Question

Consider the titration of a 22.0 mL sample of 0.105 molL−1 CH3COOH (Ka=1.8×10−5) with 0.120 molL−1 NaOH. Determine each quantity:

Part A

the initial pH Express your answer using two decimal places. pH =

Part B

the volume of added base required to reach the equivalence point. V = mL

Part C

the pH at 6.0 mL of added base. pH =

Part D

the pH at one-half of the equivalence point. pH =

Part E

the pH at the equivalence point. pH =

Part F

The pH after adding 4.0 mL of base beyond the equivalence point.

Answer 1

part A

initial pH = 1/2(pka-logC)

   pka = -logka = - log(1.8*10^-5) = 4.74

   pH = 1/2(4.74-log0.105) = 2.86

part B

no of mol of aceticacid = 22*0.105 = 2.31 mmol

the volume of added base required to reach the equivalence point. V

V = 2.31/0.12 = 19.25 ml

part C

no of mol of aceticacid = 22*0.105 = 2.31 mmol

no of mol of NaOH = 6*0.12 = 0.72 mmol

pka of aceticacid = 4.74

pH of acidic buffer = pka + log(base/(acid-base))

    = 4.74+log(0.72/(2.31-0.72))

pH = 4.4

part D

the pH at one-half of the equivalence point. pH = pka

   pH = 4.74

part E

no of mol of aceticacid = 22*0.105 = 2.31 mmol

no of mol of NaOH = 2.31 mmol

concentration of salt = 2.31/(19.25+22) = 0.056 M

pH = 7+1/2(pka+logC)

    = 7+1/2(4.74+log0.056)

    = 8.74

part F

concentration of excess base = 4*0.12/(19.25+22+4) = 0.01 M

pH = 14 - (-log(OH-))

         = 14-(-log0.01)

       = 12