Question

Pertaining to the titration of 25.0mL of 0.100 M HCN (Ka=6.2x10-10) with 0.200 M NaOH, calculate the pH at the following points. a) initial pH, 0.0mL NaOH b) after the addition of 4.0mL NaOH c) after the addition of 6.25 mL NaOH d) at the equivalence point e) after the addition of 17.0 mL NaOH

Answer 1

a)

HA <->H + + A-

Ka = [H+][A-]/[HA]

6.2*10^-10 = x*x/(0.1-x)

x = H = 7.87*10^-6

pH = -log( 7.87*10^-6) = 5.1

b)

mmol of base = MV = 4*0.2 = 0.8 mmol

mmol of acid = 25*0.1 = 2.5 mmol

mmol of conjugate formed = 0.8 mmol

mmo lof acid left = 2.5-0.8 = 1.7 mmol

this is a buffer

so

pKa = -log(Ka) = -log(6.2*10^-10) = 9.20760

pH = 9.20760+ log(0.8/1.7) = 8.88

c)

mmol of base = MV = 6.25*0.2 = 1.25 mmol

mmol of acid = 25*0.1 = 2.5 mmol

mmol of conjugate formed = 1.25 mmol

mmo lof acid left = 2.5-1.25= 1.25 mmol

this is a buffer

so

pKa = -log(Ka) = -log(6.2*10^-10) = 9.20760

pH = 9.20760+ log(1.25/1.25)

pH= 9.21

d)

in equivalence point

[A-]= M1V!/(V1+V2) = 25*0.1 / (25+12.5)= 0.0666

A- + H2O <->HA + Oh-

Kb = [HA][OH-]/ [A-]

(10^-14)/(6.2*10^-10) = x*x/(0.0666-x)

x = OH = 0.001

pOH= -log(0.001) = 3

pH = 14-3 = 11

finally

mmol of OH excess = MV = 17*0.2 = 3.4 mmol of OH-

[OH-] = 3.4 / (17+12.5+25) = 0.0623

pH = 14 +log(0.0623 = 12.79448