calculate the pH

Question

25.00 mL of a solution containing 0.0997 M acetic acid (Ka = 1.75 x 10-5) is to be titrated with 0.129 M NaOH.

CH3COOH + NaOH → CH3COONa + H2O

Calculate pH after the addition of 25.85 mL of base. (Enter your answer to three significant figures. Ignore the dissociation of water.)

Answer 1

moles CH₃COOH present = 0.025 L x 0.0997 mol/L = 0.002493 moles

moles NaOH addd = 0.02585 L x 0.129 mol/L = 0.003335 moles

CH₃COOH + NaOH ==> CH₃COONa + H2O

All 0.002493 moles CH3COOH will be converted to 0.002493 moles of CH3COONa (no buffer capacity left)

This will leave 0.003335 - 0.002493 = 0.000842 moles NaOH in a final volume of 50.85 ml (25 + 25.85)

The final [OH-] = 0.000842 mol/0.05085 L = 0.01656 M

pOH = -log [OH-] = 1.78

pH = 14 - 1.78

pH = 12.2

(NOTE: the hydrolysis of the CH3COO- that is formed will not contribute significant to the pH. If you wanted to include it, you would use the Kb for CH3COO- and then Kb = (x)(x)/[CH3COO-]-x and solve for x which wold be [OH-] which you would add to the above calculation)