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} H_{2}SO_{4}(aq) solution at 25^{o}C.
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:
H_{2}SO_{4} 
→ 
H^{+}(aq) 
+ 
HSO_{4}^{}(aq) 

K_{a1} is very large 
HSO_{4}^{}(aq) 

H^{+}(aq) 
+ 
SO_{4}^{2}(aq) 

K_{a2} = 1.2 x 10^{2} 


← NaOH (aq) 

← H_{2}SO_{4}(aq) 


0 mL NaOH(aq) added to 5.00 mL of 0.10 mol L^{1} H_{2}SO_{4} at 25^{o}C
Calculate the initial pH of the sulfuric acid in the conical flask.
Stage 1: 0.10 mol L^{1} H_{2}SO_{4} fully dissociates to produce 0.10 mol L^{1} H^{+}(aq) and 0.10 mol L^{1} HSO_{4}^{}(aq)
Stage 2: some of the HSO_{4}^{} dissociates to produce more H^{+}(aq) and some SO_{4}^{2}(aq)
Let X = change in concentration

R.I.C.E. Table 
Reaction 
HSO_{4}^{}(aq) 

H^{+}(aq) 
+ 
SO_{4}^{2}(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:
K_{a2} 
= 
[H^{+}(aq)][SO_{4}^{2}]
[HSO_{4}^{}(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^{2}X 
= 
0.10X + X^{2} 
and rearrange:
0 
= 
X^{2} + 0.112X  1.2 x 10^{3} 
Which is a quadratic equation of the form 0 = aX^{2} + bX^{2} + c
for which
a = 1
b = 0.112
c = 1.2 x 10^{3}
Write an expression to find X:
Substitute in the values to solve for X
X 
= 
0.112 ± √(0.112^{2} 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 ± 0.132 2 
X 
= 
0.112 + 0.132 2 
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 = log_{10}[H^{+}(aq)] = log_{10}[0.11] = 0.96
1.00 mL 0.10 mol L^{1} NaOH(aq) added to 5.00 mL 0.010 H_{2}SO_{4} at 25^{o}C
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:
Reaction 
H_{2}SO_{4} 
→ 
H^{+}(aq) 
+ 
HSO_{4}^{} 
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) → H_{2}O(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 HSO_{4}^{} available for second stage of dissociation:

H_{2}SO_{4} 
→ 
H^{+}(aq) 
+ 
HSO_{4}^{} 
moles /mol 
0 

4 x 10^{4} 

5.00 x 10^{4} 
^{3}total 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:

HSO_{4}^{}(aq) 

H^{+}(aq) 
+ 
SO_{4}^{2}(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:
K_{a2} 
= 
[H^{+}(aq)][SO_{4}^{2}]
[HSO_{4}^{}(aq)] 
1.2 x 10^{2} 
= 
[0.067 + X][X]
[0.083  X] 
0 
= 
X^{2} + 0.079X 9.96 x 10^{4} 
X 
= 
b ± √(b^{2} 4ac) 2a 
X 
= 
0.079 ± √(0.079^{2} 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 = log_{10}[H^{+}(aq)] = log_{10}[0.078] = 1.11
2.00 mL 0.10 mol L^{1} NaOH(aq) added to 5.00 mL 0.010 H_{2}SO_{4} at 25^{o}C
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:

H_{2}SO_{4} 
→ 
H^{+}(aq) 
+ 
HSO_{4}^{} 
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) → H_{2}O(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 HSO_{4}^{} available for second stage of dissociation:

H_{2}SO_{4} 
→ 
H^{+}(aq) 
+ 
HSO_{4}^{} 
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:

HSO_{4}^{}(aq) 

