 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) → SO42-(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

Sulfuric acid can be neutralised by adding a strong base such as sodium hydroxide, NaOH(aq).

This neutralisation reaction can be used as the basis of a titration as shown in the diagram on the right.

In this titration, the burette is filled with an aqueous solution of sodium hydroxide, NaOH(aq).

An aliquot of an aqueous solution of sulfuric acid is pipetted into the conical (erlenmeyer) flask.
An indicator would be added to the acid (to "indicate" the end point of the titration).

 ← NaOH (aq) ← H2SO4(aq)

We can think of the reactions occurring in the conical (erlenmeyer) flask during the titration as happening in 4 stages:

• First stage: moles of OH- less than moles of H2SO4
(a) dissociation of sulfuric acid goes to completion: H2SO4 → H+(aq) + HSO42-(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
SO42-(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) + SO42-(aq)
(c) species in solution (ignoring water) are:
HSO4-(aq) from dissociation of H2SO4
SO42-(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 SO42-(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 25oC)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.

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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 25oC

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 SO42-(aq)
Let X = change in concentration

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

Write the equilibrium expression for the second stage dissociation:

 Ka2 = [H+(aq)][SO42-][HSO4-(aq)]

Substitute in the values:

 1.2 x 10-2 = [0.10 + X][X][0.10 - X]

Rearrange the expression:

 1.2 x 10-2[0.10 - X] = [0.10 + X][X]

and expand

 1.2 x 10-3 - 1.2 x 10-2X = 0.10X + X2

and rearrange:

 0 = X2 + 0.112X - 1.2 x 10-3

Which is a quadratic equation of the form 0 = aX2 + bX2 + c
for which
a = 1
b = 0.112
c = -1.2 x 10-3

Write an expression to find X:

 X = -b ± √(b2 -4ac)2a

Substitute in the values to solve for X

 X = -0.112 ± √(0.1122 -4 x 1 x -1.2 x 10-3)2 x 1 X = -0.112 ± √(0.0125 + 4.8 x 10-3)2 X = -0.112 ± √(0.0173)2 X = -0.112 ± 0.1322 X = -0.112 + 0.1322 X = 9.76 x 10-3

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 25oC

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:

 Reaction H2SO4 → H+(aq) + HSO4- initial concentration mol L-1 0.10 0 0 final concentration mol L-1 0 0.10 0.10 initial moles / mol 0 0.10 x 5.00/1000 = 5.00 x 10-4 0.10 x 5.00/1000 = 5.00 x 10-4

Calculate moles of H+(aq) that will react with OH-(aq)
H+(aq) + OH-(aq) → H2O(l)

1.00 x 10-4 mol OH-(aq) will react with 1.00 x 10-4 mol H+(aq)

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:

 H2SO4 → H+(aq) + HSO4- moles /mol 0 4 x 10-4 5.00 x 10-4 3total volume /L (5+1)/1000 = 6 x 10-3 6.00 x 10-3 concentration / mol L-1 0 0.067 0.083

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:

 Ka2 = [H+(aq)][SO42-][HSO4-(aq)] 1.2 x 10-2 = [0.067 + X][X][0.083 - X] 0 = X2 + 0.079X -9.96 x 10-4

 X = -b ± √(b2 -4ac)2a X = -0.079 ± √(0.0792 -4 x 1 x -9.96 x 10-4)2 x 1 X = 0.011

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 25oC

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:

 H2SO4 → H+(aq) + HSO4- initial concentration / mol L-1 0.10 0 0 final concentration / mol L-1 0 0.10 0.10 initial moles / mol 0 0.10 x 5.00/1000 = 5.00 x 10-4 0.10 x 5.00/1000 = 5.00 x 10-4

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

H+(aq) + OH-(aq) → H2O(l)

2.00 x 10-4 mol OH-(aq) will react with 2.00 x 10-4 mol H+(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:

 H2SO4 → H+(aq) + HSO4- moles /mol 0 3 x 10-4 5.00 x 10-4 total volume /L (5+2)/1000 = 7 x 10-3 7.00 x 10-3 concentration / mol L-1 0 0.043 0.071

