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.

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.
b. What is the name of this reaction? Give the reason.
c. Calculate [HCl]
d. Calculate the pH before titration.
e. Calculate the pH of the equivalent point.
f. Determine the indicator used in this titration.
g. When the volume of NaOH(aq) which titrated to HCl(aq) is 15 mL, calculate the pH of solution.
h. If the volume of NaOH(aq) which titrated to HCl (aq) is 50 mL, calculate the pH of solution.
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 acid needed is 20 mL,

a. Calculate [NaOH]
b. Calculate the pH before titration.
c. Calculate the pH of the equivalent point.
d. Determine the indicator used in this titration.
e. When the volume of HCl(aq) which titrated to the base is 15 mL, calculate the pH of solution.
f. If the volume of HCl(aq) which titrated to the base is 25 mL, calculate the pH of solution.
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,

a. Write down the ionic equation.
b. Calculate [CH3COOH]
c. Calculate the pH before titration.
d. Explain what happens at the equivalent point. Write down the ionic equation.
e. Calculate the pH of the equivalent point.
f. Determine the indicator used in this titration.
g. When the volume of NaOH(aq) which titrated to the acid is 15 mL, calculate the pH of solution.
h. If the volume of NaOH(aq) which titrated to the acid is 50 mL, calculate the pH of solution.
i. Draw the curve of the titration above.

ACID BASE TITRATION QUIZ

1. 50 mL of HCl(aq) is titrated with 0,10 M NaOH(aq). If the volume of NaOH needed is 60 mL, the [HCl] is ….

a. 0,08 M

b. 0,01 M

c. 0,12 M

d. 0,15 M

e. 0,20 M

2. The pH of the equivalent point is ….

a. 2

b. 5

c. < 7

d. 7

e. > 7

3. When the volume of NaOH which added to HCl(aq) is 25 mL, the pH of solution is ….

a. 1 – log 3

b. 1 + log 3

c. > 1

d. < 7

e. 7 + log 3

4. The pH after the addition of 50 mL of NaOH(aq) is ….

a. > 2

b. 2

c. 5

d. 7

e. < 7

5. The pH after the addition of 70 mL of NaOH(aq) is ….

a. 3 – 3log2

b. 3 + 3log2

c. 7

d. 11 – 3log2

e. 11 + 3log2

6. Which indicator would probably not work for detecting the endpoint, when strong acid is titrated with a strong base?

a. bromocresol purple, with a color change in the range pH = 5.2 to

b. thymol blue, with a color change in the range pH = 8.0 to 9.6

c. bromophenol blue, with a color change in the range pH = 3.0 to 4.

d. bromothymol blue, with a color change in the range pH = 6.0 to 7.6

7. If the above titration uses other indicators, which indicator would be worked?

a. MO, with a color change in the range pH = 3,1 to 4,4

b. MR, with a color change in the range pH = 4,2 to 6,2

c. PP, with a color change in the range pH = 8.3 to 10.0

d. MO or PP

Consider the following indicator ranges.

MO 3,1 – 4,4 (red – yellow)

MR 4,4 – 6,2 (red – yellow)

BTB 6,2 – 7,6 (yellow – blue)

PP 8,0–10,0 (colorless–red)

8. Which indicator(s) would be the best choice for the titration of strong acid that is titrated with strong base?

a. Methyl orange and methyl red would work best.

b. Bromothymol blue and phenolphthalein would work.

c. Methyl red would be best.

d. litmus

e. Phenolphthalein would work best.

9. Which indicator(s) would be the best choice for the titration of strong base that is titrated with strong acid?

a. MO

b. MR

c. BTB

d. PP

e. litmus

10. Which indicator(s) would be the best choice for the titration of weak acid that is titrated with strong base?

a. MO

b. MR

c. BTB

d. PP

e. litmus

11. It would require ... mL of 0.30 M hydrochloric acid solution to neutralize 20.0 mL of 0.30 M sodium hydroxide solution.

a. 15.0

b. 10.0

c. 40.0

d. 30.0

e. 20.0

12. 200.0 mL of 0.200 M HCl is titrated with 0.050 M NaOH. What is the pH after the addition of 100.0 mL of the NaOH solution?

a. 1,45

b. 1,03

c. 0,93

d. 0,82

e. 0,76

13. What is the molarity of a potassium hydroxide solution if 50.0 mL of it neutralizes 25.0 mL of a 0.250 M solution of sulfuric acid?

