AP Chemistry · Topic 8.9
Henderson-Hasselbalch Equation Practice
Part of Acids and Bases.(SAP-10.C)
Practice questions
11
Sample questions
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Sample 1difficulty 2/5
A buffer with [HA] = [A⁻] has pH equal to
- A
7
- B
0
- Ccheck_circle
pKa
- D
pKw
Why
log(1) = 0 → pH = pKa.
- A
Sample 2difficulty 3/5
An NH₃/NH₄⁺ buffer is desired at pH 9.00. Kb(NH₃) = 1.8×10⁻⁵; Ka(NH₄⁺) = 5.6×10⁻¹⁰.
What ratio [NH₃]/[NH₄⁺] is needed?
- A
1.00
- B
1.80
- C
0.25
- Dcheck_circle
0.56
Why
pH = pKa + log([NH₃]/[NH₄⁺]). 9.00 = 9.25 + log(r), log r = -0.25, r = 0.56.
- A
Sample 3difficulty 3/5
At pH = pKa + 1, what is the ratio [A-]/[HA]?
- A
1
- Bcheck_circle
10
- C
0.1
- D
100
Why
pH - pKa = log([A-]/[HA]) = 1, so [A-]/[HA] = 10^1 = 10.
- A
Sample 4difficulty 3/5
Plot of fraction protonated (% HA) vs pH for a weak acid HA.
At what pH is HA exactly 50% protonated?
- Acheck_circle
pH = pKa
- B
pH = 14 - pKa
- C
pH = 7
- D
pH = pKa - 1
Why
From Henderson-Hasselbalch, when [HA] = [A-] (50% each), log(1) = 0, so pH = pKa.
- A
Sample 5difficulty 3/5
pH of a buffer is given by
- A
pH = pKa − [A⁻]/[HA]
- B
pH = pKb + log([A⁻]/[HA])
- C
pH = pKa + log([HA]/[A⁻])
- Dcheck_circle
pH = pKa + log([A⁻]/[HA])
Why
Henderson-Hasselbalch: pH = pKa + log([base]/[acid]).
- A