Calculate the pH of 1.0M solution of methylamine (𝐶𝐻3𝑁𝐻2 , 𝐾𝑏 = 4.38 × 10−4)?

Method 1 : Use ICE table

         CH₃NH₂(aq) + H₂O(l) ⇌ CH₃NH₃⁺(aq) + OH⁻(aq)   Kb = 4.38 × 10⁻⁴
Initial:    1.0 M                             0 M                0 M
Change:  -y M                           +y M              +y M
Eqm:  (1.0 - y) M                         y M                y M
             ≈ 1.0 M

At equilibrium:
Kb = [CH₃NH₃⁺] [OH⁻] / [CH₃NH₂]
4.38 × 10⁻⁴ = y² / 1.0
y = √(1.0 ×4.38 × 10⁻⁴)
y = 0.0209

pOH = -log[OH⁻] = -log(0.0209) = 1.68
pH = pKw - pOH = 14.00 - 1.68 = 12.32

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Method 2: Use short-cut formula for small Kb

pOH = -(1/2){log(Kb) + log[CH₃NH₂]} = -(1/2){log(4.38 × 10⁻⁴) + log(1.0)} = 1.68
pH = pKw - pOH = 14.00 - 1.68 = 12.32

here is a formula that we can use when the Kb is very small due to 
very little dissociation of the base . This bypasses the need to ICE tables too  

[OH-] = SQRT( Kb* c)

= SQRT( 4.38 × 10−4 * 1.0 )
= 0.02092845
pOH = 1.68
pH = 12.32 <<<<< your answer