Determine the ph of 0.57 m methylamine (ch3nh2) with kb = 4.4 x 10-4 : ch3nh2 (aq) + h2o (l) ⇌ ch3nh3 + (aq) + oh - (aq)

Question

Chemistry

Gavyn Singh

11 July, 00:14

Determine the ph of 0.57 m methylamine (ch3nh2) with kb = 4.4 x 10-4 : ch3nh2 (aq) + h2o (l) ⇌ ch3nh3 + (aq) + oh - (aq)

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Judah Meza · Answer 1

Answer is: pH of methylamine is 12,2.

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

Kb (CH₃NH₂) = 4,4·10⁻⁴.

c₀ (CH₃NH₂) = 0,57 M.

c (CH₃NH₃⁺) = c (OH⁻) = x.

c (NH₂OH) = 0,57 M - x.

Kb = c (CH₃NH₃⁺) · c (OH⁻) / c (CH₃NH₂).

0,00044 = x² / (0,57 M - x).

Solve quadratic equation: x = c (OH⁻) = 0,0156 mol/L.

pOH = - log (0,0156 mol/L.) = 1,80.

pH = 14 - 1,80 = 12,2.