Frequently Asked Question

This is a classic acid-base titration problem. Here's a breakdown of how to solve it, including the formulas used:
1. Initial pH (Va = 0 mL)

Formula: pH = pKw - (1/2)(pKb - log[Base])
Calculations:

pKw = 14 (at 25°C)

pKb = 5.0

[Base] = 0.1 M (initial concentration of the weak base)

pH = 14 - (1/2)(5 - log(0.1)) = 9.5
2. pH at Va = 5 mL

Formula: pH = pKa + log([conjugate base]/[weak acid]) (Henderson-Hasselbalch Equation)
Calculations:

pKa = 14 - pKb = 9

Moles of base initially = 0.1 mol/L
0.1 L = 0.01 mol

Moles of acid added = 1 mol/L
0.005 L = 0.005 mol

Moles of base remaining = 0.01 mol - 0.005 mol = 0.005 mol

Moles of conjugate acid formed = 0.005 mol

[conjugate base] = 0.005 mol / 0.105 L (total volume) ≈ 0.048 M

[weak acid] = 0.005 mol / 0.105 L (total volume) ≈ 0.048 M

pH = 9 + log(0.048/0.048) = 9
3. pH at Va = 9 mL

This is the equivalence point. At this point, all of the weak base has reacted with the acid, and only the conjugate acid remains. The pH is determined by the hydrolysis of the conjugate acid.
Formula: pH = (1/2)(pKa - log[conjugate acid])
Calculations:

[conjugate acid] ≈ (0.01 mol / 0.109 L) = 0.092 M (total volume = 100 mL + 9 mL)

pH = (1/2)(9 - log(0.092)) ≈ 4.5
4. pH at Va = 12 mL

Beyond the equivalence point, the pH is primarily determined by the excess strong acid.
Formula: pH = -log([H+])
Calculations:

Moles of excess acid = (1 mol/L
0.012 L) - 0.01 mol = 0.002 mol

[H+] = 0.002 mol / 0.112 L (total volume) ≈ 0.018 M

pH = -log(0.018) ≈ 1.7
Sketching the Titration Curve

The titration curve will start at a pH around 9.5 (initial pH) and gradually decrease as acid is added. It will level off near pH 9 (buffer region) until approaching the equivalence point at Va = 9 mL. At the equivalence point, the pH will jump sharply due to the formation of the conjugate acid. After the equivalence point, the pH will continue to decrease rapidly as excess strong acid is added. The curve will look like a classic S-shape.
Key Formulas Used:

pH and pOH: pH + pOH = pKw (where pKw ≈ 14 at 25°C)
Henderson-Hasselbalch Equation: pH = pKa + log([conjugate base]/[weak acid])
Hydrolysis of Conjugate Acid: pH = (1/2)(pKa - log[conjugate acid])
pH of Strong Acid Solutions: pH = -log([H+])
Remember: This is a simplified explanation. A more detailed analysis would involve considering the change in volume and using the appropriate equilibrium constants.