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:

 Ka2 = [H+(aq)][SO42-][HSO4-(aq)] 1.2 x 10-2 = [0.043 + X][X][0.071 - X] 0 = X2 + 0.055X -8.52 x 10-4

 X = -b ± √(b2 -4ac)2a X = -0.055 ± √(0.0552 -4 x 1 x -8.52 x 10-4)2 x 1 X = 0.013

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 25oC

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:

 H2SO4 → H+(aq) + HSO4- initial concentration / mol L-1 0.10 0 0 final concentration / mol L-1 0 0.10 0.10 initial moles / mol 0 0.10 x 5.00/1000 = 5.00 x 10-4 0.10 x 5.00/1000 = 5.00 x 10-4

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

H+(aq) + OH-(aq) → H2O(l)

3.00 x 10-4 mol OH-(aq) will react with 3.00 x 10-4 mol H+(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:

 H2SO4 → H+(aq) + HSO4- moles /mol 0 2 x 10-4 5.00 x 10-4 total volume /L (5+3)/1000 = 8 x 10-3 8.00 x 10-3 concentration / mol L-1 0 0.025 0.063

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:

 Ka2 = [H+(aq)][SO42-][HSO4-(aq)] 1.2 x 10-2 = [0.025 + X][X][0.063 - X] 0 = X2 + 0.037X -7.56 x 10-4

 X = -b ± √(b2 -4ac)2a X = -0.037 ± √(0.0372 -4 x 1 x -7.56 x 10-4)2 x 1 X = 0.015

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 25oC

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:

 H2SO4 → H+(aq) + HSO4- initial concentration / mol L-1 0.10 0 0 final concentration / mol L-1 0 0.10 0.10 initial moles / mol 0 0.10 x 5.00/1000 = 5.00 x 10-4 0.10 x 5.00/1000 = 5.00 x 10-4

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

H+(aq) + OH-(aq) → H2O(l)

4.00 x 10-4 mol OH-(aq) will react with 4.00 x 10-4 mol H+(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:

 H2SO4 → H+(aq) + HSO4- moles /mol 0 1 x 10-4 5.00 x 10-4 total volume /L (5+4)/1000 = 9 x 10-3 9.00 x 10-3 concentration / mol L-1 0 0.011 0.056

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:

 Ka2 = [H+(aq)][SO42-][HSO4-(aq)] 1.2 x 10-2 = [0.011 + X][X][0.056 - X] 0 = X2 + 0.023X -6.72 x 10-4

 X = -b ± √(b2 -4ac)2a X = -0.023 ± √(0.0232 -4 x 1 x -6.72 x 10-4)2 x 1 X = 0.017

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 25oC

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:

 H2SO4 → H+(aq) + HSO4- initial concentration / mol L-1 0.10 0 0 final concentration / mol L-1 0 0.10 0.10 initial moles / mol 0 0.10 x 5.00/1000 = 5.00 x 10-4 0.10 x 5.00/1000 = 5.00 x 10-4

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

H+(aq) + OH-(aq) → H2O(l)

5.00 x 10-4 mol OH-(aq) will react with 5.00 x 10-4 mol H+(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:

 H2SO4 → H+(aq) + HSO4- moles /mol 0 0 5.00 x 10-4 total volume /L (5+5)/1000 = 10 x 10-3 10.00 x 10-3 concentration / mol L-1 0 0.00 0.050

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:

 Ka2 = [H+(aq)][SO42-][HSO4-(aq)] 1.2 x 10-2 = [X][X][0.050 - X] 0 = X2 + 1.2 x 10-2X -6.00 x 10-4

 X = -b ± √(b2 -4ac)2a X = -1.2 x 10-2 ± √((1.2 x 10-2)2 -4 x 1 x -6.00 x 10-4)2 x 1 X = 0.019

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 25oC

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:

 H2SO4 → H+(aq) + HSO4- initial concentration / mol L-1 0.10 0 0 final concentration / mol L-1 0 0.10 0.10 initial moles / mol 0 0.10 x 5.00/1000 = 5.00 x 10-4 0.10 x 5.00/1000 = 5.00 x 10-4