H^{+}(aq) 
+ 
SO_{4}^{2}(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:
K_{a2} 
= 
[H^{+}(aq)][SO_{4}^{2}]
[HSO_{4}^{}(aq)] 
1.2 x 10^{2} 
= 
[0.043 + X][X]
[0.071  X] 
0 
= 
X^{2} + 0.055X 8.52 x 10^{4} 
X 
= 
b ± √(b^{2} 4ac) 2a 
X 
= 
0.055 ± √(0.055^{2} 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 = log_{10}[H^{+}(aq)] = log_{10}[0.056] = 1.25
3.00 mL 0.10 mol L^{1} NaOH(aq) added to 5.00 mL 0.010 H_{2}SO_{4} at 25^{o}C
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:

H_{2}SO_{4} 
→ 
H^{+}(aq) 
+ 
HSO_{4}^{} 
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) → H_{2}O(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 HSO_{4}^{} available for second stage of dissociation:

H_{2}SO_{4} 
→ 
H^{+}(aq) 
+ 
HSO_{4}^{} 
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:

HSO_{4}^{}(aq) 

H^{+}(aq) 
+ 
SO_{4}^{2}(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:
K_{a2} 
= 
[H^{+}(aq)][SO_{4}^{2}]
[HSO_{4}^{}(aq)] 
1.2 x 10^{2} 
= 
[0.025 + X][X]
[0.063  X] 
0 
= 
X^{2} + 0.037X 7.56 x 10^{4} 
X 
= 
b ± √(b^{2} 4ac) 2a 
X 
= 
0.037 ± √(0.037^{2} 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 = log_{10}[H^{+}(aq)] = log_{10}[0.040] = 1.40
4.00 mL 0.10 mol L^{1} NaOH(aq) added to 5.00 mL 0.010 H_{2}SO_{4} at 25^{o}C
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:

H_{2}SO_{4} 
→ 
H^{+}(aq) 
+ 
HSO_{4}^{} 
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) → H_{2}O(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 HSO_{4}^{} available for second stage of dissociation:

H_{2}SO_{4} 
→ 
H^{+}(aq) 
+ 
HSO_{4}^{} 
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:

HSO_{4}^{}(aq) 

H^{+}(aq) 
+ 
SO_{4}^{2}(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:
K_{a2} 
= 
[H^{+}(aq)][SO_{4}^{2}]
[HSO_{4}^{}(aq)] 
1.2 x 10^{2} 
= 
[0.011 + X][X]
[0.056  X] 
0 
= 
X^{2} + 0.023X 6.72 x 10^{4} 
X 
= 
b ± √(b^{2} 4ac) 2a 
X 
= 
0.023 ± √(0.023^{2} 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 = log_{10}[H^{+}(aq)] = log_{10}[0.028] = 1.55
5.00 mL 0.10 mol L^{1} NaOH(aq) added to 5.00 mL 0.010 H_{2}SO_{4} at 25^{o}C
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:

H_{2}SO_{4} 
→ 
H^{+}(aq) 
+ 
HSO_{4}^{} 
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) → H_{2}O(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 HSO_{4}^{} available for second stage of dissociation:

H_{2}SO_{4} 
→ 
H^{+}(aq) 
+ 
HSO_{4}^{} 
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:

HSO_{4}^{}(aq) 

H^{+}(aq) 
+ 
SO_{4}^{2}(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:
K_{a2} 
= 
[H^{+}(aq)][SO_{4}^{2}]
[HSO_{4}^{}(aq)] 
1.2 x 10^{2} 
= 
[X][X]
[0.050  X] 
0 
= 
X^{2} + 1.2 x 10^{2}X 6.00 x 10^{4} 
X 
= 
b ± √(b^{2} 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 = log_{10}[H^{+}(aq)] = log_{10}[0.019] = 1.72
6.00 mL 0.10 mol L^{1} NaOH(aq) added to 5.00 mL 0.010 H_{2}SO_{4} at 25^{o}C
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:

