Buffer equations & buffer capacity MCQs With Answer

Buffer Equations & Buffer Capacity MCQs with Answer for B. Pharm

Mastering buffer equations and buffer capacity is essential for B. Pharm students to handle pH control in pharmaceutical formulations, dosage stability, and biopharmaceutics. This introduction covers the Henderson–Hasselbalch equation, pKa-based pH calculations, buffer selection for target pH, and the quantitative concept of buffer capacity (β), including factors affecting it such as total buffer concentration, ratio of acid/base, ionic strength, and temperature. Learn how to compute pH for weak acid–base pairs, optimize pharmaceutical buffers for injections and ophthalmics, and estimate capacity needs for process robustness. Build confidence for the B. Pharm exam with practical calculations and theory-driven reasoning. Now let’s test your knowledge with 50 MCQs on this topic.

Q1. What is a buffer solution?

• A solution that resists pH change upon dilution only
• A solution that maintains constant ionic strength
• A solution that resists changes in pH when small amounts of acid or base are added
• A solution of strong acid and strong base at equivalence

Correct Answer: A solution that resists changes in pH when small amounts of acid or base are added

Q2. The Henderson–Hasselbalch equation for a weak acid buffer is:

• pH = pKa − log([A−]/[HA])
• pH = pKa + log([A−]/[HA])
• pH = pKb + log([B]/[BH+])
• pH = pKw − pKb + log([BH+]/[B])

Correct Answer: pH = pKa + log([A−]/[HA])

Q3. For a weak base buffer, the convenient Henderson–Hasselbalch form is:

• pH = pKa + log([BH+]/[B])
• pH = pKa + log([B]/[BH+])
• pH = pKb + log([BH+]/[B])
• pH = pKa − log([B]/[BH+])

Correct Answer: pH = pKa + log([B]/[BH+])

Q4. Buffer capacity (β) is best defined as:

• The pH of a buffer at equivalence
• The amount of strong acid or base required to change the pH of 1 L of solution by one unit
• The total concentration of buffer components
• The ionic strength of a buffer

Correct Answer: The amount of strong acid or base required to change the pH of 1 L of solution by one unit

Q5. What is the pH of an acetic acid/acetate buffer with pKa = 4.76 when [A−] = [HA]?

• 3.76
• 4.76
• 5.76
• 6.76

Correct Answer: 4.76

Q6. For pKa = 4.76, if [A−]/[HA] = 10, what is the buffer pH?

• 3.76
• 4.26
• 4.76
• 5.76

Correct Answer: 5.76

Q7. For pKa = 4.76, if [A−]/[HA] = 0.1, the pH is approximately:

• 3.76
• 4.46
• 4.76
• 5.06

Correct Answer: 3.76

Q8. A phosphate buffer using pKa2 = 7.20 with equimolar H2PO4− and HPO42− will have pH:

• 6.20
• 7.20
• 7.40
• 8.20

Correct Answer: 7.20

Q9. For pKa = 7.20, to achieve pH = 7.40, [base]/[acid] should be approximately:

• 0.63
• 1.00
• 1.58
• 2.00

Correct Answer: 1.58

Q10. A buffer is generally most effective when pH is within what range of its pKa?

• ±0.2 pH units
• ±0.5 pH units
• ±1.0 pH unit
• ±2.0 pH units

Correct Answer: ±1.0 pH unit

Q11. The buffer capacity (β) for a weak acid/conjugate base buffer is given by:

• β = 2.303 C (Ka + [H+])
• β = 2.303 C Ka[H+]/(Ka + [H+])²
• β = C/2.303
• β = Ka/C

Correct Answer: β = 2.303 C Ka[H+]/(Ka + [H+])²

Q12. At pH = pKa, the maximum buffer capacity equals approximately:

• 0.100 × C
• 0.288 × C
• 0.500 × C
• 0.576 × C

Correct Answer: 0.576 × C

Q13. For total buffer concentration C = 0.50 M at pH = pKa, β is approximately:

• 0.0576
• 0.115
• 0.288
• 0.500

Correct Answer: 0.288

Q14. If both [HA] and [A−] are doubled while keeping their ratio constant, what happens?

