Question

What is the pH equivalence point in the titration of 25.0 mL of 0.750 M CH3COOH with 2.00 M KOH?

Please show step by step instructions

Answer 1

at equivalence point the amount of acid present equals to the amount of base added

moles of CH3COOH present = 0.750 M x 0.025 L = 0.01875 mols

So, moles of KOH added = 0.01875 mols

Volume of base added = moles/molarity = 0.01875/2 = 0.009375 L

moles of salt generated by reaction: CH3COOH + KOH ---> CH3COOK + H2O

= 0.01875 mols

concentration of salt in solution = moles/total volume of solution = 0.01875/(0.025 + 0.009375) = 0.545 M

salt hydrolyzes in water as,

CH3COO- + H2O <==> CH3COOH + OH-

let x be the amount of salt hydrolyzed then,

Kb = [CH3COOH][OH-]/[CH3COO-]

Kw/Ka = 1 x 10^-14/1.8 x 10^-5 = x^2/0.545

x = [OH-] = 1.74 x 10^-5 M

pOH = -log[OH-] = 4.76

pH at equivalence point = 14 - pOH = 14 - 4.76 = 9.24