Question

A) A 29.0 mL sample of 0.274 M methylamine, CH₃NH₂, is titrated with 0.238 M hydroiodic acid.

The pH before the addition of any hydroiodic acid is ______.

B) A 24.5 mL sample of 0.220 M ethylamine, C₂H₅NH₂, is titrated with 0.399 M hydroiodic acid.

At the titration midpoint, the pH is ______.

C) A 28.2 mL sample of 0.293 M dimethylamine, (CH₃)₂NH, is titrated with 0.273 M hydrochloric acid.

After adding 13.6 mL of hydrochloric acid, the pH is ______.  

Answer 1

A)  

CH3NH2 + H2O -----------------> CH3NH3+ + OH-

       which gives OH-  

Kb of CH3NH2 = 5.0 x 10^-4

[OH-] = sqrt (Kb x C)

            = sqrt (5.0 x 10^-4 x 0.274)

             = 0.0117 M

pOH = -log [OH-] = -log (0.0117) = 1.93

pH + pOH = 14

pH = 12.07

B)

Kb of C2H5NH2 = 6.3 x 10^-4

pKb = -log Kb = -log (6.3 x 10^-4)

pKb = 3.20

at mid point : pOH = pKb

so

pOH = 3.20

pH + pOH = 14

pH = 10.80

C)

millimoles of base = 28.2 x 0.293 = 8.26

millimoles of acid = 0.273 x 13.6 = 3.71

base + acid --------------> salt

8.26         3.71                   0 ---------------> initial

4.55            0                   3.71 -------------> after reaction

pOH = pKb + log [salt /base]

pOH = 3.27 + log (3.71 /4.55)

pOH = 3.18

pH + pOH = 14

pH = 10.82