• pH decreases; buffer capacity unchanged
• pH unchanged; buffer capacity doubles
• pH increases; buffer capacity halves
• pH and buffer capacity both unchanged

Correct Answer: pH unchanged; buffer capacity doubles

Q15. Upon diluting a buffer tenfold, the expected outcome is:

• pH decreases by 1 unit; capacity unchanged
• pH unchanged; capacity decreases tenfold
• pH increases by 1 unit; capacity increases
• Both pH and capacity remain constant

Correct Answer: pH unchanged; capacity decreases tenfold

Q16. A 1.0 L buffer has 0.10 M HA and 0.10 M A− (pKa = 4.76). After adding 0.01 mol HCl, the pH is approximately:

• 4.57
• 4.67
• 4.76
• 4.86

Correct Answer: 4.67

Q17. The same buffer in Q16 receives 0.01 mol NaOH instead. The new pH is approximately:

• 4.66
• 4.76
• 4.85
• 5.00

Correct Answer: 4.85

Q18. An NH3/NH4+ buffer has pKb(NH3) = 4.75 at 25°C. For [NH3]/[NH4+] = 3, pH is approximately:

• 8.73
• 9.25
• 9.73
• 10.25

Correct Answer: 9.73

Q19. For parenteral formulations, buffers should generally:

• Have very high buffer capacity to prevent any pH change in vivo
• Have minimal capacity compatible with stability to reduce tissue irritation
• Use organic buffers exclusively to mimic blood
• Be avoided entirely

Correct Answer: Have minimal capacity compatible with stability to reduce tissue irritation

Q20. Best practice for selecting a buffer to achieve a target pH is to choose:

• A buffer with pKa far from the target pH
• A buffer with pKa close to the target pH
• A strong acid/base pair
• Any salt that increases ionic strength

Correct Answer: A buffer with pKa close to the target pH

Q21. The Henderson–Hasselbalch equation is least accurate when:

• Solutions are very dilute
• Activity coefficients deviate significantly from unity (high ionic strength)
• Temperature is 25°C
• pH = pKa

Correct Answer: Activity coefficients deviate significantly from unity (high ionic strength)

Q22. For the phosphate buffer around neutrality, the appropriate equation is:

• pH = pKa1 + log([H2PO4−]/[H3PO4])
• pH = pKa2 + log([HPO42−]/[H2PO4−])
• pH = pKa3 + log([PO43−]/[HPO42−])
• pH = pKb + log([base]/[acid])

Correct Answer: pH = pKa2 + log([HPO42−]/[H2PO4−])

Q23. If a process may add 5 mmol strong acid per liter and you allow a pH drop of 0.05, the minimum β required is:

• 0.01 mol·L−1·pH−1
• 0.05 mol·L−1·pH−1
• 0.10 mol·L−1·pH−1
• 0.50 mol·L−1·pH−1

Correct Answer: 0.10 mol·L−1·pH−1

Q24. Which total buffer concentration at pH = pKa meets β ≥ 0.10 mol·L−1·pH−1?

• C = 0.05 M
• C = 0.10 M
• C = 0.15 M
• C = 0.20 M

Correct Answer: C = 0.20 M

Q25. A 500 mL buffer with β = 0.12 requires how many millimoles of NaOH to raise pH by 0.20?

• 6 mmol
• 10 mmol
• 12 mmol
• 20 mmol

Correct Answer: 12 mmol

Q26. Why does a buffer have pH = pKa when [A−] = [HA]?

• Because Ka = 1
• Because [A−] = [HA] makes the log term zero
• Because [H+] = 0
• Because pKw = 14

Correct Answer: Because [A−] = [HA] makes the log term zero

Q27. At pH = pKa + 0.50, the fraction of base form (A−) is approximately:

• 24%
• 50%
• 76%
• 90%

Correct Answer: 76%

Q28. If [A−]/[HA] = 0.10, approximately what percent is in base form?

• 9.1%
• 25%
• 50%
• 90.9%

Correct Answer: 9.1%

Q29. At 37°C, pKw ≈ 13.6. For NH4+/NH3 with pKb(NH3) = 4.75, pKa at 37°C is approximately:

• 8.35
• 8.85
• 9.25
• 9.85

Correct Answer: 8.85

Q30. Which action increases buffer capacity without changing pH?

• Increasing [A−] only
• Increasing [HA] only
• Increasing both [A−] and [HA] proportionally
• Decreasing total buffer concentration

Correct Answer: Increasing both [A−] and [HA] proportionally

Q31. Which pair is suitable to buffer near pH ≈ 3?

