BUFFER AND PH LABORATORY EXAM PRACTICE
QUESTIONS AND VERIFIED ANSWERS WITH RATIONALES
This comprehensive practice examination is designed to assess and reinforce mastery of core
concepts related to buffer solutions, pH calculations, and the Henderson-Hasselbalch equation,
which are fundamental to clinical laboratory science, biochemistry, and physiology. The exam
covers a range of topics including buffer capacity, the effects of strong acid/base addition,
acid-base disturbances, and the selection of appropriate buffer systems. The following
questions are presented in a test bank format to help students and professionals prepare for
certifications, course exams, or to simply solidify their understanding of these essential
principles.
DOMAINS COVERED:
1. Buffer Fundamentals and the Henderson-Hasselbalch Equation
2. Buffer Capacity and its Calculation
3. Impact of Strong Acid or Base Addition on Buffer pH
4. Clinical Acid-Base Disturbances and Compensation
5. Buffer System Selection and Preparation
1. A research buffer is prepared by mixing 0.20 M NaH2PO4 and 0.20 M Na2HPO4 to
achieve pH 7.00. Using the Henderson-Hasselbalch equation with pKa2 = 7.21, what is
the buffer capacity (i) in mol/L per pH unit at this pH?
A. 0.057
B. 0.115
C. 0.230
D. 0.460
Correct Answer: B
Rationale: β = 2.303 × Ctotal × ([HA][A⁻])/([HA]+[A⁻])². At pH 7.00, [A⁻]/[HA] =
10^(7.00 - 7.21) = 0.617. Ctotal = 0.40 M, so [HA] = 0.247 M and [A⁻] = 0.153 M. β =
2.303 × 0.40 × (0.247 × 0.153)/0.40 = 0.115.
2. A solution contains 0.10 M acetic acid (pKa 4.76) and 0.20 M sodium acetate. If 5.0
mL of 1.0 M HCl is added to 100 mL of this buffer, what is the new pH?
A. 4.76
B. 4.96
C. 5.06
D. 5.16
, Correct Answer: A
Rationale: Initial moles: acid = 0.010, acetate = 0.020. HCl added = 0.005 mol,
converting acetate to acid. Final: acid = 0.015, acetate = 0.015. pH = 4.76 +
log(0.015/0.015) = 4.76.
3. In a clinical laboratory, a blood gas analyzer reports pH = 7.32, pCO2 = 60 mmHg,
and [HCO3⁻] = 30 mmol/L. Using the Henderson-Hasselbalch equation with pKa' = 6.1
and solubility coefficient 0.030, which acid-base disturbance is present?
A. Acute respiratory acidosis with renal compensation
B. Chronic respiratory acidosis with metabolic compensation
C. Metabolic alkalosis with respiratory compensation
D. Mixed respiratory and metabolic acidosis
Correct Answer: B
Rationale: The elevated pCO2 (60 mmHg) indicates respiratory acidosis. The HCO3⁻ is
elevated (30 mmol/L), which is a compensatory metabolic response. In acute respiratory
acidosis, HCO3⁻ typically rises by only 1 mmol/L per 10 mmHg pCO2 above normal
(expected ~24+1=25), but here it is 30, suggesting chronic compensation (expected
~24+4=28-30). Therefore, chronic respiratory acidosis with metabolic compensation is
correct.
4. A buffer is designed to maintain pH 9.0. Which of the following weak acids would
provide the highest buffer capacity at this pH?
A. Boric acid (pKa 9.24)
B. Ammonium ion (pKa 9.25)
C. Glycine (pKa2 = 9.60)
D. Tris (pKa 8.07)
Correct Answer: A
Rationale: Buffer capacity is maximal when pH = pKa. At pH 9.0, boric acid (pKa
9.24) is closest to the target pH, giving the highest capacity.
5. A buffer solution is prepared by mixing 50 mL of 0.10 M NaHCO3 with 50 mL of 0.05
M Na2CO3. What is the pH of this solution? (For carbonic acid, pKa1 = 6.35, pKa2 =
10.33)
A. 6.35
B. 9.90
C. 10.03
D. 10.63
Correct Answer: C
Rationale: This buffer involves the second dissociation: HCO3⁻ / CO3²⁻. Moles HCO3⁻
= 0.005, moles CO3²⁻ = 0.0025. Ratio = 0.5. pH = 10.33 + log(0.5) = 10.03.
