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. Show all work.

Answer 1

HCN is a wek acid

so

a) initially, V = 0

HCN <-> H+ + CN-

Ka = [H+][CN-]/[HC]

Ka = 6.17*10^-10

pKa = -log(KA) = -log(6.17*10^-10) = 9.2097

6.17*10^-10 = x*x/(0.175-x)

x = 1.039*10^-5

pH = -log(1.039*10^-5) = 4.9833

b)

V = 7.29 mL

mmol of base = MV = 0.3*7.29 = 2.187 mmol

mmol of acid = MV = 25*0.175 = 4.375 mmol

after reaction

mmol of conjguate formed = 2.187

mmol of acid left = 4.375 -2.187 = 2.188

so.. this is a buffer

pH = pKa + log(A-/HA)

pH =9.2097 + log(2.187/2.188) = 9.21

c)

V = 14.5833

mmol of base = MV = 0.3*14.5833= 4.3749mmol

mmol of acid = MV = 25*0.175 = 4.375 mmol

this is neutralization

so

CN- + H2O <-> HCN + OH

Kb = [HCN][OH-]/[CN-]

Kb = Kw/Ka = (10^-14)/(1.039*10^-5) = 9.624*10^-10

M = mol/ (V1+V2) = 4.375 / (25+14.5833) =

9.624*10^-10 = x*x/ ( 0.11052 -x)

x = 1.03*10^-5

pOH = -log( 1.03*10^-5) = 4.987

pH = 14-4.987 = 9.013

d)

excess base

mmol base = MV = 0.3*20 = 6 mmol

mmol acid = 4.375   

after reaction

mmol of base left = 6-4.375   

6 - 4.37500 =1.625

Vtotal = 25+20 = 45

[OH-] = 1.625/45 = v

pOH = -log(0.0361) = 1.44249

pH = 14- 1.44249 = 12.55751