Practice Titration in Chemistry

Use these practice problems to test your method after reviewing the concept explanation and worked examples.

Quick Recap

A lab technique for finding an unknown solution concentration by gradually adding a solution of known concentration until the reaction is complete.

Slowly adding a known solution to an unknown one until the reaction is just complete — the volume used reveals the concentration.

Showing a random 20 of 50 problems.

Example 1

medium
How many moles of HClHCl are in 40 mL40\text{ mL} of 0.25 M0.25\text{ M} HClHCl?

Example 2

medium
15.0 mL15.0\text{ mL} of H2SO4H_2SO_4 requires 40.0 mL40.0\text{ mL} of 0.150 M0.150\text{ M} NaOHNaOH to fully neutralize. Find the acid concentration.

Example 3

medium
A 0.250 g0.250\text{ g} sample of KHPKHP (M=204 g/molM = 204\text{ g/mol}) is titrated with NaOHNaOH, requiring 22.0 mL22.0\text{ mL} (1:11:1). Find [NaOH][NaOH].

Example 4

medium
28.5 mL28.5\text{ mL} of 0.120 M0.120\text{ M} NaOHNaOH neutralizes 25.0 mL25.0\text{ mL} of acetic acid (1:11:1). Find [CH3COOH][CH_3COOH].

Example 5

easy
40 mL40\text{ mL} of 0.10 M0.10\text{ M} HClHCl needs how many mL of 0.20 M0.20\text{ M} NaOHNaOH (1:11:1)?

Example 6

medium
25 mL25\text{ mL} of H2SO4H_2SO_4 is neutralized by 50 mL50\text{ mL} of 0.10 M0.10\text{ M} NaOHNaOH. Find the acid concentration. (H2SO4+2NaOHNa2SO4+2H2OH_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O)

Example 7

easy
True or false: the endpoint and equivalence point of a titration are always identical.

Example 8

medium
15 mL15\text{ mL} of 0.20 M0.20\text{ M} NaOHNaOH neutralizes 30 mL30\text{ mL} of acetic acid (1:11:1). Find the acid concentration.

Example 9

medium
32.0 mL32.0\text{ mL} of 0.250 M0.250\text{ M} H2SO4H_2SO_4 requires VV mL of 0.400 M0.400\text{ M} KOHKOH at equivalence (1:21:2). Find VV.

Example 10

easy
A titration uses 30 mL30\text{ mL} of 0.10 M0.10\text{ M} NaOHNaOH to neutralize 20 mL20\text{ mL} of HClHCl. Find the HClHCl concentration (1:11:1).

Example 11

medium
What volume of 0.200 M0.200\text{ M} HClHCl is needed to neutralize 30.0 mL30.0\text{ mL} of 0.100 M0.100\text{ M} Ba(OH)2Ba(OH)_2? (Ba(OH)2+2HClBaCl2+2H2OBa(OH)_2 + 2HCl \rightarrow BaCl_2 + 2H_2O)

Example 12

medium
A 0.500 g0.500\text{ g} sample of KHPKHP (M=204 g/molM=204\text{ g/mol}) is titrated with NaOHNaOH. How many moles of NaOHNaOH are needed (1:11:1)?

Example 13

hard
30.0 mL30.0\text{ mL} of 0.150 M0.150\text{ M} HClHCl is mixed with 20.0 mL20.0\text{ mL} of 0.200 M0.200\text{ M} NaOHNaOH. Find the pH of the resulting solution.

Example 14

easy
In MAVA=MBVBM_AV_A = M_BV_B, if both volumes are equal, how do the concentrations compare (1:11:1)?

Example 15

medium
50.0 mL50.0\text{ mL} of 0.0500 M0.0500\text{ M} H2SO4H_2SO_4 needs how many mL of 0.100 M0.100\text{ M} NaOHNaOH at equivalence (1:21:2)?

Example 16

easy
20 mL20\text{ mL} of 0.10 M0.10\text{ M} acid is titrated with 0.10 M0.10\text{ M} base (1:11:1). What volume of base reaches equivalence?

Example 17

easy
The ____ point is where the indicator just changes color; the ____ point is where moles of acid equal moles of base.

Example 18

easy
In a 1:11:1 acid-base titration, 25 mL25\text{ mL} of 0.10 M0.10\text{ M} acid is neutralized by 0.20 M0.20\text{ M} base. What volume of base is needed?

Example 19

easy
At the equivalence point of a weak acid - strong base titration, is the pH less than, equal to, or greater than 7?

Example 20

hard
A 0.450 g sample of impure KHPKHP (M=204M = 204) requires 20.5 mL20.5\text{ mL} of 0.100 M0.100\text{ M} NaOHNaOH (1:11:1). Find the percent purity.