Answer to Question #350290 in General Chemistry for cath

Question

(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)

Answer 1

a.)

C_M (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

C_M (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) = C_M * V = 0.1 * 0.1 = 0.01 mol

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

n(KOH) = C_M * 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

V_new = 0.1 + 0.08 = 0.18 L

C_M(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) = C_M * V = 0.1 * 0.1 = 0.01 mol

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

n(HNO₃) = C_M * 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

V_new = 0.1 + 0.08 = 0.18 L

C_M(CH₃NH₂) =n/V=0.018/0.18 =0.1 M

n(CH₃NHNa) = 0.04 - 0.008= 0.032 mol

V_new = 0.1 + 0.08 = 0.18 L

C_M(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

C_M (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