Question

Calculate the pH during titration of 25 mL of 0.175 M HCN with 0.3 M NaOH after the addition of 0 mL, 7.29mL, 14.5833 mL, and 20 mL base. Kb of CN- is 2.03 x 10^-5. Show all work.

Answer 1

pKb of CN^- = -log(2.03*10^-5) = 4.7

pka of HCN = 14-4.7 = 9.3

after addition og 0 ml

pH = 1/2(pka-logC)

    = 1/2(9.3-log0.175)

    = 5.02

after 7.29 ml NaOH

no of mol of HCN taken = 25*0.175 = 4.375 mmol

no of mol of NaOH = 7.29*0.3 = 2.187 mmol

pH = pka + log(salt/acid)

    = 9.3+log(2.187/4.375)

    = 9

after 14.5833 ml NaOH

no of mol of HCN taken = 25*0.175 = 4.375 mmol

no of mol of NaOH = 14.5833*0.3 = 4.375 mmol

so that ,

concentration of salt formed = n/V in ml = 4.375/(25+14.5833) = 0.11 M

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

    = 7+1/2(9.3+log0.11)

    = 11.7

after 20 ml NaOH

no of mol of HCN taken = 25*0.175 = 4.375 mmol

no of mol of NaOH = 20*0.3 = 6 mmol

so that ,

no of mol of excess NaOH added = 6 - 4.375 = 1.625 mmol

concentration of excess NaOH added = 1.625/(25+20) = 0.0361 M

pOH = -log(OH-)

     = -log(0.0361)

    = 1.44

pH = 14 - 1.44 = 12.56