AP Chemistry Practice Quiz: Henderson-Hasselbalch Equation
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 7 questions to check your progress.
Question 1 of 7
All Questions (7)
A) 4.75
B) 7.00
C) Less than 4.75
D) Greater than 4.75
Correct Answer: A
According to the Henderson-Hasselbalch equation, pH = pKa + log([A⁻]/[HA]). When the concentrations of the conjugate base [A⁻] and the weak acid [HA] are equal, the ratio [A⁻]/[HA] is 1. The logarithm of 1 is 0. Therefore, the equation simplifies to pH = pKa.
A) 4.0
B) 5.0
C) 6.0
D) 7.0
Correct Answer: C
Using the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]). Given pKa = 5.0 and [A⁻]/[HA] = 10. The equation becomes pH = 5.0 + log(10). Since log(10) = 1, the pH is 5.0 + 1 = 6.0.
A) The concentration of the conjugate base is significantly greater than the concentration of the weak acid.
B) The concentration of the weak acid is significantly greater than the concentration of the conjugate base.
C) The concentrations of the weak acid and conjugate base are approximately equal.
D) The pKa of the weak acid has decreased.
Correct Answer: B
The equation is pH = pKa + log([A⁻]/[HA]). For the pH to be lower than the pKa, the term log([A⁻]/[HA]) must be negative. This occurs when the ratio [A⁻]/[HA] is less than 1, meaning the concentration of the weak acid [HA] is greater than the concentration of the conjugate base [A⁻].
A) The ratio [A⁻]/[HA] increases, and the pH increases significantly.
B) The ratio [A⁻]/[HA] decreases, and the pH decreases slightly.
C) The ratio [A⁻]/[HA] remains unchanged, and the pH is constant.
D) The pKa of the weak acid decreases, causing a drop in pH.
Correct Answer: B
Adding a strong acid (H⁺) will cause it to react with the conjugate base (A⁻) to form more weak acid (HA). This decreases the concentration of A⁻ and increases the concentration of HA. As a result, the ratio [A⁻]/[HA] decreases, and the log of this ratio becomes more negative, causing a slight decrease in the overall pH. The buffer's function is to ensure this change is not significant.
A) The molar concentration of the strong acid.
B) The molar concentration of the weak acid.
C) The molar concentration of the conjugate base.
D) The acid dissociation constant.
Correct Answer: C
The Henderson-Hasselbalch equation relates the pH of a buffer to the pKa of the weak acid and the ratio of the concentrations of its components. In this equation, [HA] represents the molar concentration of the weak acid, and [A⁻] represents the molar concentration of its conjugate base.
A) An acid with a pKa of 4.1 and a [base]/[acid] ratio of approximately 1:1.
B) An acid with a pKa of 6.0 and a [base]/[acid] ratio of approximately 1:100.
C) An acid with a pKa of 4.1 and a [base]/[acid] ratio of approximately 10:1.
D) An acid with a pKa of 7.0 and a [base]/[acid] ratio of approximately 1:1.
Correct Answer: A
An effective buffer has a pH close to the pKa of the weak acid, which occurs when the ratio of the conjugate base to the weak acid, [A⁻]/[HA], is close to 1. To create a buffer with a pH of 4.0, it is best to choose a weak acid with a pKa close to 4.0. An acid with a pKa of 4.1 and a ratio near 1:1 (where pH ≈ pKa) is the most suitable choice for maintaining a stable buffer system.
A) The added acid or base is completely neutralized by water.
B) The pKa of the weak acid component changes to counteract the added substance.
C) The ratio of the concentrations of the conjugate acid-base pair does not change significantly.
D) The total volume of the solution increases, diluting the added acid or base.
Correct Answer: C
A buffer works because it contains both a weak acid (to neutralize added base) and its conjugate base (to neutralize added acid). When a small amount of acid or base is added, it reacts with one of the buffer components. This causes a small change in the individual concentrations of [HA] and [A⁻], but the ratio [A⁻]/[HA] does not change drastically. According to the Henderson-Hasselbalch equation, since the logarithm of this ratio changes only slightly, the pH also changes only slightly.