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):

OH-(aq) + HSO4-(aq) → H2O(l) + SO42-(aq)

1.00 x 10-4 moles OH-(aq) reacts with 1.00 x 10-4 moles HSO4-(aq)
moles HSO4- in solution = 5.00 x 10-4 - 1.00 x 10-4 = 4.00 x 10-4 mol
 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

Calculate changes in concentration as a result of dissociation of HSO4-(aq):
 HSO4-(aq) H+(aq) + SO42-(aq) concentration / mol L-1 0.036 - X 0 + X = X 0 + X = X

Calculate the value of X:
 Ka2 = [H+(aq)][SO42-][HSO4-(aq)] 1.2 x 10-2 = [X][X][0.036 - X] 0 = X2 + 1.2 x 10-2X -4.32 x 10-4

 X = -b ± √(b2 -4ac)2a X = -1.2 x 10-2 ± √((1.2 x 10-2)2 -4 x 1 x -4.32 x 10-4)2 x 1 X = 0.016

Calculate the concentration of H+(aq) in the final solution:
[H+(aq)] = 0.00 + 0.016 = 0.016 mol L-1

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 25oC

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:

 H2SO4 → H+(aq) + HSO4- initial concentration / mol L-1 0.10 0 0 final concentration / mol L-1 0 0.10 0.10 initial moles / mol 0 0.10 x 5.00/1000 = 5.00 x 10-4 0.10 x 5.00/1000 = 5.00 x 10-4

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)) = 7.00 x 10-4 - 5.00 x 10-4 = 2.00 x 10-4 mol

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

OH-(aq) + HSO4-(aq) → H2O(l) + SO42-(aq)

2.00 x 10-4 moles OH-(aq) reacts with 2.00 x 10-4 moles HSO4-(aq)
moles HSO4- in solution = 5.00 x 10-4 - 2.00 x 10-4 = 3.00 x 10-4 mol
 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

Calculate changes in concentration as a result of dissociation of HSO4-(aq):
 HSO4-(aq) H+(aq) + SO42-(aq) concentration / mol L-1 0.025 - X 0 + X = X 0 + X = X

Calculate the value of X:
 Ka2 = [H+(aq)][SO42-][HSO4-(aq)] 1.2 x 10-2 = [X][X][0.025 - X] 0 = X2 + 1.2 x 10-2X -3.00 x 10-4

 X = -b ± √(b2 -4ac)2a X = -1.2 x 10-2 ± √((1.2 x 10-2)2 -4 x 1 x -3.00 x 10-4)2 x 1 X = 0.012

Calculate the concentration of H+(aq) in the final solution:
[H+(aq)] = 0.00 + 0.012 = 0.012 mol L-1

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 25oC

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:

 H2SO4 → H+(aq) + HSO4- initial concentration / mol L-1 0.10 0 0 final concentration / mol L-1 0 0.10 0.10 initial moles / mol 0 0.10 x 5.00/1000 = 5.00 x 10-4 0.10 x 5.00/1000 = 5.00 x 10-4

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)) = 8.00 x 10-4 - 5.00 x 10-4 = 3.00 x 10-4 mol

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

OH-(aq) + HSO4-(aq) → H2O(l) + SO42-(aq)

3.00 x 10-4 moles OH-(aq) reacts with 3.00 x 10-4 moles HSO4-(aq)
moles HSO4- in solution = 5.00 x 10-4 - 3.00 x 10-4 = 2.00 x 10-4 mol
 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

Calculate changes in concentration as a result of dissociation of HSO4-(aq):
 HSO4-(aq) H+(aq) + SO42-(aq) concentration / mol L-1 0.015 - X 0 + X = X 0 + X = X

Calculate the value of X:
 Ka2 = [H+(aq)][SO42-][HSO4-(aq)] 1.2 x 10-2 = [X][X][0.015 - X] 0 = X2 + 1.2 x 10-2X -1.8 x 10-4

