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^{o}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.


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) 
+ 
H_{2}O(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) → H_{2}O(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 = log_{10}[OH^{}(aq)] = log_{10}[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}) x volume (L)
n(HCl) = c(HCl) x V(HCl)
c(HCl) = 0.10 mol L^{1}
V(HCl) = 1.00 mL = 1.00 x 10^{3} L
n(HCl) = 0.10 x 1.00 x 10^{3} = 1.00 x 10^{4} mol
 Calculate moles NaOH unreacted = initial moles NaOH  moles NaOH reacted
initial moles NaOH = concentration (mol L^{1}) x volume (L)
initial moles NaOH = c(NaOH) x V(NaOH)
c(NaOH) = 0.10 mol L^{1}
V(NaOH) = 10.00 mL = 10.00 x 10^{3} L
initial moles NaOH = 0.10 x 10.00 x 10^{3} = 1.00 x 10^{3} mol
moles NaOH reacted = moles HCl added = 1.00 x 10^{4} mol
moles NaOH unreacted = 1.00 x 10^{3}  1.00 x 10^{4} = 9.00 x 10^{4} mol
 Calculate [OH^{}] = moles(unreacted OH^{}) ÷ total volume of solution
moles(unreacted OH^{}) = moles(unreacted NaOH) = 9.00 x 10^{4} mol
total volume of solution = 10.00 mL + 1.00 mL = 11.00 mL = 11.00 x 10^{3} L
[OH^{}] = 9.00 x 10^{4} ÷ 11.00 x 10^{3} = 0.082 mol L^{1}
 Calculate pOH:
pOH = log_{10}[OH^{}] = log_{10}[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 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.
 moles HCl: n(HCl) = concentration (mol L^{1}) x volume (L)
n(HCl) = c(HCl) x V(HCl)
c(HCl) = 0.10 mol L^{1}
V(HCl) = 11.00 mL = 11.00 x 10^{3} L
n(HCl) = 0.10 x 11.00 x 10^{3} = 1.10 x 10^{3} mol
 Calculate moles HCl in excess
moles(HCl) unreacted = total moles(HCl)  moles(HCl) reacted
total moles(HCl) = 1.10 x 10^{3} mol
moles(HCl) reacted = moles(NaOH) = 1.00 x 10^{3} mol
moles(HCl) unreacted = 1.10 x 10^{3}  1.00 x 10^{3} = 1.00 x 10^{4} mol
 Calculate [H^{+}]:
[H^{+}] = moles(H^{+} unreacted) ÷ total volume of solution
n(H^{+}) unreacted = n(HCl) unreacted = 1.00 x 10^{4} mol
total volume = 10.00 mL + 11.00 mL = 21.00 mL = 21.00 x 10^{3} L
[H^{+}] = 1.00 x 10^{4} ÷ 21.00 x 10^{3} = 4.76 x 10^{3} mol L^{1}
 Calculate pH of the solution
pH = log_{10}[H^{+}] = log_{10}[4.76 x 10^{3}] = 2.32
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 = log_{10}[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 = log_{10}[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)} + H_{2}O_{(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.

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^{o}C 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^{o}C.
3. We will assume additivity of volumes.