AP®ChemistryCollege BoardExam QuestionsUnit 8: Acids & BasesBuffers

Buffers (College Board AP® Chemistry): Exam Questions

59 mins17 questions

Select a download format for Buffers

Scroll for more

Select an answer set to view for
Buffers

Easy
Medium
Hard
11 mark

Which mixture will not act as a buffer?

  • CH3COOH (aq) and CH3COONa (aq) 

  • CH3CH2COOH (aq) and CH3CH2COONa (aq) 

  • HNO3 (aq) and NaOH (aq)

  • NH3 (aq) and NH4Cl (aq) 

Choose your answer

21 mark

A buffer at pH 4.40 is prepared from a weak acid with a pKa = 4.60. If 50 mL of this buffer is diluted to 100 mL with distilled water, what is the pH of the dilute solution?

  • 5.40

  • 4.50

  • 4.60

  • 4.40

Choose your answer

11 mark

Which statement about buffer solutions is not correct?

  • A buffer resists small changes in pH.

  • A buffer can be formed from sodium hydroxide and an excess of ethanoic acid.

  • The pH during a titration between a strong acid and weak base changes slowly in the buffer region.

  • When a strong acid is added to a buffer, hydrogen ions react with salt ions. 

Choose your answer

21 mark

A buffer with a pH of 5.00 is needed. Which of the following should be used?

  • Ammonia with a Kb of 1.8 x 10-5.

  • Ethanoic acid with a Ka of 1.8 x 10-5.

  • Carbonic(II) acid with a Ka of 4.7 x 10-11.

  • Nitrous acid with a Ka of 4.0 x 10-4.

Choose your answer

31 mark

Adding which of the following to a 50. mL sample of 0.10 M CH3COOH (aq) (pKa = 4.76) would result in the formation of a buffer solution with a pH of 4.76?

  • 50 mL of 0.10 M NaOH (aq)

  • 25 mL of 0.10 M CH3COONa (aq)

  • 50 mL of 0.10 M CH3COONa (aq)

  • 25 mL of 0.10 M NaOH (aq)

Choose your answer

11 mark

The weak acid CH3COOH has a pKa of 4.76.

A solution is prepared by mixing 250 mL of 0.200 M CH3COOH (aq) and 0.8 g of NaOH (s).

Which of the following can be used to calculate the pH of the solution?

  • pH = 4.76 + log10 fraction numerator stretchy left square bracket 0.03 stretchy right square bracket over denominator stretchy left square bracket 0.02 stretchy right square bracket end fraction = 4.93

  • pH = 4.76 + log10 fraction numerator stretchy left square bracket 0.08 stretchy right square bracket over denominator stretchy left square bracket 0.02 stretchy right square bracket end fraction = 6.36

  • pH = 4.76 + log10 fraction numerator stretchy left square bracket 0.02 stretchy right square bracket over denominator stretchy left square bracket 0.03 stretchy right square bracket end fraction = 4.58

  • pH = 4.76 + log10 fraction numerator stretchy left square bracket 0.02 stretchy right square bracket over denominator stretchy left square bracket 0.05 stretchy right square bracket end fraction = 4.36

Choose your answer

21 mark

Acid

Ka at 298 K

acetic acid

1.7 x 10-5

formic acid

1.8 x 10-4

The correct solution to form a buffer with a pH of 4.77 is:

  • 0.10 M acetic acid 0.50 L with 0.10 M sodium hydroxide 0.25 L

  • 0.10 M acetic acid 0.50 L with 0.20 M sodium hydroxide 0.25 L

  • 0.10 M formic acid 0.50 L with 0.10 M sodium hydroxide 0.25 L

  • 0.10 M formic acid 0.50 L with 0.20 M sodium hydroxide 0.25 L

Choose your answer