H_{2}SO_{4} 
→ 
H^{+}(aq) 
+ 
HSO_{4}^{} 
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 H_{2}SO_{4}
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 HSO_{4}^{}(aq) produced from the dissociation of H_{2}SO_{4} and OH^{}(aq):
OH^{}(aq) + HSO_{4}^{}(aq) → H_{2}O(l) + SO_{4}^{2}(aq)
1.00 x 10^{4} moles OH^{}(aq) reacts with 1.00 x 10^{4} moles HSO_{4}^{}(aq)
moles HSO_{4}^{} in solution = 5.00 x 10^{4}  1.00 x 10^{4} = 4.00 x 10^{4} mol

HSO_{4}^{}(aq) 

H^{+}(aq) 
+ 
SO_{4}^{2}(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 HSO_{4}^{}(aq):

HSO_{4}^{}(aq) 

H^{+}(aq) 
+ 
SO_{4}^{2}(aq) 
concentration / mol L^{1} 
0.036  X 

0 + X = X 

0 + X = X 
Calculate the value of X:
K_{a2} 
= 
[H^{+}(aq)][SO_{4}^{2}]
[HSO_{4}^{}(aq)] 
1.2 x 10^{2} 
= 
[X][X]
[0.036  X] 
0 
= 
X^{2} + 1.2 x 10^{2}X 4.32 x 10^{4} 
X 
= 
b ± √(b^{2} 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 = log_{10}[H^{+}(aq)] = log_{10}[0.016] = 1.80
7.00 mL 0.10 mol L^{1} NaOH(aq) added to 5.00 mL 0.010 H_{2}SO_{4} at 25^{o}C
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:

H_{2}SO_{4} 
→ 
H^{+}(aq) 
+ 
HSO_{4}^{} 
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 H_{2}SO_{4}
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 HSO_{4}^{}(aq) produced from the dissociation of H_{2}SO_{4} and OH^{}(aq):
OH^{}(aq) + HSO_{4}^{}(aq) → H_{2}O(l) + SO_{4}^{2}(aq)
2.00 x 10^{4} moles OH^{}(aq) reacts with 2.00 x 10^{4} moles HSO_{4}^{}(aq)
moles HSO_{4}^{} in solution = 5.00 x 10^{4}  2.00 x 10^{4} = 3.00 x 10^{4} mol

HSO_{4}^{}(aq) 

H^{+}(aq) 
+ 
SO_{4}^{2}(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 HSO_{4}^{}(aq):

HSO_{4}^{}(aq) 

H^{+}(aq) 
+ 
SO_{4}^{2}(aq) 
concentration / mol L^{1} 
0.025  X 

0 + X = X 

0 + X = X 
Calculate the value of X:
K_{a2} 
= 
[H^{+}(aq)][SO_{4}^{2}]
[HSO_{4}^{}(aq)] 
1.2 x 10^{2} 
= 
[X][X]
[0.025  X] 
0 
= 
X^{2} + 1.2 x 10^{2}X 3.00 x 10^{4} 
X 
= 
b ± √(b^{2} 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 = log_{10}[H^{+}(aq)] = log_{10}[0.012] = 1.91
8.00 mL 0.10 mol L^{1} NaOH(aq) added to 5.00 mL 0.010 H_{2}SO_{4} at 25^{o}C
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:

H_{2}SO_{4} 
→ 
H^{+}(aq) 
+ 
HSO_{4}^{} 
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 H_{2}SO_{4}
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 HSO_{4}^{}(aq) produced from the dissociation of H_{2}SO_{4} and OH^{}(aq):
OH^{}(aq) + HSO_{4}^{}(aq) → H_{2}O(l) + SO_{4}^{2}(aq)
3.00 x 10^{4} moles OH^{}(aq) reacts with 3.00 x 10^{4} moles HSO_{4}^{}(aq)
moles HSO_{4}^{} in solution = 5.00 x 10^{4}  3.00 x 10^{4} = 2.00 x 10^{4} mol

HSO_{4}^{}(aq) 

H^{+}(aq) 
+ 
SO_{4}^{2}(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 HSO_{4}^{}(aq):

HSO_{4}^{}(aq) 

