Consider the titration of a24.0 mL sample of 0.180 mol L^-1 CH_3NH_2 (K_b = 4.4

Question

Consider the titration of a24.0 mL sample of 0.180 mol L^-1 CH_3NH_2 (K_b = 4.4 times 10^-4) with 0.145 mol L^-1 HBr. Determine each quantity the initial pH Express your answer using two decimal places the volume of added acid required to reach the equivalence point the pH at 4.0 mL of added acid Express your answer using two decimal places. the pH at one-half of the equivalence point. Express your answer using two decimal places. the pH at the equivalence point Express your answer using two decimal places.

Explanation / Answer

Q1

a)

initial pH for base:

B + H2O <> HB+ + OH-

Kb = [HB+][OH-]/[B]

4.4*10^-4 = x*x/(0.18-x)

x = OH = 0.00868

pH = 14 + log(OH)= 14 + log(0.00868)

pH = 11.938

B)

find volume required for equivalence point

mmol of base = mmol of acid

so

M1V1 = M2V2

for

Vbase = M1/M2*V1 = (0.18/0.145)*24 = 29.7931 mL

C)

pH for 4 mL of acid

this is a buffer

so

pOH = pKb + log(HB+/B)

pKb = -log(Kb) = -log(4.4*10^-4) = 3.35654

mmol of acid = MV = 0.145*4 = 0.58

mmol of base left = 24*0.18 - 0.58 = 3.74

pOH = 3.35654 + log(0.58/3.74) = 2.5470

pH = 14- 2.5470

pH = 11.453

D)

pH at half equivalence point:

this is a buffer, and special one

since

HB+ = B (half and half)

so

log(HB+/B) = 1

then buffer equation

pOH = pKb + log(1)

pOH = 3.35654

pH = 14-3.35654 = 10.6434

E)

pH in equivalnce point

HB+ + H2O <-> B + H3O+

Ka = [B][H3O+]/[HB+]

Ka = Kw/Kb = (10^-14)/(4.4*10^-4) = 2.2727*10^-11

M in equivalence point

Vtotal = V1+V2 = 24+29.7931 = 53.7931

[M] = M1V1/(V1+V2) = (0.18+24)/(53.7931) = 0.0803076

2.2727*10^-11 = x*x/(0.0803076 -x)

x = H3O+ = 1.347*10^-6

pH = -log(X) = -log(1.347*10^-6

pH = 5.8706

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