Question

# Consider the titration of a 26.0-mL sample of 0.180 M CH3NH2 with 0.145 M HBr. (The...

Consider the titration of a 26.0-mL sample of 0.180 M CH3NH2 with 0.145 M HBr. (The value of Kb for CH3NH2 is 4.4×10−4.)

Part A

Determine the initial pH.

Part B

Determine the volume of added acid required to reach the equivalence point. answer to 3 sig figs

Part C

Determine the pH at 6.0 mL of added acid

Part D

Determine the pH at one-half of the equivalence point.

Part E

Determine the pH at the equivalence point.

Part F

Determine the pH after adding 5.0 mL of acid beyond the equivalence point.

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

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

Kb = x^2 / (0.180-x)

4.4*10^-4 = x^2 / (0.180-x)
x = ~7.92*10^(-3)

[OH-] = ~7.92*10^(-3)
pOH = -log(~7.92*10^(-3))= 0.101

pH = 14 - pOH
pH = 13.8

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

M1V1 = M2V2
(0.180)(26) = (0.145)(V2)
32.276 mL = V2

c) When you add 4mL of acid, the volume changes.
Total volume = 26mL + 6mL

Initial mol CH3NH2
0.180 mol CH3NH2 / L * 0.0256L = 0.00468 mol CH3NH2

0.145 mol HBr / L * 0.004L = .00058 mol HBr

Determine the Ka value
Kw = Ka * Kb
Ka = (1.0*10^-14) / (4.4 *10^-4) = ~ 2.27*10^-11

pH = pKa + log ([base]/[acid])
pH = -log(2.27*10^-11) + log [(1.466*10^-4)/(2.0690*10^-5)]
pH = 10.64 + 0.8503
pH = ~11.49

d) Half of the equivalence point.
The mol ratio of base : acid will be equal.
Therefore it's 1:1. pH = pKa

pH = pKa + log ([base]/[acid])
pH = 10.64 + log (1/1)
pH = 10.64

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