Question

Consider the titration of a 26.0 −mL sample of 0.180 M CH3NH2 with 0.150 M HBr. Determine each of the following.

A the initial

pH

B the volume of added acid required to reach the equivalence point

C   the

pH at 6.0 mL of added acid

D   the

pH at one-half of the equivalence point

E   the

pH at the equivalence point

F the

pH after adding 5.0 mL of acid beyond the equivalence point

Answer 1

a) there is only base in the solution at the initIal value.

Make an I.C.E Table

CH3NH2(aq) <--> CH3NH3+(aq) + OH-(aq)

I...........0.170 0 0
C...........-x +x +x
E........0.180-x x x

Kb = x² / (0.180-x)

4.4*10^-4 = x² / (0.180-x)
x = ~8.88*10^-3

[OH-] = ~8.88*10^-3
pOH = -log(8.88*10^-3)

P^{OH =}   2.05063740921

pH = 14 - pOH
pH = 11.94 .

b) the volume of added acid required to reach the equivalence point

The equivalence point is reached when moles of acid equals moles of base.

M1V1 = M2V2
(0.180)(26) = (0.150)(V2)
31.2mL = V2

HBr --> H+ + Br-.....CH3NH2 --> CH3NH3+ + OH-

Moles CH3NH2 = 0.18 M x 0.026L = 0.00468moles
At stoichiometric point moles HBr = 0.00468moles
Volume HBr = 0.00468moles / 0.15M = 0.0325L
Total volume = 0.026L + 0.0325L = 0.0585L
[CH3NH3+] = 0.00468moles / 0.0585L =0.08 M

CH3NH3+ + H2O >>CH3NH2 + H3O+
Ka = Kw / Kb = 2.27 x 10^-11 = x² / 0.08-x
x² + 2.27x10^-11x - 1.88x10^-12
x = [H+] = 1.37x10^-6M
pH = ~5.8

c) with a volume 6ml beyond the equivalence point, the total volume would be 0.0645L
0.00468moles / 0.0645L = 0.072M
2.27x10^-11 = x² / 0.078-x
x² + 2.27x10^-11x - 1.77x10^-12 = 0
x = [H+] = 1.33x10^-6M
pH = 5.88

In the same way please do for the remaining stages of the addition of excess acid is added.

And you have asked more than one question at a time.

please dont mind and thank you.