Question

Consider the titration of a 30.0 mL sample of 0.150 M CH3COOH (Ka=1.8×10−5) with 0.200 M NaOH. Determine each quantity: Part A: What is the volume of base required to reach the equivalence point? Express your answer using one decimal place. Part B: What is the pH after 10.0 mL of base have been added? Express your answer using two decimal places. Part C What is the pH at the equivalence point? Express your answer using two decimal places. Part D What is the pH after adding 5.0 mL of base beyond the equivalence point? Express your answer using two decimal places.

Answer 1

weak acid vs strong base

a) no of mol of CH3COOH = M*V

            = 0.15*30

                         = 4.5 mmol

no of mol of NaOH required = 4.5 mmol

volume of base(NaOH) required to reach the equivalence point = n/M

                           = 4.5/0.2

                           = 22.5 ml

B) no of mol of CH3COOH = M*V

            = 0.15*30

                         = 4.5 mmol

no of mol of NaOH required = M*V

                           = 0.2*10

                           = 2 mmol

pH of acidic buffer = pka + log(ch3cooNa/CH33COOH)

pka of CH3COOH = -logKa = -log(1.8*10^-5) = 4.74

no of mol of CH3COOH = 4.5 mmol

No of mol of CH3COONa = NaOH = 2 mmol

pH = 4.74+log(2/(4.5-2))

    = 4.64

C) at equivalence point.

no of mol of CH3COOH = 4.5 mmol

No of mol of CH3COONa = NaOH = 4.5 mmol

concentration of CH3COONa = n/v

                           = 4.5/(30+22.5)

                           = 0.086 M

D) beyond equivalence point

concentration of excess NaOH = n/v

                               = (5*0.2)/(30+27.5)

                               = 0.0174 M

pOH = -log(OH-)

     = -log0.0174

     = 1.76

pH = 14-1.76

    = 12.24