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AP Chemistry Practice Quiz: pH and pOH of Strong Acids and Bases

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 9 questions to check your progress.

Question 1 of 9

What is the pH of a 0.010 M solution of hydrochloric acid (HCl)?

All Questions (9)

What is the pH of a 0.010 M solution of hydrochloric acid (HCl)?

A) 1.0

B) 2.0

C) 12.0

D) 13.0

Correct Answer: B

HCl is a strong acid, so it completely ionizes. Therefore, the hydronium ion concentration [H3O+] is equal to the initial acid concentration, 0.010 M. The pH is calculated as -log[H3O+], so pH = -log(0.010) = 2.0.

A solution is prepared with 0.1 M potassium hydroxide (KOH). What is the pOH of the solution?

A) 1.0

B) 7.0

C) 13.0

D) 14.0

Correct Answer: A

KOH is a strong base (a Group I hydroxide) and dissociates completely. The hydroxide ion concentration [OH-] is equal to the initial base concentration, 0.1 M. The pOH is calculated as -log[OH-], so pOH = -log(0.1) = 1.0.

What is the pH of a 0.01 M solution of sodium hydroxide (NaOH)?

A) 2.0

B) 7.0

C) 12.0

D) 14.0

Correct Answer: C

NaOH is a strong base, so [OH-] = 0.01 M. First, calculate the pOH: pOH = -log[OH-] = -log(0.01) = 2.0. Then, use the relationship pH + pOH = 14 to find the pH: pH = 14 - pOH = 14 - 2.0 = 12.0.

Calculate the pOH of a 0.001 M solution of nitric acid (HNO3).

A) 3.0

B) 7.0

C) 11.0

D) 14.0

Correct Answer: C

HNO3 is a strong acid, so [H3O+] = 0.001 M. First, calculate the pH: pH = -log[H3O+] = -log(0.001) = 3.0. Then, use the relationship pH + pOH = 14 to find the pOH: pOH = 14 - pH = 14 - 3.0 = 11.0.

What is the pH of a 0.05 M solution of barium hydroxide, Ba(OH)2?

A) 1.3

B) 1.0

C) 12.7

D) 13.0

Correct Answer: D

Ba(OH)2 is a strong base and a Group II hydroxide, so it releases two OH- ions for each formula unit. The [OH-] is double the initial base concentration: [OH-] = 2 * 0.05 M = 0.1 M. The pOH = -log(0.1) = 1.0. The pH = 14 - pOH = 14 - 1.0 = 13.0.

Which of the following strong acids, all at 0.1 M concentration, will result in a solution with a pH of 1.0?

A) HCl only

B) H2SO4 only

C) HCl and HNO3 only

D) HCl, HBr, HI, HClO4, and HNO3

Correct Answer: D

All listed acids (HCl, HBr, HI, HClO4, HNO3) are strong monoprotic acids. They completely ionize in solution, meaning a 0.1 M solution of any of them will produce an [H3O+] of 0.1 M. The pH is -log(0.1) = 1.0 for all of them. H2SO4 is a strong acid, but it is diprotic, which complicates the calculation.

What is the concentration of hydroxide ions, [OH-], in a 0.01 M solution of perchloric acid (HClO4)?

A) 1.0 x 10^-12 M

B) 1.0 x 10^-7 M

C) 1.0 x 10^-2 M

D) 12.0 M

Correct Answer: A

HClO4 is a strong acid, so [H3O+] = 0.01 M. The pH is -log(0.01) = 2.0. The pOH is 14 - pH = 14 - 2.0 = 12.0. The hydroxide concentration is found by [OH-] = 10^-pOH = 10^-12 M.

A solution of strontium hydroxide, Sr(OH)2, has a pH of 13.0. What is the molar concentration of the Sr(OH)2 solution?

A) 0.1 M

B) 0.05 M

C) 1.0 M

D) 0.2 M

Correct Answer: B

If the pH is 13.0, the pOH is 14.0 - 13.0 = 1.0. The hydroxide concentration [OH-] is 10^-pOH = 10^-1.0 = 0.1 M. Since Sr(OH)2 is a Group II hydroxide, it dissociates to produce two OH- ions per formula unit. Therefore, the initial concentration of Sr(OH)2 is half the [OH-], which is 0.1 M / 2 = 0.05 M.

A student incorrectly calculates the pH of a 0.005 M Ca(OH)2 solution as 11.7. What was the most likely error in the calculation?

A) The student calculated pH directly from the initial base concentration.

B) The student calculated pOH but reported it as pH.

C) The student forgot to double the hydroxide concentration for the Group II base.

D) The student used 10 instead of 14 for the pH + pOH relationship.

Correct Answer: C

The correct calculation is: [OH-] = 2 * [Ca(OH)2] = 2 * 0.005 M = 0.01 M. pOH = -log(0.01) = 2. pH = 14 - 2 = 12. The student's incorrect calculation was likely: [OH-] = 0.005 M (error). pOH = -log(0.005) ≈ 2.3. pH = 14 - 2.3 = 11.7. The error was failing to account for the two hydroxide ions produced by Ca(OH)2.