Question

Part A

Find the [OH−][OH−] of a 0.49 M M  methylamine (CH3NH2CH3NH2) solution. (The value of KbKb for methylamine (CH3NH2CH3NH2) is 4.4×10−44.4×10−4.)

Part B

Find the pHpH of a 0.49 M M  methylamine (CH3NH2CH3NH2) solution.

Answer 1

Answer

part A :[OH-] = 0.46 M

,part B pH= 12.17

explanation:

We have been given , concentration of methyl amine = 0.49 M

Kb = 4.4*10^-4.

Writing the equilibrium equation,

CH3NH2(aq) + H₂O(l) ⇌ CH₃NH₃⁺(aq) + OH⁻(aq) .

Kb = [ CH3NH3(+)][OH(-)] /[CH3NH2]

since, [CH3NH3+] = [ OH-]

Kb= [OH-]^2/ [CH3NH2]

[OH-]^2 = Kb *[CH3NH2]

[OH-] =√4.4*10^-4*0.49

[OH-] = 1.46*10^-2

[OH-] = 0.0146 M.

Explanation

part B : since OH - = 0.0146 M

pOH = -log [OH-]

= -log [0.0146 ]

=1.83

So,

pH = 14- pOH

p​​​​​H= 14 - 1.83 = 12.17

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