Question

Construct and calculate a theoretical titration curve, showing and discussing all key areas, for the titration of 0.100 M NaOH with 50.0 mL 0.100 M HNO_2 (K_aHNO_2 = 7.1 times 10^-4)

Answer 1

Key areas:

a) pure acid

b) buffer region --> Acid = Conjugate

c) equivalence point

d) excess base

so...

a)

for pure acid:

HNO2 <->H+ + NO2-

Ka = [H+][NO2-]/[HNO2]

7.1*10^-4 = x*x/(0.1-x)

x = 0.0081

pH = -log(x) = -log(0.0081) = 2.0915; at V = 0

b)

at equivalence point

mmol of acid = mmol of conjugate

so.. use henderson haselbach equation

pH = pKa + log(A-/HA)

pKa 0 -log(Ka) = -log(7.1*10^-4) = 3.148

pH = 3.148 + log(1/1)

the volume of base:

equivalence point:

Vacid*Macid = Mbase*Vbase

50*0.1 = 0.1*V

V = 50 mL of base required

half equivalence point:

V = 1/2*V

V = 50/2 = 25 mL of base has pH = 3.148

c)

in equivalnece point:

Vtotal = 50+50 = 100 mL

New concentration of NO-2 in solution:

[NO2-] = M1V1/(Vtotal) =50*0.1 / 100 = 0.05 M

so

hydrolysis ocurss

NO2- + H2O <-> HNO2 + OH-

Kb = [HNO2][OH-] /[ NO2-]

Kb = Kw/KA = (10^-14) /(7.1*10^-4) = 1.4084*10^-11

1.4084*10^-11 = x*x/(0.05-x)

x = 8.39*10^-7

pOH= -log(OH) = -log(8.39*10^-7 = 6.07

pH = 14- 6.07 = 7.93

d)

finally... after 10 mL exces base.

Vtotal = 50+60 = 110 mL

base added = Mbase*Vbae = 0.1*60 = 6 mmol

acid neutralized = Macid*Vacid = 0.1*50 = 5 mmol

so

mmol of OH- left = 6-5 = 1 mmol left

[OH-] = 1 / 60 = 0.0166 M

pOH = -log(0.0166 ) = 1.77989

pH = !4-1.77989 = 12.22

so...

the diagram:

el C1