(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)
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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