Answer to Question #350284 in General Chemistry for Cath

Question

Calculate the pH of 50 ml of 0.1000 M methylamine solution after the following volume of 0.1000 M HNO3, where added 25

Answer 1

n(CH₃NH₂) = C_M * V = 0.1 * 0.05 = 0.005mol

n(HNO₃) = C_M * V = 0.1 * 0.025 = 0.0025mol

CH₃NH₂ + HNO₃ = CH₃OH + N₂ + H₂O

1mol — 1mol

X — 0.0025mol

X = 0.0025 mol

n(CH₃NH₂) = 0.005 - 0.0025 = 0.0025 mol

V_new = 0.05 + 0.025 = 0.075 L

C_M(CH₃NH₂) =n/V =0.0025 /0.075 =0.033 M

CH₃NH₂ + H₂O = CH₃NH₃⁺ + OH^-

1mol CH₃NH₂ => 1mol OH^-

0.033 M => X = 0.033 M OH^-

Methylamine, CH₃NH₂, is a weak base that reacts with water according to the above equation. For this reaction at equilibrium, Kb = 4.4 x 10^-4

C_M (OH^-) = 0.033 * 4.4*10^-4 = 1.47*10^-5 M

pOH = -log [OH^-] = - log (1.47*10^-5) = 4.833

Using pH + pOH = 14, we find the pH value:

pH = 14 - pOH = 14 - 4.833 = 9.167

Answer: pH = 9.167