6. A buffer solution has a pH of 4.50 and contains 0.20 M of a weak acid HA (pKa =
4.20). What is the concentration of the conjugate base A⁻?
A. 0.20 M
B. 0.40 M
, C. 0.10 M
D. 0.30 M
Correct Answer: B
Rationale: 4.50 = 4.20 + log([A⁻]/[HA]). log([A⁻]/[HA]) = 0.30, [A⁻]/[HA] = 10^0.30
≈ 2.0. Thus [A⁻] = 2 × 0.20 = 0.40 M.
7. What is the pH of a 0.1 M solution of HCl?
A. 1.0
B. 7.0
C. 13.0
D. 0.1
Correct Answer: A
Rationale: HCl is a strong acid that dissociates completely. pH = -log(0.1) = 1.0.
8. Which of the following is NOT a property of an effective buffer?
A. Its pH is resistant to change upon addition of small amounts of acid or base.
B. It is composed of a weak acid and its conjugate base.
C. Its pH is equal to the pKa of the weak acid.
D. Its capacity is determined by the absolute concentrations of the acid and conjugate
base.
Correct Answer: C
Rationale: While buffer capacity is highest when pH = pKa, a buffer is effective within
a range of pH = pKa ± 1. A buffer does not require its pH to be exactly equal to its pKa.
9. A 1.0 L buffer solution contains 0.10 mol of acetic acid and 0.10 mol of sodium
acetate. If 0.01 mol of NaOH is added, what is the change in pH? (pKa = 4.76)
A. 0.00
B. +0.08
C. -0.08
D. +0.04
Correct Answer: B
Rationale: Initial pH = pKa = 4.76. After adding 0.01 mol NaOH, acid decreases to
0.09 mol, acetate increases to 0.11 mol. New pH = 4.76 + log(0.11/0.09) = 4.76 +
0.087 = 4.847. Change ≈ +0.09.
10. What is the relationship between pH and pKa in the Henderson-Hasselbalch equation?
A. pH = pKa + log([acid]/[base])
B. pH = pKa - log([base]/[acid])
C. pH = pKa + log([base]/[acid])
D. pH = pKa × log([base]/[acid])
Correct Answer: C
Rationale: The Henderson-Hasselbalch equation is pH = pKa + log([A⁻]/[HA]), where
[A⁻] is the conjugate base and [HA] is the weak acid.
, 11. If the pH of a buffer is equal to the pKa of the weak acid, what is the ratio of the
concentration of conjugate base to the acid?
A. 0:1
B. 1:1
C. 10:1
D. 1:10
Correct Answer: B
Rationale: When pH = pKa, log([A⁻]/[HA]) = 0, meaning [A⁻]/[HA] = 1.
12. What is the pH of a solution that has a hydrogen ion concentration [H+] of 1.0 × 10⁻⁸
M?
A. 6.0
B. 7.0
C. 8.0
D. 9.0
Correct Answer: C
Rationale: pH = -log(1.0 × 10⁻⁸) = 8.0.
13. Which of the following acid-base pairs would be best for preparing a buffer at pH
7.4?
A. H2CO3/HCO3⁻ (pKa = 6.1)
B. H2PO4⁻/HPO4²⁻ (pKa = 7.21)
C. NH4⁺/NH3 (pKa = 9.25)
D. CH3COOH/CH3COO⁻ (pKa = 4.76)
Correct Answer: B
Rationale: The most effective buffer is one whose pKa is close to the desired pH. The
pKa of H2PO4⁻/HPO4²⁻ (7.21) is closest to 7.4.
14. The buffer capacity of a solution is increased by:
A. Diluting the buffer solution.
B. Increasing the concentration of the buffer components.
C. Adding a strong acid.
D. Changing the ratio of the buffer components to 10:1.
Correct Answer: B
Rationale: Buffer capacity is directly proportional to the total concentration of the
buffer components.
