the concentration of H⁺(aq) in the final solution:
[H⁺(aq)] = 0.011 + 0.017 = 0.028 mol L^-1

Calculate the pH of the final solution:
pH = -log₁₀[H⁺(aq)] = -log₁₀[0.028] = 1.55

5.00 mL 0.10 mol L^-1 NaOH(aq) added to 5.00 mL 0.010 H₂SO₄ at 25^oC

Calculate moles,n, of OH^- added:
n(OH^-) = n(NaOH) = c(NaOH) x V(NaOH in L) = 0.10 x 5.00/1000 = 5.00 x 10^-4 mol

Calculate moles of H⁺(aq) available to react:

Calculate moles of H⁺(aq) that will react with OH^-(aq)

Calculate moles of H⁺(aq) in excess = 5.00 x 10^-4 - 5.00 x 10^-4 = 0.00 mol

Calculate concentrations of H⁺(aq) and HSO₄^- available for second stage of dissociation:

Calculate the change in concentrations as a result of the second stage dissociation:

	HSO4-(aq)		H+(aq)	+	SO42-(aq)
initial concentration / mol L-1	0.050		0.00		0
final concentration / mol L-1	0.050 - X		0.00 + X = X		0 + X = X

Calculate the value of X:

Calculate the concentration of H⁺(aq) in the final solution:
[H⁺(aq)] = 0.00 + 0.019 = 0.019 mol L^-1

Calculate the pH of the final solution:
pH = -log₁₀[H⁺(aq)] = -log₁₀[0.019] = 1.72

6.00 mL 0.10 mol L^-1 NaOH(aq) added to 5.00 mL 0.010 H₂SO₄ at 25^oC

Calculate moles,n, of OH^- added:
n(OH^-) = n(NaOH) = c(NaOH) x V(NaOH in L) = 0.10 x 6.00/1000 = 6.00 x 10^-4 mol

Calculate moles of H⁺(aq) available to react:

OH^-(aq) is now in excess of H⁺(aq) produced as a result of the first dissociation of H₂SO₄
moles OH^-(aq) in excess = moles OH^-(aq) added - moles OH^-(aq) reacted with H⁺(aq)
n(OH^-(aq)) = 6.00 x 10^-4 - 5.00 x 10^-4 = 1.00 x 10^-4 mol

The relevant reaction is now between HSO₄^-(aq) produced from the dissociation of H₂SO₄ and OH^-(aq):

	HSO4-(aq)		H+(aq)	+	SO42-(aq)
moles / mol	4.00 x 10-4		0		0
total volume / L	(5.00 + 6.00)/1000 =		11 x 10-3		11 x 10-3
concentration / mol L-1	0.036		0		0

	HSO4-(aq)		H+(aq)	+	SO42-(aq)
concentration / mol L-1	0.036 - X		0 + X = X		0 + X = X

Calculate the pH of the final solution:
pH = -log₁₀[H⁺(aq)] = -log₁₀[0.016] = 1.80

7.00 mL 0.10 mol L^-1 NaOH(aq) added to 5.00 mL 0.010 H₂SO₄ at 25^oC

Calculate moles,n, of OH^- added:
n(OH^-) = n(NaOH) = c(NaOH) x V(NaOH in L) = 0.10 x 7.00/1000 = 7.00 x 10^-4 mol

Calculate moles of H⁺(aq) available to react:

The relevant reaction is now between HSO₄^-(aq) produced from the dissociation of H₂SO₄ and OH^-(aq):

	HSO4-(aq)		H+(aq)	+	SO42-(aq)
moles / mol	3.00 x 10-4		0		0
total volume / L	(5.00 + 7.00)/1000 =		12 x 10-3		12 x 10-3
concentration / mol L-1	0.025		0		0

	HSO4-(aq)		H+(aq)	+	SO42-(aq)
concentration / mol L-1	0.025 - X		0 + X = X		0 + X = X

Calculate the pH of the final solution:
pH = -log₁₀[H⁺(aq)] = -log₁₀[0.012] = 1.91

8.00 mL 0.10 mol L^-1 NaOH(aq) added to 5.00 mL 0.010 H₂SO₄ at 25^oC

Calculate moles,n, of OH^- added:
n(OH^-) = n(NaOH) = c(NaOH) x V(NaOH in L) = 0.10 x 8.00/1000 = 8.00 x 10^-4 mol

Calculate moles of H⁺(aq) available to react:

The relevant reaction is now between HSO₄^-(aq) produced from the dissociation of H₂SO₄ and OH^-(aq):

	HSO4-(aq)		H+(aq)	+	SO42-(aq)
moles / mol	2.00 x 10-4		0		0
total volume / L	(5.00 + 8.00)/1000 =		13 x 10-3		13 x 10-3
concentration / mol L-1	0.015		0		0

	HSO4-(aq)		H+(aq)	+	SO42-(aq)
concentration / mol L-1	0.015 - X		0 + X = X		0 + X = X