• Acetic acid/sodium acetate (pKa 4.76)
• Formic acid/sodium formate (pKa 3.75)
• Phosphate (H2PO4−/HPO42−, pKa2 7.2)
• Borate (pKa ~9.2)

Correct Answer: Formic acid/sodium formate (pKa 3.75)

Q32. Approximate β at pH = pKa ± 1 for C = 0.25 M (using β ≈ 0.190 × C) is:

• 0.024
• 0.038
• 0.048
• 0.144

Correct Answer: 0.048

Q33. Changing [A−]/[HA] by a factor of 10 will change buffer pH by approximately:

• 0.10 unit
• 0.30 unit
• 1.00 unit
• 2.00 units

Correct Answer: 1.00 unit

Q34. Which combination is not a buffer system?

• Acetic acid and sodium acetate
• Ammonia and ammonium chloride
• Hydrochloric acid and sodium chloride
• Sodium dihydrogen phosphate and disodium hydrogen phosphate

Correct Answer: Hydrochloric acid and sodium chloride

Q35. Mixing equal volumes of 0.20 M HA and 0.10 M A− gives pH relative to pKa of approximately:

• pH = pKa − 0.30
• pH = pKa
• pH = pKa + 0.30
• pH = pKa + 1.00

Correct Answer: pH = pKa − 0.30

Q36. For phosphate buffer with pKa2 = 7.20, to make pH = 7.00, the ratio [HPO42−]/[H2PO4−] should be:

• 0.40
• 0.63
• 1.00
• 1.58

Correct Answer: 0.63

Q37. The common symbol used for buffer capacity is:

• α
• β
• γ
• λ

Correct Answer: β

Q38. If 0.01 mol HCl is added to 0.50 L of buffer with β = 0.10, the approximate pH change is:

• 0.05 pH units
• 0.10 pH units
• 0.20 pH units
• 0.50 pH units

Correct Answer: 0.20 pH units

Q39. A commonly used buffer system for ophthalmic solutions around pH 7.4 is:

• Borate buffer only
• Citrate buffer only
• Phosphate buffer
• Tartarate buffer

Correct Answer: Phosphate buffer

Q40. Which factor does not directly influence buffer capacity?

• Total buffer concentration
• Ratio of conjugate base to acid
• Temperature (via Ka and pKw)
• Container color

Correct Answer: Container color

Q41. For pKa = 5.00, [A−] = 0.08 M, [HA] = 0.02 M, the buffer pH is approximately:

• 5.30
• 5.48
• 5.60
• 6.00

Correct Answer: 5.60

Q42. Upon 10-fold dilution of a buffer, which statement is correct?

• pH decreases by 1 unit
• pH increases by 1 unit
• pH remains essentially constant; buffer capacity decreases tenfold
• Both pH and capacity increase

Correct Answer: pH remains essentially constant; buffer capacity decreases tenfold

Q43. In the bicarbonate buffer (pKa ≈ 6.10), at pH 7.40, [HCO3−]/[H2CO3] is approximately:

• 5
• 10
• 15
• 20

Correct Answer: 20

Q44. At 25°C, the relation between pKa and pKb is:

• pKa = pKb
• pKa = 7 − pKb
• pKa = 10 − pKb
• pKa = 14 − pKb

Correct Answer: pKa = 14 − pKb

Q45. For phosphate at pH 7.40 (pKa2 = 7.20), the fraction present as HPO42− is about:

• 39%
• 50%
• 61%
• 80%

Correct Answer: 61%

Q46. A 0.10 M buffer at pH = pKa (β ≈ 0.0576) in 200 mL is to be lowered by 0.50 pH units. Approximate millimoles of HCl needed:

• 2.88 mmol
• 5.76 mmol
• 8.00 mmol
• 10.0 mmol

Correct Answer: 5.76 mmol

Q47. Excessive buffer capacity in parenteral products is undesirable primarily because:

• It reduces drug solubility
• It increases osmolarity above isotonic levels always
• It may overwhelm physiological pH control and irritate tissues
• It accelerates drug hydrolysis

Correct Answer: It may overwhelm physiological pH control and irritate tissues

Q48. If the pKa used in calculations is off by ±0.10, the calculated pH (for a fixed ratio) will be off by approximately:

• ±0.01 pH unit
• ±0.05 pH unit
• ±0.10 pH unit
• ±0.50 pH unit

Correct Answer: ±0.10 pH unit

Q49. A convenient weak base buffer relation using pKb is:

• pH = pKb + log([B]/[BH+])
• pOH = pKb + log([BH+]/[B])
• pOH = pKb − log([BH+]/[B])
• pH = pKb + log([BH+]/[B])

Correct Answer: pOH = pKb + log([BH+]/[B])

Q50. In differential form, buffer capacity is defined as:

• β = d(pH)/dB
• β = dB/d(pH)
• β = dB × d(pH)
• β = B × pH

Correct Answer: β = dB/d(pH)