Calculate pH of a solution of given mixture 0.1 mol CH₃COOH + 0.2 mol CH³COO Na in 100ml of mixture. Ka = 2×10⁻⁵ .

Given:

• Acetic Acid (CH₃COOH): 0.1 mol
• Sodium Acetate (CH₃COONa): 0.2 mol
• Volume of solution: 100 mL = 0.1 L
• Ka (for CH₃COOH): 2 × 10^-5

We know that:

pKa = -log(Ka)

pKa = -log(2 × 10^-5) = 4.70

• [HA] = 0.1 mol / 0.1 L = 1 M
• [A^-] = 0.2 mol / 0.1 L = 2 M

The Henderson-Hasselbalch equation is given by:

pH = pKa + log([A^-]/[HA])

Substitute the values:

pH = 4.70 + log(2)

We know that log(2) ≈ 0.3010, so:

pH = 4.70 + 0.3010 = 5.0

Final Answer

The correct pH of the solution is 5