Answer to Question #350289 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 60

Answer 1

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

n(HNO₃) = C_M * V = 0.1 * 0.06 = 0.006mol

CH₃NH₂ + HNO₃ = CH₃OH + N₂ + H₂O

1mol — 1mol

0.005mol — X

X = 0.005 mol

n(HNO₃) = 0.006 - 0.005 = 0.001 mol

V_new = 0.05 + 0.06 = 0.11 L

C_M(HNO₃) =n/V=0.001/0.11 =0.0091M

HNO₃ => H⁺ + NO₃^-

1mol HNO₃ => 1mol H⁺

0.0091 M => X = 0.0091 M H⁺

pH = -log [H⁺] = - log (9.1*10^-3) = 2.04

Answer: pH = 2.04