Answer to Question #350290 in General Chemistry for cath

Question #350290

(a.) Calculate the pH of the solution that is 0.100 M methylamine (NH2CH3) and 0.400 M of


sodium methyalamine


(b.) Calculate the pH change when 80 mL of 0.15 KOH is added to the 100 mL solution stated


above (a)


(c.) Calculate the pH change when 80 mL of 0.10 HNO3 is added to the 100 mL solution


stated above (a)

1
Expert's answer
2022-06-16T21:21:04-0400

a.)

CM (CH3NH2) = 0.1 M

CH3NH2 + H2O = CH3NH3+ + OH-

1mol CH3NH2 => 1mol OH-

0.1 M => X = 0.1 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-) = 0.1 * 4.4*10-4 = 4.4*10-6 M


pOH = -log [OH-] = - log (4.4*10-6) = 5.36

pH = 14 - 5.36 = 8.64


b.)

n(CH3NH2) = CM * V = 0.1 * 0.1 = 0.01 mol

n(CH3NHNa) = CM * V = 0.4 * 0.1 = 0.04 mol

n(KOH) = CM * V = 0.15 * 0.08 = 0.012 mol


CH3NH2 + KOH = CH3NHK + H2O

1mol — 1mol KOH

0.01mol — X


X = 0.01 mol KOH


n(KOH) = 0.012 + 0.01= 0.002 mol

Vnew = 0.1 + 0.08 = 0.18 L


CM(KOH) =n/V=0.002/0.18 =0.1 M


KOH => K+ + OH-

1mol KOH => 1mol OH-

0.002 M => X = 0.002 M OH-


pOH = -log [OH-] = - log (0.002) = 2.7

pH = 14 - 2.7 = 11.3


c.)

n(CH3NH2) = CM * V = 0.1 * 0.1 = 0.01 mol

n(CH3NHNa) = CM * V = 0.4 * 0.1 = 0.04 mol

n(HNO3) = CM * V = 0.1 * 0.08 = 0.008 mol


CH3NHNa + HNO3 = CH3NH2 + NaNO3

1mol — 1mol HNO3 — 1mol CH3NH2

X — 0.008 mol — Y


X = 0.008 mol CH3NHNa

Y = 0.008 mol CH3NH2


n(CH3NH2) = 0.01 + 0.008= 0.018 mol

Vnew = 0.1 + 0.08 = 0.18 L


CM(CH3NH2) =n/V=0.018/0.18 =0.1 M


n(CH3NHNa) = 0.04 - 0.008= 0.032 mol

Vnew = 0.1 + 0.08 = 0.18 L


CM(CH3NHNa) =n/V=0.032/0.18 =0.178 M


CH3NH2 + H2O = CH3NH3+ + OH-

1mol CH3NH2 => 1mol OH-

0.1 M => X = 0.1 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-) = 0.1 * 4.4*10-4 = 4.4*10-6 M


pOH = -log [OH-] = - log (4.4*10-6) = 5.36

pH = 14 - 5.36 = 8.64



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