 X = -b ± √(b2 -4ac)2a X = -1.2 x 10-2 ± √((1.2 x 10-2)2 -4 x 1 x -1.8 x 10-4)2 x 1 X = 0.0087

Calculate the concentration of H+(aq) in the final solution:
[H+(aq)] = 0.00 + 0.0087 = 0.0087 mol L-1

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 25oC

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:

 H2SO4 → H+(aq) + HSO4- initial concentration / mol L-1 0.10 0 0 final concentration / mol L-1 0 0.10 0.10 initial moles / mol 0 0.10 x 5.00/1000 = 5.00 x 10-4 0.10 x 5.00/1000 = 5.00 x 10-4

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)) = 9.00 x 10-4 - 5.00 x 10-4 = 4.00 x 10-4 mol

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

OH-(aq) + HSO4-(aq) → H2O(l) + SO42-(aq)

4.00 x 10-4 moles OH-(aq) reacts with 4.00 x 10-4 moles HSO4-(aq)
moles HSO4- in solution = 5.00 x 10-4 - 4.00 x 10-4 = 1.00 x 10-4 mol
 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

Calculate changes in concentration as a result of dissociation of HSO4-(aq):
 HSO4-(aq) H+(aq) + SO42-(aq) concentration / mol L-1 0.0071 - X 0 + X = X 0 + X = X

Calculate the value of X:
 Ka2 = [H+(aq)][SO42-][HSO4-(aq)] 1.2 x 10-2 = [X][X][0.0071 - X] 0 = X2 + 1.2 x 10-2X -8.52 x 10-5

 X = -b ± √(b2 -4ac)2a X = -1.2 x 10-2 ± √((1.2 x 10-2)2 -4 x 1 x -8.52 x 10-5)2 x 1 X = 0.0050

Calculate the concentration of H+(aq) in the final solution:
[H+(aq)] = 0.00 + 0.0050 = 0.0050 mol L-1

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 25oC

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:

 H2SO4 → H+(aq) + HSO4- initial concentration / mol L-1 0.10 0 0 final concentration / mol L-1 0 0.10 0.10 initial moles / mol 0 0.10 x 5.00/1000 = 5.00 x 10-4 0.10 x 5.00/1000 = 5.00 x 10-4

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)) = 1.00 x 10-3 - 5.00 x 10-4 = 5.00 x 10-4 mol

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

OH-(aq) + HSO4-(aq) → H2O(l) + SO42-(aq)

5.00 x 10-4 moles OH-(aq) reacts with 5.00 x 10-4 moles HSO4-(aq)
moles HSO4- in solution = 5.00 x 10-4 - 5.00 x 10-4 = 0 mol
This represents the equivalence point for the reaction between sulfuric acid and hydroxide ions.
The pH of this solution is dependent on the hydrolysis (reaction with water) of the SO42-(aq) in the solution.

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 25oC

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

H2SO4(aq) + 2NaOH(aq) → 2H2O(l) + Na2SO4(aq)

Calculate moles and concentrations:
 H2SO4(aq) NaOH(aq) available moles / mol 0.10 x 5.00/1000 = 5.00 x 10-4 0.10 x 11.00/1000 = 1.10 x 10-3 moles reacted / mol 5.00 x 10-4 2 x 5.00 x 10-4 = 1.00 x 10-3 moles in solution / mol 5.00 x 10-4 - 5.00 x 10-4 = 0 1.10 x 10-3 - 1.00 x 10-3 = 1.00 x 10-4 total volume / L (11.00 + 5.00)/1000 = 16.00 x 10-3 concentration / mol L-1 0 0.0063 pOH -log10[OH-(aq)] = 2.20 pH 14 - pOH = 11.80

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

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

H2SO4(aq) + 2NaOH(aq) → 2H2O(l) + Na2SO4(aq)