15. In the bicarbonate buffer system, the Henderson-Hasselbalch equation is often written
as pH = pKa + log([HCO3⁻]/[CO2]). A patient has a plasma pH of 7.50. This condition is
referred to as:
A. Acidemia
B. Alkalemia
C. Acidosis
D. Alkalosis
QUESTIONS AND VERIFIED ANSWERS WITH RATIONALES
This comprehensive practice examination is designed to assess and reinforce mastery of core
concepts related to buffer solutions, pH calculations, and the Henderson-Hasselbalch equation,
which are fundamental to clinical laboratory science, biochemistry, and physiology. The exam
covers a range of topics including buffer capacity, the effects of strong acid/base addition,
acid-base disturbances, and the selection of appropriate buffer systems. The following
questions are presented in a test bank format to help students and professionals prepare for
certifications, course exams, or to simply solidify their understanding of these essential
principles.
DOMAINS COVERED:
1. Buffer Fundamentals and the Henderson-Hasselbalch Equation
2. Buffer Capacity and its Calculation
3. Impact of Strong Acid or Base Addition on Buffer pH
4. Clinical Acid-Base Disturbances and Compensation
5. Buffer System Selection and Preparation
1. A research buffer is prepared by mixing 0.20 M NaH2PO4 and 0.20 M Na2HPO4 to
achieve pH 7.00. Using the Henderson-Hasselbalch equation with pKa2 = 7.21, what is
the buffer capacity (i) in mol/L per pH unit at this pH?
A. 0.057
B. 0.115
C. 0.230
D. 0.460
Correct Answer: B
Rationale: β = 2.303 × Ctotal × ([HA][A⁻])/([HA]+[A⁻])². At pH 7.00, [A⁻]/[HA] =
10^(7.00 - 7.21) = 0.617. Ctotal = 0.40 M, so [HA] = 0.247 M and [A⁻] = 0.153 M. β =
2.303 × 0.40 × (0.247 × 0.153)/0.40 = 0.115.
2. A solution contains 0.10 M acetic acid (pKa 4.76) and 0.20 M sodium acetate. If 5.0
mL of 1.0 M HCl is added to 100 mL of this buffer, what is the new pH?
A. 4.76
B. 4.96
C. 5.06
D. 5.16
, Correct Answer: A
Rationale: Initial moles: acid = 0.010, acetate = 0.020. HCl added = 0.005 mol,
converting acetate to acid. Final: acid = 0.015, acetate = 0.015. pH = 4.76 +
log(0.015/0.015) = 4.76.
3. In a clinical laboratory, a blood gas analyzer reports pH = 7.32, pCO2 = 60 mmHg,
and [HCO3⁻] = 30 mmol/L. Using the Henderson-Hasselbalch equation with pKa' = 6.1
and solubility coefficient 0.030, which acid-base disturbance is present?
A. Acute respiratory acidosis with renal compensation
B. Chronic respiratory acidosis with metabolic compensation
C. Metabolic alkalosis with respiratory compensation
D. Mixed respiratory and metabolic acidosis
Correct Answer: B
Rationale: The elevated pCO2 (60 mmHg) indicates respiratory acidosis. The HCO3⁻ is
elevated (30 mmol/L), which is a compensatory metabolic response. In acute respiratory
acidosis, HCO3⁻ typically rises by only 1 mmol/L per 10 mmHg pCO2 above normal
(expected ~24+1=25), but here it is 30, suggesting chronic compensation (expected
~24+4=28-30). Therefore, chronic respiratory acidosis with metabolic compensation is
correct.
4. A buffer is designed to maintain pH 9.0. Which of the following weak acids would
provide the highest buffer capacity at this pH?
A. Boric acid (pKa 9.24)
B. Ammonium ion (pKa 9.25)
C. Glycine (pKa2 = 9.60)
D. Tris (pKa 8.07)
Correct Answer: A
Rationale: Buffer capacity is maximal when pH = pKa. At pH 9.0, boric acid (pKa
9.24) is closest to the target pH, giving the highest capacity.
5. A buffer solution is prepared by mixing 50 mL of 0.10 M NaHCO3 with 50 mL of 0.05
M Na2CO3. What is the pH of this solution? (For carbonic acid, pKa1 = 6.35, pKa2 =
10.33)
A. 6.35
B. 9.90
C. 10.03
D. 10.63
Correct Answer: C
Rationale: This buffer involves the second dissociation: HCO3⁻ / CO3²⁻. Moles HCO3⁻
= 0.005, moles CO3²⁻ = 0.0025. Ratio = 0.5. pH = 10.33 + log(0.5) = 10.03.
