Question

A)

Kb=6.9×10^{-4 for} diethylamine

A 26.8 mL sample of 0.330 M diethylamine, (C₂H₅)₂NH, is titrated with 0.333 M hydrobromic acid.

The pH before the addition of any hydrobromic acid is ___ .

B)

Kb=6.3×10^{-5 for} trimethylamine

A 22.2 mL sample of 0.275 M trimethylamine, (CH₃)₃N, is titrated with 0.293 M nitric acid.

At the equivalence point, the pH is ___.

C)

Kb=5.9×10^{-4 for} dimethylamine

A 25.9 mL sample of 0.239 M dimethylamine, (CH₃)₂NH, is titrated with 0.213 M hydroiodic acid.

At the titration midpoint, the pH is ___.

Answer 1

A)

Diethyl amine = (C2H5)2NH = 26.8 mL of 0.330M                  Kb= 6.9x10^-4

for weak bases

[OH-] = square root of KbxC

[OH-] = square root of [ 6.9x10^-4 x 0.330]

[OH-] = 1.51x10^-2M

-log[OH-] = -log( 1.51x10^-2)

POH= 1.82

PH+POH= 14

PH= 14 - POH

PH= 14 - 1.82

PH= 12.18.

b)

Trimethyl amine = (CH3)3N = 22.2 mL of 0.275M

number of moles of (CH3)3N = 0.275Mx0.0222L = 0.006105 moles

HNO3 = 0.293M

at equivalent point

Volume of HNO3 = 0.275x22.2/0.293 = 20.836 mL = 20.84 mL

Total volume of the solution = 22.2+20.84 =43.04 mL= 0.04304L

at end point

PH = 7 - 1/2[ PKb + log C]

kb= 6.3x10^-5

-log(Kb) = -log( 6.3x10^-5)

Pkb= 4.20

C= number of moles/total volume in L.

C= 0.006105/0.04304 = 0.1418 M

PH= 7 - 1/2[ 4.20 + log(0.1418)]

PH= 5.32

c)

Dimethyl amine = (CH3)2NH = 25.9ml of 0.239 M

Kb= 5.9x10^-4

-log(Kb) = -log(5.9x10^-4)

Pkb= 3.23

at mid point

POH= PKb

POH= 3.23

PH= 14 - POH

PH= 14-3.23

PH= 10.77