Question

A solution is prepared by placing 0.55 moles of acetic acid (HC2H3O2) and 0.35 moles of...

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. Ka for acetic acid is 1.8 x 10-5.)

Homework Answers

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

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