6. A buffer solution has a pH of 4.50 and contains 0.20 M of a weak acid HA (pKa =
4.20). What is the concentration of the conjugate base A⁻?
A. 0.20 M
B. 0.40 M
, C. 0.10 M
D. 0.30 M
Correct Answer: B
Rationale: 4.50 = 4.20 + log([A⁻]/[HA]). log([A⁻]/[HA]) = 0.30, [A⁻]/[HA] = 10^0.30
≈ 2.0. Thus [A⁻] = 2 × 0.20 = 0.40 M.
7. What is the pH of a 0.1 M solution of HCl?
A. 1.0
B. 7.0
C. 13.0
D. 0.1
Correct Answer: A
Rationale: HCl is a strong acid that dissociates completely. pH = -log(0.1) = 1.0.
8. Which of the following is NOT a property of an effective buffer?
A. Its pH is resistant to change upon addition of small amounts of acid or base.
B. It is composed of a weak acid and its conjugate base.
C. Its pH is equal to the pKa of the weak acid.
D. Its capacity is determined by the absolute concentrations of the acid and conjugate
base.
Correct Answer: C
Rationale: While buffer capacity is highest when pH = pKa, a buffer is effective within
a range of pH = pKa ± 1. A buffer does not require its pH to be exactly equal to its pKa.
9. A 1.0 L buffer solution contains 0.10 mol of acetic acid and 0.10 mol of sodium
acetate. If 0.01 mol of NaOH is added, what is the change in pH? (pKa = 4.76)
A. 0.00
B. +0.08
C. -0.08
D. +0.04
Correct Answer: B
Rationale: Initial pH = pKa = 4.76. After adding 0.01 mol NaOH, acid decreases to
0.09 mol, acetate increases to 0.11 mol. New pH = 4.76 + log(0.11/0.09) = 4.76 +
0.087 = 4.847. Change ≈ +0.09.
10. What is the relationship between pH and pKa in the Henderson-Hasselbalch equation?
A. pH = pKa + log([acid]/[base])
B. pH = pKa - log([base]/[acid])
C. pH = pKa + log([base]/[acid])
D. pH = pKa × log([base]/[acid])
Correct Answer: C
Rationale: The Henderson-Hasselbalch equation is pH = pKa + log([A⁻]/[HA]), where
[A⁻] is the conjugate base and [HA] is the weak acid.
, 11. If the pH of a buffer is equal to the pKa of the weak acid, what is the ratio of the
concentration of conjugate base to the acid?
A. 0:1
B. 1:1
C. 10:1
D. 1:10
Correct Answer: B
Rationale: When pH = pKa, log([A⁻]/[HA]) = 0, meaning [A⁻]/[HA] = 1.
12. What is the pH of a solution that has a hydrogen ion concentration [H+] of 1.0 × 10⁻⁸
M?
A. 6.0
B. 7.0
C. 8.0
D. 9.0
Correct Answer: C
Rationale: pH = -log(1.0 × 10⁻⁸) = 8.0.
13. Which of the following acid-base pairs would be best for preparing a buffer at pH
7.4?
A. H2CO3/HCO3⁻ (pKa = 6.1)
B. H2PO4⁻/HPO4²⁻ (pKa = 7.21)
C. NH4⁺/NH3 (pKa = 9.25)
D. CH3COOH/CH3COO⁻ (pKa = 4.76)
Correct Answer: B
Rationale: The most effective buffer is one whose pKa is close to the desired pH. The
pKa of H2PO4⁻/HPO4²⁻ (7.21) is closest to 7.4.
14. The buffer capacity of a solution is increased by:
A. Diluting the buffer solution.
B. Increasing the concentration of the buffer components.
C. Adding a strong acid.
D. Changing the ratio of the buffer components to 10:1.
Correct Answer: B
Rationale: Buffer capacity is directly proportional to the total concentration of the
buffer components.
15. In the bicarbonate buffer system, the Henderson-Hasselbalch equation is often written
as pH = pKa + log([HCO3⁻]/[CO2]). A patient has a plasma pH of 7.50. This condition is
referred to as:
A. Acidemia
B. Alkalemia
C. Acidosis
D. Alkalosis
