Question

12. Consider the titration of 40.0 mL of 0.500 M NH3 with 1.00 M HCl a) What is the initial pH of the NH3(aq)? b) What is the pH halfway to the equivalence point? c) What is the volume of HCl needed to reach the equivalence point? d) What is the pH at the equivalence point? e) Sketch the titration curve. Label the point(s) where there is a A) a weak base B) weak acid C) Buffer D) Strong acid in excess

Answer 1

a)

This is a base in water so, let the base be CH3NH2 = "B" and CH3NH3+ = 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 + (1.8*10^-5)x - (0.5)(1.8*10^-5= 0

solve for x

x = 0.00299

substitute:

[HB+] = 0 + x = 0.00299M

[OH-] = 0 + x = 0.00299M

pH = 14 + pOH = 14 + log(0.00299) = 11.48

b)

pH half way is given by

pH = pKa + log(NH3/NH4+)

pKa = 14 ´Kb = 14-4.25 = 9.25

pH = 9.25+ log(1)

pH = 9.25

c)

V o fHcl required

mmol of base = MV = 40*0.5 = 20

mmol of acid = 20

V acid =mol/M = 20/1 = 20 mL required

d)

pH in equivalence

[NH4+] = 20/(40+20) = 0.33

the next equilibrium is formed, the conjugate acid and water

BH+(aq) + H2O(l) <-> B(aq) + H3O+(aq)

The equilibrium is best described by Ka, the acid constant

Ka by definition since it is an base:

Ka = [H3O][B]/[BH+]

Ka can be calculated as follows:

Ka = Kw/Kb = (10^-14)/(1.8*10^-555) = 5.55*10^-10

get ICE table:

Initially

[H3O+] = 0

[B] = 0

[BH+] = M

the Change

[H3O+] = 0 + x

[B] = 0 + x

[BH+] = -x

in Equilibrium

[H3O+] = 0 + x

[B] = 0 + x

[BH+] = M - x

substitute in Kb expression

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

5.55*10^-10 = x*x/(0.333-x)

solve for x

x^2 + Ka*x - M*Ka= 0

solve for x with quadratic equation

x = H3O+ =1.359*10^-5

[H3O+]  =1.359*10^-5

pH = -log([H3O+]) = -log(1.359*10^-5) = 4.87

pH = 4.87