Answer to Question #350284 in General Chemistry for Cath

Question #350284

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

1
Expert's answer
2022-06-14T20:42:03-0400

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

n(HNO3) = CM * 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

Vnew = 0.05 + 0.025 = 0.075 L


CM(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

CM (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

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