Question

Dissociation constants of acids:

HOAc Ka=1.78x10^-5

H2SO4 Ka1=1x10⁷ Ka2=1.0x10^-2

HCNO Ka=2x10^-4

H2CO3 Ka1=4.3x10^-7 Ka2=5.6x10^-11

H2O Ka=1.0x10^-14

HIO Ka=2.3x10^-11

HIO3 Ka = 1.7x10^-4

HNO2 Ka=4.5x10^-4

HF Ka = 3.5x10^-4

H2C2O4 Ka1 = 6.0x10^-2 Ka2 = 6.1x10^-5

Dissociation constants of bases:

NH3 Kb=1.76x10^^-5

CH3NH2 Kb=4.4x10^-4

NH2OH Kb=1.1x10^-8

I titrate 50.00 mL of a 0.100 M CH3NH2 solution with 0.075 M HCl.

A) What is the initial pH of the solution?

B) What is the pH after the addition of 10.00 mL of the HCl?

C) What is the pH at the equivalence point?

D) What is the pH 10.00 mL past the equivalence point?

Answer 1

A) What is the initial pH of the solution?

CH3NH2 Kb=4.4x10^-4

pKb = -log Kb = 3.36

pOH = 1/2 [pKb - logC]

pOH = 1/2 [3.36 - log 0.1]

pOH = 2.18

pH + pOH = 14

pH = 11.82

B) What is the pH after the addition of 10.00 mL of the HCl?

millimoles of CH3NH2 = 50 x0.1 = 5

millimoles of HCl = 10 x 0.075 = 0.75

CH3NH2 + HCl -----------------------> CH3NH3+

5                0.75                                  0

4.25                                                     0.75

pOH = pKb + log [CH3NH3+ / CH3NH2]

pOH =3.36 + log (0.75 / 4.25)

pOH = 2.61

pH + pOH =14

pH = 11.39

C) What is the pH at the equivalence point?

volume of HCl at equiavlence point = 5 / 0.075 = 66.7 mL

salt only remains here. salt concentration = 5 / (66.7 + 50) = 0.0429 M

pH = 7 - 1/2 [pKb + logC]

pH = 7 - 1/2 [3.36 + log 0.0429]

pH = 6.00

D) What is the pH 10.00 mL past the equivalence point?

volume of HCl = 76.7 mL

millimoles of HCl = 0.075 x 76.7 = 5.75

[H+] = 5.75 - 5 / (76.7 + 50) =5.94 x 10^-3 M

pH = -log [H+]

pH = 2.23