pH Calculations - UNSOLVED PRACTICE SET
Chapter: Equilibrium | Topic: pH Calculations
PH CALCULATIONS - UNSOLVED PRACTICE SET
Topic: pH Calculations
Multiple Choice Questions
Q1. The pH of a solution is defined as:
- log[H⁺]
- –log[H⁺]
- ln[H⁺]
- –ln[H⁺]
Q2. The pH of 0.001 M HCl solution is:
- 1
- 2
- 3
- 11
Q3. A solution has pH = 9. The solution is:
- Acidic
- Basic
- Neutral
- Amphoteric
Q4. The pH of pure water at 60°C (Kw = 9.6 × 10⁻¹⁴) is:
- 7.0
- Less than 7.0
- Greater than 7.0
- 14.0
Q5. When pH of a solution decreases by 2 units, the H⁺ concentration:
- Decreases by 2 times
- Increases by 2 times
- Decreases by 100 times
- Increases by 100 times
Q6. The pOH of a 0.01 M NaOH solution at 25°C is:
- 2
- 12
- 7
- 14
Short Answer Questions
Q7. Define pH and pOH. Write the relationship between pH, pOH, and pKw at 25°C.
Q8. Calculate the pH of a solution containing 0.005 M H₂SO₄. (Assume complete ionisation of both H⁺ ions.)
Q9. The pH of a solution is 4.5. Calculate the H⁺ ion concentration and the OH⁻ ion concentration at 25°C.
Q10. Explain why the pH of pure water is 7 at 25°C but less than 7 at higher temperatures, even though the water remains neutral.
Q11. A student mixes equal volumes of pH 2 and pH 4 solutions. Is the resulting pH equal to 3? Explain with calculation.
Q12. Calculate the pH of 10⁻⁸ M HCl solution. Explain why the pH is not simply 8, even though the acid is very dilute.
Long Answer Questions
Q13. (a) Define pH and derive the relationship pH + pOH = pKw.
(b) Calculate the pH of the following solutions at 25°C:
(i) 0.01 M HNO₃
(ii) 0.001 M Ba(OH)₂
(iii) A solution obtained by mixing 50 mL of 0.1 M HCl with 50 mL of 0.1 M NaOH
(c) A student claims that a solution with pH = 6.5 at 60°C is acidic. Another student says it is basic. Who is correct? Explain using the concept of Kw.
Q14. (a) Explain why the pH of a very dilute strong acid solution (less than 10⁻⁶ M) cannot be calculated simply by taking the negative logarithm of the acid concentration.
(b) Calculate the pH of 5.0 × 10⁻⁸ M HCl solution at 25°C. (Hint: Consider the contribution of H⁺ ions from water.)
(c) Derive the expression for the pH of a mixture of a strong acid and a strong base when they are not in stoichiometric amounts.
Q15. (a) Derive the Henderson-Hasselbalch equation for a weak acid and its conjugate base:
pH = pKa + log([A⁻]/[HA])
(b) A buffer solution contains 0.1 M CH₃COOH and 0.1 M CH₃COONa. Calculate its pH. (pKa of CH₃COOH = 4.74)
(c) Explain how this buffer resists pH change when a small amount of acid or base is added.
Numerical / Application-Based Problems
Q16. Calculate the pH of the following solutions at 25°C:
(a) 0.05 M H₂SO₄ (diprotic strong acid, assume both H⁺ completely ionised)
(b) 0.02 M Ca(OH)₂ (strong base)
(c) 0.1 M CH₃COOH (Ka = 1.8 × 10⁻⁵)
(d) A solution obtained by mixing 25 mL of 0.2 M HCl with 75 mL of 0.1 M NaOH
Q17. The pH of blood is maintained between 7.35 and 7.45.
(a) Calculate the H⁺ ion concentration range in blood.
(b) If the pH of blood drops to 7.20, calculate the percentage increase in H⁺ ion concentration compared to normal blood (pH = 7.40).
(c) A patient with acidosis has blood pH of 7.10. Calculate the ratio of [H⁺] in this patient's blood to [H⁺] in normal blood (pH 7.40).
(d) Explain why maintaining blood pH within a narrow range is crucial for enzyme function and overall health.
Q18. In India, water quality is a major concern. The pH of drinking water should ideally be between 6.5 and 8.5.
(a) A water sample from a village well has a pH of 5.2. Calculate the H⁺ ion concentration. Is this water safe for drinking? Explain.
(b) To neutralise 100 L of this acidic water, how many grams of CaO (quicklime) would be required? (Assume the acid present is HCl for simplicity.)
(c) Another water sample from a different source has a pH of 9.5. Calculate the OH⁻ ion concentration. What could be the possible cause of this alkalinity?
(d) A student tests the pH of rainwater in an industrial area and finds it to be 4.2. Explain the cause of this 'acid rain' and its effects on soil pH and agriculture.