a. 0.10 M

b. 0.15 M

c. 0.20 M

d. 0.25 M

e. 0.30 M

14. It would require ... mL of 0.300 N sulfuric acid solution to neutralize 20.0 mL of 0.300 N sodium hydroxide solution.

a. 40.0

b. 10.0

c. 20.0

d. 30.0

e. 15.0

15. How many grams of solid sodium hydroxide may be neutralized by 150 mL of 0.150 HCl?

a. 90g

b. 1.11g

c. 9g

d. 11.2g

e. 0.9g

16. Calculate the concentration of a 25.0 mL sample of mono-protic acid if it took 31.6mL of 0.110 M NaOH to neutralize it completely.

a. 0.139

b. 0.159

c. 0.087

d. 0.87

e. 11.5

17. The stoichiometric point in an acid-base titration ….

a. will always have a pH of 7.00

b. will have a pH of 7.00 when it involves a strong acid and strong base

c. will have a pH of 7.00 when a weak acid and a weak base are used

d. will have a pH above 7.00 when a weak base and a strong acid are used

e. will have a pH below 7.00 when a weak acid and a strong base are used

18. A 25.00 mL sample of 0.100 M HCl is titrated with 0.100 M NaOH. What is the pH of the solution at the points where 24.5 and 25.5 mL of NaOH have been added.

a. 3.30, 10.70

b. 3.30, 10.00

c. 3.30, 11.00

d. 3.00, 11.00

e. 3.00, 10.00

19. What is the pH of 100.0 mL of 0.100 M HCl after 20.0 mL of 0.100 M NaOH have been added?

a. 0.98

b. 1.17

c. 1.86

d. 2.02

e. 2.42

20. What is the pH of 100.0 mL of 0.100 M HCl after 100.0 mL of 0.100 M NaOH have been added?

a.0.00

b.1.00

c.6.86

d.7.00

e.7.25

21.What is the pH of 100.0 mL of 0.100 M HCl after 150.0 mL of 0.100 M NaOH have been added?

a. 1.70

b. 5.89

c. 8.98

d. 10.78

e. 12.30

22.The pH of a titration of 0.10M HCl by 0.150M NaOH at equivalence point is ....

a. 6.85

b. 7.00

c. 7.15

d. 7.36

e. 8.20

23. What is the pH of a solution in which 25.0 mL of 0.010 M Sr(OH)2 are added to 20.0 mL of 0.020 M H2SO4?

a. 11.82

b. 11.52

c. 2.52

d. 2.48

e. 1.76

24. A 1.20 g sample of fumaric acid is dissolved in 100.0 mL of water and titrated with 0.300 M NaOH to the second equivalence point. The volume of base used is 69.0 mL. What is the molecular mass of fumaric acid which contains two dissociable protons?

a. 84

b. 116

c. 142

d. 58

e. 232

25. A 1.50 g sample of Vitamin C is dissolved in 100.0 mL of water and titrated with 0.250 M NaOH to the methyl orange equivalence point. The volume of the base used is 34.1 mL. What is the molecular mass of Vitamin C assuming one dissociable proton per molecule?

a. 139

b. 176

c. 146

d. 164

e. 152

26. 0.550 g sample of butyric acid is dissolved in 100.0 mL of water and titrated with 0.100 M NaOH to the phenolphthalein equivalence point. The volume of the base used is 62.4 mL. What is the molecular mass of butyric acid assuming one dissociable proton per molecule?

a. 140

b. 88

c. 122

d. 160

e. 46

27. A 25.00 mL sample of 0.100 M CH3CO2H is titrated with 0.100 M NaOH. What is the pH of the solution at the points where 24.0 and 24.5 mL of NaOH have been added. (Ka = 1.8x10-5)

a. 5.32, 6.13

b. 5.61, 6.44

c. 5.61, 6.13

d. 6.13, 7.00

e. 6.13, 6.43

28. A 25.00 mL sample of 0.100 M CH3CO2H is titrated with 0.100 M NaOH. What is the pH of the solution at the points where 24.0 and 26.5 mL of NaOH have been added. (Ka=1.8 x 10-5)

a. 5.61, 10.30

b. 5.61, 11.30

c. 6.44, 10.7

d. 6.13, 11.46

e. 6.13, 10.3