Calculate the pH of the final solution:
pH = -log₁₀[H⁺(aq)] = -log₁₀[0.0087] = 2.06

9.00 mL 0.10 mol L^-1 NaOH(aq) added to 5.00 mL 0.010 H₂SO₄ at 25^oC

Calculate moles,n, of OH^- added:
n(OH^-) = n(NaOH) = c(NaOH) x V(NaOH in L) = 0.10 x 9.00/1000 = 9.00 x 10^-4 mol

Calculate moles of H⁺(aq) available to react:

The relevant reaction is now between HSO₄^-(aq) produced from the dissociation of H₂SO₄ and OH^-(aq):

	HSO4-(aq)		H+(aq)	+	SO42-(aq)
moles / mol	1.00 x 10-4		0		0
total volume / L	(5.00 + 9.00)/1000 =		14 x 10-3		14 x 10-3
concentration / mol L-1	0.0071		0		0

	HSO4-(aq)		H+(aq)	+	SO42-(aq)
concentration / mol L-1	0.0071 - X		0 + X = X		0 + X = X

Calculate the pH of the final solution:
pH = -log₁₀[H⁺(aq)] = -log₁₀[0.0050] = 2.30

10.00 mL 0.10 mol L^-1 NaOH(aq) added to 5.00 mL 0.010 H₂SO₄ at 25^oC

Calculate moles,n, of OH^- added:
n(OH^-) = n(NaOH) = c(NaOH) x V(NaOH in L) = 0.10 x 10.00/1000 = 1.00 x 10^-3 mol

Calculate moles of H⁺(aq) available to react:

The relevant reaction is now between HSO₄^-(aq) produced from the dissociation of H₂SO₄ and OH^-(aq):

Any addition of OH^-(aq) beyond 10.00 mL will result in the pH of the solution reflecting the concentration of the excess hydroxide ions in the solution.

11.00 mL 0.10 mol L^-1 NaOH(aq) added to 5.00 mL sulfuric acid at 25^oC

Since we are now past the equivalence point for the reaction, we can simplify the reaction to:

12.00 mL 0.10 mol L^-1 NaOH(aq) added to 5.00 mL sulfuric acid at 25^oC

Since we are now past the equivalence point for the reaction, we can simplify the reaction to:

13.00 mL 0.10 mol L^-1 NaOH(aq) added to 5.00 mL sulfuric acid at 25^oC

Since we are now past the equivalence point for the reaction, we can simplify the reaction to:

Titration Curve

Notice that although sulfuric acid is a diprotic acid, we can only see one equivalence point not two.⁴
While the first dissociation (ionisation) reaction goes to completion, the second dissociation constant for sulfuric acid is still quite large, too large for us to detect a separation of the two equivalence points.

Also notice that the pH of the solution at the equivalence point is not 7.00, but is actually slightly greater than 7.00.
At the equivalence point, the predominant ionic species in solution are Na⁺(aq) and SO₄^2-(aq).
Na₂SO₄ is the salt of a strong base (NaOH) and a weak acid (HSO₄^-) so it is basic (refer to pH of Salt Solutions).
In the next section we calculate the pH of this salt solution.

Calculating pH at the Equivalence Point

H₂SO₄(aq) + 2NaOH(aq) → 2H₂O(l) + 2Na⁺(aq) + SO₄^2-(aq)

and the resulting solution contains Na⁺(aq) and SO₄^2-(aq).
Na⁺ does not hydrolyse (react with water), but SO₄^2- acts as a weak base, accepting a proton from a water molecule:

SO₄^2- + H₂O HSO₄^- + OH^-

Which is equivalent to K_w/K_a(HSO4^-),

At 25^oC, K_a(HSO4^-) = 1.2 x 10^-2 and K_w = 1.0 x 10^-14, so,

At the equivalence point: H₂SO₄(aq) + 2NaOH(aq) → 2Na⁺(aq) + SO₄^2-(aq) + 2H₂O(l)
moles(SO₄^2-) = n(SO₄^2-) = n(H₂SO₄) = c(H₂SO₄) x V(H₂SO₄) = 5.00/1000 x 0.10 = 5.00 x 10^-4 mol
total volume of solution = mL H₂SO₄ + mL NaOH added = 5.00 mL + 10.00 mL = 15.00 mL = 0.0150 L
concentration(SO₄^2-) = c(SO₄^2-) = moles(SO₄^2-)/total volume = 5.00 x 10^-4/0.0150 = 0.033 mol L^-1
Let X = [OH^-] produced from the hydrolysis reaction
then X = [HSO₄^-] (since HSO₄^- is produced in the same hydrolysis reaction)
and [SO₄^2-] = 0.033 - X (concentration of SO₄^2- decreases as a result of reaction with water)

[OH^-] = X = 1.66 x 10^-7
pOH = -log₁₀[OH^-] = -log₁₀[1.66 x 10^-7] = 6.78
pH = 14 - pOH = 14 - 6.78 = 7.22