H^{+}(aq) 
+ 
SO_{4}^{2}(aq) 
concentration / mol L^{1} 
0.015  X 

0 + X = X 

0 + X = X 
Calculate the value of X:
K_{a2} 
= 
[H^{+}(aq)][SO_{4}^{2}]
[HSO_{4}^{}(aq)] 
1.2 x 10^{2} 
= 
[X][X]
[0.015  X] 
0 
= 
X^{2} + 1.2 x 10^{2}X 1.8 x 10^{4} 
X 
= 
b ± √(b^{2} 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 = log_{10}[H^{+}(aq)] = log_{10}[0.0087] = 2.06
9.00 mL 0.10 mol L^{1} NaOH(aq) added to 5.00 mL 0.010 H_{2}SO_{4} at 25^{o}C
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:

H_{2}SO_{4} 
→ 
H^{+}(aq) 
+ 
HSO_{4}^{} 
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 H_{2}SO_{4}
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 HSO_{4}^{}(aq) produced from the dissociation of H_{2}SO_{4} and OH^{}(aq):
OH^{}(aq) + HSO_{4}^{}(aq) → H_{2}O(l) + SO_{4}^{2}(aq)
4.00 x 10^{4} moles OH^{}(aq) reacts with 4.00 x 10^{4} moles HSO_{4}^{}(aq)
moles HSO_{4}^{} in solution = 5.00 x 10^{4}  4.00 x 10^{4} = 1.00 x 10^{4} mol

HSO_{4}^{}(aq) 

H^{+}(aq) 
+ 
SO_{4}^{2}(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 HSO_{4}^{}(aq):

HSO_{4}^{}(aq) 

H^{+}(aq) 
+ 
SO_{4}^{2}(aq) 
concentration / mol L^{1} 
0.0071  X 

0 + X = X 

0 + X = X 
Calculate the value of X:
K_{a2} 
= 
[H^{+}(aq)][SO_{4}^{2}]
[HSO_{4}^{}(aq)] 
1.2 x 10^{2} 
= 
[X][X]
[0.0071  X] 
0 
= 
X^{2} + 1.2 x 10^{2}X 8.52 x 10^{5} 
X 
= 
b ± √(b^{2} 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 = log_{10}[H^{+}(aq)] = log_{10}[0.0050] = 2.30
10.00 mL 0.10 mol L^{1} NaOH(aq) added to 5.00 mL 0.010 H_{2}SO_{4} at 25^{o}C
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:

H_{2}SO_{4} 
→ 
H^{+}(aq) 
+ 
HSO_{4}^{} 
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 H_{2}SO_{4}
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 HSO_{4}^{}(aq) produced from the dissociation of H_{2}SO_{4} and OH^{}(aq):
OH^{}(aq) + HSO_{4}^{}(aq) → H_{2}O(l) + SO_{4}^{2}(aq)
5.00 x 10^{4} moles OH^{}(aq) reacts with 5.00 x 10^{4} moles HSO_{4}^{}(aq)
moles HSO_{4}^{} 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 SO_{4}^{2}(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 25^{o}C
Since we are now past the equivalence point for the reaction, we can simplify the reaction to:
H_{2}SO_{4}(aq) + 2NaOH(aq) → 2H_{2}O(l) + Na_{2}SO_{4}(aq)
Calculate moles and concentrations:

H_{2}SO_{4}(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 
log_{10}[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 25^{o}C
Since we are now past the equivalence point for the reaction, we can simplify the reaction to:
H_{2}SO_{4}(aq) + 2NaOH(aq) → 2H_{2}O(l) + Na_{2}SO_{4}(aq)
Calculate moles and concentrations:

H_{2}SO_{4}(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 
log_{10}[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 25^{o}C
Since we are now past the equivalence point for the reaction, we can simplify the reaction to:
H_{2}SO_{4}(aq) + 2NaOH(aq) → 2H_{2}O(l) + Na_{2}SO_{4}(aq)
Calculate moles and concentrations:

H_{2}SO_{4}(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 
log_{10}[OH^{}(aq)] 
= 
1.78 
pH 
14  pOH 
= 
12.22 