Question

1.For the diprotic weak acid H2A, Ka1 = 2.5 × 10-6 and Ka2 = 7.5 × 10-9. What is the pH of a 0.0400 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?

2.HClO is a weak acid (Ka = 4.0 × 10–8) and so the salt NaClO acts as a weak base. What is the pH of a solution that is 0.026 M in NaClO at 25 °C?

3.NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.057 M in NH4Cl at 25 °C?

Answer 1

Ques 1. Since Ka1 >> Ka2, then nearly all of the H3O+ produced comes from the first hydrolysis reaction.

Molarity . . . . .H2A + H2O <==> H3O+ + HA-
Initial . . . . . .0.0400 . . . . . . . . . . . . .0 . . . . . .0
Change . . . . . .-x . . . . . . . . . . . . . . .x . . . . . .x
Equilibrium . .0.0400-x . . . . . . . . . . . x . . . . . .x

Ka1 = [H3O+][HA-] / [H2A] = (x)(x) / (0.0400-x) = 2.5 x 10^-6

x^2 / (0.0400-x) = 2.5 x 10^-6 . . .because Ka1 is small compared to [H2A]o, we can delete the -x term.
x^2 / 0.0400 = 2.5 x 10^-6
x^2 = 1 x 10^-7
x = 0.0003162 M = 3.1 x 10^-4 M = [H3O+]

pH = -log [H3O+] = -log [3.1 x 10^-4] = 3.508

Also [H2A]eq = 0.0400 - x = 0.0400 - (3.1 x 10^-4) = 0.03969 M

[A2-]eq = ?

HA^-    <=> H⁺    + A^2-

^I 3.1 x 10^-4 M ^{0 0}

^C    - x +x +x

E    3.1 x 10^-4 - x x x

Ka2 = [H+] [A^2-] / [HA^-]

7.5 × 10-9 = x^2 / (3.1 x 10^-4 M)

7.5 × 10-9 x  (3.1 x 10^-4 M) =  x^2

  74.99 =  x^2

x = 8.659 M = [A^2-]

Ques 2.   Kb = 1 x 10^-14 / Ka
Kb = 1 x 10^-14 / 4.0 x 10^-8
Kb = 2.5 x 10^-7

Kb = [HOCl][OH-] / [OCl-]
Kb = [OH-]^2 / [OCl-]
[OH-] = SQRT (Kb x [OCl-])
[OH-] = SQRT [2.5 x 10^-7 x (0.026)]
[OH-] = 3.25 X 10^-9 molL-1

pOH = -log[OH-]
pOH = -log(3.25 X 10^-9)
pOH = 8.48

pH = 14 - pOH
pH = 14 - 8.48
pH = 5.52

Ques 3. For NH4Cl

Ka =Kw/Kb = 1 x 10^-14 / 1.8 x 10^-5 = 5.56 x 10^-10
The NH4Cl dissociates completely and then the NH4+ reacts with water according to Ka as follows:
NH4+ +H2O-----> NH3 + H3O+

[NH3][H3O+]/[NH4+]=5.56 x 10^-10

H3O+ is the conjugate acid
NH3 and H3O+ are equal but unknown so the both equal X and are insignificant to 0.057 so no subtraction from 0.057 is necessary

X^2/[0.057] = 5.56 x 10^-10
X^2 = 3.169 x 10^-11
X = 1.5846 x 10^-11 M = H3O+ or H+

so pH=-log[H+] = -log[ 1.5846 x 10^-11]

pH=10.8