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Henderson–Hasselbalch worksheet

Grade 12 to college · CCSS HSF.BF.B.5

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Solve for: Base-to-acid ratio for a target pH
- pKa of the weak acid: 5
- Target pH: 5
Find: Ratio [A⁻] ÷ [HA]
Answer
2.
Problem 2.
- Solve for: Concentrations for a target pH
- pKa of the weak acid: 4
- Target pH: 4
- Total buffer concentration ([HA] + [A⁻]): 0.1 mol/L
Find: Conjugate base [A⁻]
Answer
3.
Problem 3.
- Solve for: Buffer pH
- pKa of the weak acid: 5.25
- Conjugate base [A⁻]: 0.15
- Weak acid [HA]: 0.1
Find: Buffer pH
Answer
4.
Problem 4.
- Solve for: Concentrations for a target pH
- pKa of the weak acid: 4.25
- Target pH: 4
- Total buffer concentration ([HA] + [A⁻]): 0.1 mol/L
Find: Conjugate base [A⁻]
Answer
5.
Problem 5.
- Solve for: Base-to-acid ratio for a target pH
- pKa of the weak acid: 6
- Target pH: 5
Find: Ratio [A⁻] ÷ [HA]
Answer
6.
Problem 6.
- Solve for: Concentrations for a target pH
- pKa of the weak acid: 6.5
- Target pH: 4
- Total buffer concentration ([HA] + [A⁻]): 0.1 mol/L
Find: Conjugate base [A⁻]
Answer
7.
Problem 7.
- Solve for: Buffer pH
- pKa of the weak acid: 5.5
- Conjugate base [A⁻]: 0.11
- Weak acid [HA]: 0.1
Find: Buffer pH
Answer
8.
Problem 8.
- Solve for: Buffer pH
- pKa of the weak acid: 5.6
- Conjugate base [A⁻]: 0.15
- Weak acid [HA]: 0.1
Find: Buffer pH
Answer
9.
Problem 9.
- Solve for: Buffer pH
- pKa of the weak acid: 4
- Conjugate base [A⁻]: 0.13
- Weak acid [HA]: 0.1
Find: Buffer pH
Answer
10.
Problem 10.
- Solve for: Concentrations for a target pH
- pKa of the weak acid: 2.75
- Target pH: 6
- Total buffer concentration ([HA] + [A⁻]): 0.1 mol/L
Find: Conjugate base [A⁻]
Answer

