Question

# 1. A 100 mL sample of 0.100 M weak base, B (Kb = 3.7 x 10-4)...

1. A 100 mL sample of 0.100 M weak base, B (Kb = 3.7 x 10-4) is titrated with 0.250 M HNO3. Calculate the pH after the addition of: a) 0 mL acid b) 20 mL acid

a)

This is a base in water so, let the base be "B" and HB+ the protonated base "HB+"

there are free OH- ions so, expect a basic pH

B + H2O <-> HB+ + OH-

The equilibrium Kb:

Kb = [HB+][OH-]/[B]

initially:

[HB+] = 0

[OH-] = 0

[B] = M

the change

[HB+] = x

[OH-] = x

[B] = - x

in equilibrium

[HB+] = 0 + x

[OH-] = 0 + x

[B] = M - x

Now substitute in Kb

Kb = [HB+][OH-]/[B]

Kb = x*x/(M-x)

x^2 + Kbx - M*Kb = 0

x^2 + (3.70*10^-4)x - (0.10)(3.70*10^-4) = 0

solve for x

x = 0.0059

substitute:

[HB+] = 0 + x = 0.0059M

[OH-] = 0 + x = 0.0059 M

pH = 14 + pOH = 14 + log(0.0059) = 11.77

b)

mmol of acid = MV = 0.25*20 = 5

mmol of base = MV = 0.1*100 = 10

this is a buffer

after reaction

mmol of base left = 10-5 = 5

mmol fo conjguate acid formed = 5

then

pOH = pKb + log(BH+/B)

pOH = -log(3.7*10^-4) + log(5/5) = 3.43179

pH = 14 - pOH = 14 -3.43179 = 10.57

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