Calculate moles and concentrations:
 H2SO4(aq) NaOH(aq) available moles / mol 0.10 x 5.00/1000 = 5.00 x 10-4 0.10 x 12.00/1000 = 1.20 x 10-3 moles reacted / mol 5.00 x 10-4 2 x 5.00 x 10-4 = 1.00 x 10-3 moles in solution / mol 5.00 x 10-4 - 5.00 x 10-4 = 0 1.20 x 10-3 - 1.00 x 10-3 = 2.00 x 10-4 total volume / L (12.00 + 5.00)/1000 = 17.00 x 10-3 concentration / mol L-1 0 0.012 pOH -log10[OH-(aq)] = 1.93 pH 14 - pOH = 12.07

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

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

H2SO4(aq) + 2NaOH(aq) → 2H2O(l) + Na2SO4(aq)

Calculate moles and concentrations:
 H2SO4(aq) NaOH(aq) available moles / mol 0.10 x 5.00/1000 = 5.00 x 10-4 0.10 x 13.00/1000 = 1.30 x 10-3 moles reacted / mol 5.00 x 10-4 2 x 5.00 x 10-4 = 1.00 x 10-3 moles in solution / mol 5.00 x 10-4 - 5.00 x 10-4 = 0 1.30 x 10-3 - 1.00 x 10-3 = 3.00 x 10-4 total volume / L (13.00 + 5.00)/1000 = 18.00 x 10-3 concentration / mol L-1 0 0.017 pOH -log10[OH-(aq)] = 1.78 pH 14 - pOH = 12.22

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Titration Curve

The above calculations are shown in the table below, and are also plotted as a graph:

volume NaoH added in mL pH Graph
0 0.96
 pH of solution 0.10 M NaOH - 0.10 M H2SO4 Titration Curvevolume of NaOH added (mL)
1.00 1.11
2.00 1.25
3.00 1.40
4.00 1.55
5.00 1.72
6.00 1.80
7.00 1.91
8.00 2.06
9.00 2.30
11.00 11.80
12.00 12.07
13.00 12.22

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 SO42-(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.

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Calculating pH at the Equivalence Point

At the equivalence point for the reaction between aqueous sodium hydroxide solution, NaOH(aq), and aqueous sulfuric acid solution, H2SO4(aq), we can represent the overall reaction as:

H2SO4(aq) + 2NaOH(aq) → 2H2O(l) + 2Na+(aq) + SO42-(aq)

and the resulting solution contains Na+(aq) and SO42-(aq).
Na+ does not hydrolyse (react with water), but SO42- acts as a weak base, accepting a proton from a water molecule:

SO42- + H2O HSO4- + OH-

The hydrolysis constant for this reaction is

 Kh = [HSO4-][OH-][SO42-][H2O]

Which is equivalent to Kw/Ka(HSO4-),

 Kw = [H+][OH-][H2O] Ka(HSO4-) = [H+][SO42-][HSO4-] Kh = KwKa = [OH-][H+][HSO4-][H2O][H+][SO42-] = [OH-][HSO4-][H2O][SO42-]

At 25oC, Ka(HSO4-) = 1.2 x 10-2 and Kw = 1.0 x 10-14, so,

 Kh = KwKa = 1.0 x 10-141.2 x 10-2 = 8.3 x 10-13

At the equivalence point: H2SO4(aq) + 2NaOH(aq) → 2Na+(aq) + SO42-(aq) + 2H2O(l)
moles(SO42-) = n(SO42-) = 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(SO42-) = c(SO42-) = moles(SO42-)/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 [SO42-] = 0.033 - X (concentration of SO42- decreases as a result of reaction with water)

 Kh = [HSO4-][OH-][SO42-] 8.33 x 10-13 = X20.033 - X 8.33 x 10-13(0.033 - X) = X2 X2 + 8.33 x 10-13X - 2.75 x 10-14 = 0

 X = -b ± √(b2 -4 x a x -c)2 x a X = -8.33 x 10-13 ± √((8.33 x 10-13)2 -4 x 1 x - 2.75 x 10-14)2 x 1 X = 1.66 x 10-7

[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

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

2.If the temperature of the solution has not been given in the question, assume 25oC, then Kw=10-14 and pH+pOH=14.

3. We will assume additivity of volumes.

4. 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.
On the titration curve plotted above this point is the point midway between the two plateau regions of the graph.