ph of acetid acid and potassium hydroxide titration

Question

Consider the titration of 17.4 mL of 0.454 M HCOOH (pKa=4.76) with 0.232 M KOH. What is the pH of 120% titration?

Answer 1

When you titrate a weak acid (acetic acid) with a strong base (potassium hydroxide), you create a buffer until you reach equivalence. At equivalence, when there is no more acid or base (only salt and water), the pH will be greater than 7. After equivalence (>100% titration), the pH will mostly be determined by the excess KOH, so will be very alkaline. By the way, the acid you give (HCOOH) is not acetic acid, but rather is formic acid, but I'll use acetic acid in the problem.

CH3COOH + KOH ==> CH3COOK + H2O ... balanced equation

moles CH3COOH present = 0.0174 L x 0.454 mol/L = 0.0078996 moles

moles KOH needed for 100% titration = 0.0078996 moles (1:1 mole ratio in balanced equation)

At 100% titration, this is the equivalence point and moles CH3COO^- = 0.0078996 (1:1 mol ratio)

Volume KOH used @ equivalence: 0.0078996 mol KOH x 1 L/0.232 mol = 0.03405 L

Total volume @ equivalence = 0.0174 L + 0.03405 L = 0.05145 L

Final [CH3COO^-] = 0.0078996 mol/0.05145 L = 0.1535 M

@120% titration, that would be equal to 1.2 x 0.03405 L KOH = 0.0414 L KOH

total moles KOH added = 0.0414 L x 0.232 mol/L = 0.00960 moles KOH

excess moles KOH present = 0.00960 mol - 0078996 mol = 0.00171 moles excess KOH

Total volume = 0.0414 L + 0.0174 L = 0.0588 L

Final [OH^-] from excess KOH = 0.00171 mol/0.0588 L = 0.029 M

pOH = 1.54

pH = 12.46

(note: the OH^- contributed by the hydrolysis of CH3COO^- will be very small relative to 0.029 M and so has been ignored in the calculation)