Answer key: Henderson–Hasselbalch worksheet

1. Ratio [A⁻] ÷ [HA]: 1
1. Subtract the pKa from the target pH. This is how far the log of the ratio must shift
  $log₁₀ (\frac{[A⁻]}{[HA]})$ $= 5 - 5$ $= 0$
2. Raise 10 to that power to undo the logarithm and get the base-to-acid ratio
  $\frac{[A⁻]}{[HA]}$ $= 10^{0}$ $= 1$
2. Conjugate base [A⁻]: 0.05 mol/L
1. Subtract the pKa from the target pH. This is how far the log of the ratio must shift
  $log₁₀ (\frac{[A⁻]}{[HA]})$ $= 4 - 4$ $= 0$
2. Raise 10 to that power to undo the logarithm and get the base-to-acid ratio
  $\frac{[A⁻]}{[HA]}$ $= 10^{0}$ $= 1$
3. Split the total (C) into 1 + ratio shares. The acid [HA] gets one share
  $[HA]$ $= \frac{0.1}{1 + 1}$ $= 0.05$
4. The base form [A⁻] gets the remaining shares
  $[A⁻]$ $= \frac{0.1 \times 1}{1 + 1}$ $= 0.05$
3. Buffer pH: 5.43
1. Divide the base concentration by the acid concentration, because the pH depends only on this ratio
  $\frac{[A⁻]}{[HA]}$ $= \frac{0.15}{0.1}$ $= 1.5$
2. Take the base-10 logarithm of the ratio, which turns it into a pH shift
  $log₁₀ (1.5)$ $= 0.176091$
3. Add that shift to the pKa to get the buffer's pH
  $pH$ $= 5.25 + 0.176091$ $= 5.426091$
4. Conjugate base [A⁻]: 0.04 mol/L
1. Subtract the pKa from the target pH. This is how far the log of the ratio must shift
  $log₁₀ (\frac{[A⁻]}{[HA]})$ $= 4 - 4.25$ $= −0.25$
2. Raise 10 to that power to undo the logarithm and get the base-to-acid ratio
  $\frac{[A⁻]}{[HA]}$ $= 10^{−0.25}$ $= 0.562341$
3. Split the total (C) into 1 + ratio shares. The acid [HA] gets one share
  $[HA]$ $= \frac{0.1}{1 + 0.562341}$ $= 0.064006$
4. The base form [A⁻] gets the remaining shares
  $[A⁻]$ $= \frac{0.1 \times 0.562341}{1 + 0.562341}$ $= 0.035994$
5. Ratio [A⁻] ÷ [HA]: 0.1
1. Subtract the pKa from the target pH. This is how far the log of the ratio must shift
  $log₁₀ (\frac{[A⁻]}{[HA]})$ $= 5 - 6$ $= −1$
2. Raise 10 to that power to undo the logarithm and get the base-to-acid ratio
  $\frac{[A⁻]}{[HA]}$ $= 10^{−1}$ $= 0.1$
6. Conjugate base [A⁻]: 3.15e-4 mol/L
1. Subtract the pKa from the target pH. This is how far the log of the ratio must shift
  $log₁₀ (\frac{[A⁻]}{[HA]})$ $= 4 - 6.5$ $= −2.5$
2. Raise 10 to that power to undo the logarithm and get the base-to-acid ratio
  $\frac{[A⁻]}{[HA]}$ $= 10^{−2.5}$ $= 0.003162$
3. Split the total (C) into 1 + ratio shares. The acid [HA] gets one share
  $[HA]$ $= \frac{0.1}{1 + 0.003162}$ $= 0.099685$
4. The base form [A⁻] gets the remaining shares
  $[A⁻]$ $= \frac{0.1 \times 0.003162}{1 + 0.003162}$ $= 0.000315$
7. Buffer pH: 5.54
1. Divide the base concentration by the acid concentration, because the pH depends only on this ratio
  $\frac{[A⁻]}{[HA]}$ $= \frac{0.11}{0.1}$ $= 1.1$
2. Take the base-10 logarithm of the ratio, which turns it into a pH shift
  $log₁₀ (1.1)$ $= 0.041393$
3. Add that shift to the pKa to get the buffer's pH
  $pH$ $= 5.5 + 0.041393$ $= 5.541393$
8. Buffer pH: 5.78
1. Divide the base concentration by the acid concentration, because the pH depends only on this ratio
  $\frac{[A⁻]}{[HA]}$ $= \frac{0.15}{0.1}$ $= 1.5$
2. Take the base-10 logarithm of the ratio, which turns it into a pH shift
  $log₁₀ (1.5)$ $= 0.176091$
3. Add that shift to the pKa to get the buffer's pH
  $pH$ $= 5.6 + 0.176091$ $= 5.776091$
9. Buffer pH: 4.11
1. Divide the base concentration by the acid concentration, because the pH depends only on this ratio
  $\frac{[A⁻]}{[HA]}$ $= \frac{0.13}{0.1}$ $= 1.3$
2. Take the base-10 logarithm of the ratio, which turns it into a pH shift
  $log₁₀ (1.3)$ $= 0.113943$
3. Add that shift to the pKa to get the buffer's pH
  $pH$ $= 4 + 0.113943$ $= 4.113943$
10. Conjugate base [A⁻]: 0.1 mol/L
1. Subtract the pKa from the target pH. This is how far the log of the ratio must shift
  $log₁₀ (\frac{[A⁻]}{[HA]})$ $= 6 - 2.75$ $= 3.25$
2. Raise 10 to that power to undo the logarithm and get the base-to-acid ratio
  $\frac{[A⁻]}{[HA]}$ $= 10^{3.25}$ $= 1,778.27941$
3. Split the total (C) into 1 + ratio shares. The acid [HA] gets one share
  $[HA]$ $= \frac{0.1}{1 + 1,778.27941}$ $= 0.000056$
4. The base form [A⁻] gets the remaining shares
  $[A⁻]$ $= \frac{0.1 \times 1,778.27941}{1 + 1,778.27941}$ $= 0.099944$