THE ANSWER OF ACID BASE TITRATION PROBLEMS

1. 25 mL of HCl(aq) is titrated with 0,10 M NaOH(aq). If the volume of NaOH needed is 30 mL,
a. Write down the ionic equation.
H+(aq) + OH-(aq) --> H2O(l)
b. What is the name of this reaction? Give the reason.
Neutralisation reaction. Because the H+ ions that give the
properties of acid react with the OH- ions that give the
properties of base form H2O which is neutral.
c. Calculate [HCl]
M1V1 = M2V2
M(HCl) x 25 = 0,10 x 30
M(HCl) = 0,12 mol/L
d. Calculate the pH before titration.
pH (HCl) = - log 0,12 = 2 – log (3 x 4)
= 2 – (log3 + 2log2)
= 2 – 1,1 = 0,9.
e. Calculate the pH of the equivalent point.
Strong acid titrated with strong base,
[H+] = [OH-]= √Kw = 10-7. pH = 7 (neutral)
f. Determine the indicator used in this titration.
The pH changes from 0,9 to 7, because at the equivalent
point the solution is neutral. So the appropriate indicator is
MM, pH range is 4,2 – 6,3. The color changes from red to yellow.
g. When the volume of NaOH(aq) which titrated to HCl(aq) is
15 mL, calculate the pH of solution.
H+(aq) + OH—(aq) --> H2O(l)
Initial 25 mL 0,12M 15 mL 0,10 M
3 mmol 1,5 mmol
Reaction 1,5 mmol 1,5 mmol
Final 1,5mmol 0
[H+] = 1,5 mmol/ (25 + 15) mL = 15/4 x 10-2 M
pH = - log 15/4 x 10-2 = 2 – (log 3 + log 5 – 2log2)
= 2 – (0,5 + 0,7 – 0,6) = 1,4.
h. If the volume of NaOH(aq) which titrated to HCl(aq) is 50
mL, calculate the pH of solution.
H+(aq) + OH—(aq) --> H2O(l)
Initial 25 mL 0,12M 50 mL 0,10 M
3 mmol 5 mmol
Reaction 3 mmol 3 mmol
Final 0 2 mmol
[OH--] = 2 mmol/ (25 + 50) mL = 2/75 M
pOH = - log 2/75 = – (log 2 - log 3 – 2log5)
= - 0,3 + 0,5 + 1,4 = 1,6.
pH = 14 – 1,6 = 12,4.
i. Draw the curve of the titration above.
















2. 25 mL of NaOH(aq) is titrated with 0,10 M HCl(aq). If the volume of NaOH needed is 20 mL,
a. Calculate [NaOH]
M1V1 = M2V2
M(NaOH) x 25 = 0,10 x 20
M(NaOH) = 0,08 mol/L
b. Calculate the pH before titration.
pOH(NaOH) = - log 0,08 = 2 – 3log2 = 2 – 0,9 = 1,1
pH(NaOH) = 14 – 1,1 = 12,9.
c. Calculate the pH of the equivalent point.
Strong base titrated with strong acid, at the equivalent point [H+] = [OH-]= √Kw = 10-7.
pH = 7 (neutral)
d. Determine the indicator used in this titration.
The pH changes from 12,9 to 7, because at the equivalent point the solution is neutral. So the appropriate indicator is PP, pH range is 8,3 – 10,0. The color changes from red to colorless.
e. When the volume of HCl(aq) which titrated to the base is 15 mL, calculate the pH of solution.
H+(aq) + OH—(aq) --> H2O(l)
Initial 10 mL 0,10M 25 mL 0,08 M
1 mmol 2 mmol
Reaction 1 mmol 1 mmol
Final 0 1 mmol
[OH-] = 1 mmol/ (25 + 15) mL = 1/4 x 10-1 M
pOH = - log 1/4 x 10-1 = 1 – (– 2log2) = 1 + 0,6 = 1,6.
pH = 14 – 1,6 = 12,4.
f. If the volume of HCl(aq) which titrated to the base is 25 mL, calculate the pH of solution.
H+(aq) + OH—(aq) --> H2O(l)
Initial 25 mL 0,10M 25 mL 0,08 M
2,5 mmol 2 mmol
Reaction 2 mmol 2 mmol
Final 0,5 mmol 0
[H+] = 0,5 mmol/ (25 + 25) mL = 0,5/50 M = 10-2 M
pH = - log 10-2 = 2.
g. Draw the curve of the titration above.





























3. 25 mL of CH3COOH(aq) is titrated with 0,10 M NaOH(aq). If the volume of NaOH needed is 30 mL. Ka = 2 x 10-5
a. Write down the ionic equation.
CH3COOH(aq) + OH-(aq) --> CH3COO-(aq) + H2O(l)
b. Calculate [CH3COOH]
M1V1 = M2V2
M(CH3COOH) x 25 = 0,10 x 30
M(CH3COOH) = 0,12 mol/L
c. Calculate the pH before titration.
[H+] = √Ka.Ma
[H+]= √2.10-5.0,12 = √24.10-7
pH = - ½ log 24.10-7
= 3,5 – ½ (log3 + 3 log2)
= 3,5 – ½ (0,5 + 3.0,3)
= 3,5 – 0,7
pH = 2,8.
d. Explain what happens at the equivalent point. Write down the ionic equation.
At the equivalent CH3COO-(aq), as a product, is hydrolyzed to form CH3COOH which is more stable than acetic ion.
CH3COO-(aq) + H2O(l) ↔ CH3COOH(aq) + OH-(aq)
e. Calculate the pH of the equivalent point.
n(CH3COO-) = n(NaOH) = 0,10 mmol/mL. 30 mL = 3 mmol. Vol. = 55 mL
[OH-] = √Kw/Ka x [CH3COOH]
= √ 10-14/2.10-5 x 3/55 = √ 10-10 x 3/11
pOH = - log √ 3.10-11 = 5,5 -1/2 log 3 = 5,5 – 0,25 = 5,25
pH = 14 – 5,25 = 8,75.
f. Determine the indicator used in this titration.
The pH changes from 2,8 to 8,75. So the appropriate indicator is PP which the pH range 8,3 – 10,0. The color changes from colorless to pink.
g. When the volume of NaOH(aq) which titrated to the acid is 15 mL, calculate the pH of solution.
CH3COOH(aq) + OH—(aq) --> CH3COO-(aq)+H2O(l)
Initial 25 mL 0,12M 15 mL 0,10M
3 mmol 1,5 mmol
Reaction 1,5 mmol 1,5 mmol 1,5 mmol
Final 1,5 mmol 0 1,5 mmol
[H+] = Ka x n acid / n base
[H+] = 2 x 10-5 x 1,5 / 1,5 = 2 x 10-5
pH = 5 – log 2 = 5 – 0,3 = 4,7.
h. If the volume of NaOH(aq) which titrated to the acid is 50 mL, calculate the pH of solution.
CH3COOH(aq)+OH—(aq) -->CH3COO-(aq)+H2O(l)
Initial 25 mL 0,12M 50 mL 0,10M
3 mmol 5 mmol
Reaction 3 mmol 3 mmol
Final 0 2 mmol
[OH-] = 2 / 75
pOH = – log 2 / 75
= – log2 + log 3 + 2 log5
= - 0,3 + 0,5 + 2 x 0,7 = 1,6.
pH = 14 – 1,6 = 12,4.
i. Draw the curve of the titration above.