Question

A solution is prepared by placing 0.55 moles of acetic acid (HC₂H₃O₂) and 0.35 moles of sodium acetate (NaC₂H₃O₂) into enough water to create a solution with a total volume of 1.50 liters. Then, 0.05 moles of sodium hydroxide (NaOH) are added to the solution. What is the pH of the solution after the addition of the sodium hydroxide? (Assume the volume remains constant throughout. K_a for acetic acid is 1.8 x 10^-5.)

Answer 1

PH = PKa + log[salt]/[acid]

Ka = 1.8*10^-5

Pka = -logKa

       = -log1.8*10^-5

     = 4.75

no of moles of CH3COONa = 0.35 moles

no of moles of CH3COOH = 0.55 moles

PH   = 4.75 + log0.35/0.55

     = 4.75-0.1962 = 4.5538

By the addition of NaOH

no of moles of CH3COONa = 0.35 +0.05 = 0.4 moles

no of moles of CH3COOH = 0.55-0.05 = 0.5 moles

PH = 4.75 + log0.4/0.5

    = 4.75-0.09691 = 4.65309 >>>> answer