Question

2. Calculate [OH-] of a 0.45 M solution of methylamine, CH3NH2 (Kb = 4.38 × 10-4). Then determine the pH and extent of dissociation.

Answer 1

# Calculation of [OH-] :

Let us consider reaction of methylamine

CH₃NH₂ + H₂O ----> CH₃NH₃⁺ + OH^-

Kb = [CH₃NH₃⁺] [OH^-] / [CH₃NH₂ ]

Lets consider x to be the amount of dissociation, we already know the value of Kb. Substituting them, we get Kb = x² /0.45-x = 4.38 * 10^-4

since the value of Kb is small,we consider the value of x to me very small and can be omitted for now in the denominator of the previous equation.

x² = 4.38*10^-4*0.45

= 1.971* 10^-4

x = 0.014

Let us substitute this value of x in the first equation to find out how small the variations would be because of elimination of x from the denominator.

Kb = x² /0.45-0.014 = 4.38 * 10^-4

x² = 0.436*4.38*10^-4

= 1.909*10^-4

x = 0.0138~ This is rounded off to 0.014

Hence [OH-] = 0.014 mol/L

# Calculation of pH :

pOH = -log[OH-] = - log[0.014]

pOH = 1.85

We know that pH + pOH = 14

1.85 + pH = 14

Thus pH = 12.15

# Extent of dissociation :

We know that pKb = -logKb. In this stated equation note the minus sign. Because of this sign, the inversely proportional relationship between value of Kb and extent of dissociation happens. The smaller the value of Kb ,the greater the value of pKb. The value of kb is very low and hence the extent of dissociation is high which makes methylamine a strong base.