Question

Determine the pH of each solution.

Dissociation Constants at 25 ∘C

Name	Formula	Ka
Formic acid	HCHO2	1.8×10−4
Name	Formula	Kb
Methylamine	CH3NH2	4.4×10−4

a) 0.21 M KCHO2

b) 0.21 M CH3NH3I

c)0.21 M KI

Answer 1

a) 0.21 M KCHO2

KCHO2 --> K+(aq) + CH2O-(aq)

expect hydrolysis

CH2O-(aq) + H2O(l) <->CH2OH + OH-(aq)

Let HA --> CH2OH and A- = CH2O- for simplicity

since A- is formed

the next equilibrium is formed, the conjugate acid and water

A- + H2O <-> HA + OH-

The equilibrium s best described by Kb, the base constant

Kb by definition since it is an base:

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

Ka can be calcualted as follows:

Kb = Kw/Ka = (10^-14)/(1.8*10^-4) = 5.55*10^-11

get ICE table:

Initially

[OH-] = 0

[HA] = 0

[A-] = M

the Change

[OH-] = + x

[HA] = + x

[A-] = - x

in equilibrium

[OH-] = 0 + x

[HA] = 0 + x

[A-] = M - x

substitute in Kb expression

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

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

solve for x

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

solve for x with quadratic equation

x = OH- =3.41*10^-6

[OH-]  =3.41*10^-6

pOH = -log(OH-) = -log(3.41*10^-6 = 5.4672

pH = 14-5.4672= 8.5328

pH = 8.5328

b) 0.21 M CH3NH3I

ionizes in CH3NH3I --> CH3NH3+ + I-

Let B --> CH3NH2 and BH+ = CH3NH3+ for simplicity

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)/(4.4*10^-4) = 2.2727*10^-11

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-]

2.2727*10^-11 = x*x/(M-x)

solve for x

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

solve for x with quadratic equation

x = H3O+ = 2.18*10^-6

[H3O+]  = 2.18*10^-6

pH = -log([H3O+]) = -log(2.18*10^-6) = 5.66

pH = 5.66

c)0.21 M KI

ionizaiton of the salt:

KI --> K+(aq) + I-(aq)

there will not be any kind of hydrolysis or any side equilibrium

therefore

pH = 7 approx, i.e. neutral