Question

A 29.6 mL sample of 0.260 M diethylamine, (C₂H₅)₂NH, is titrated with 0.338 M hydrochloric acid.

After adding 33.7 mL of hydrochloric acid, the pH is

.

Use the Tables link in the References for any equilibrium constants that are required.

Answer 1

Kb = 1.3 * 10^-3 (this is what I found, please correct you if have something different in your table)

Ka = Kw/Kb = 10^-14/1.3 * 10^-3 = 7.7 *10^-12

Moles of diethylamine= 0.260 M * 0.0296 L = 0.007696 moles

Moles of HCl = 0.338 M * 0.0337 L = 0.0114 moles

         Et2NH     +    HCl     <-->       Et2NH2+    +     Cl-

I           0.007696         0.0114                          0

C       -0.007696        -0.007696                 0.007696

E               0                 0.0037                   0.007696

Et2NH2+ + H2O ---> H3O+ + + Et2NH

In the equation of Ka I’m going to add the moles of HCl too because that’s take part on the acid.

Ka = (0.0037+x)(x) / (0.007696-x)

(7.7 *10^-12) * (0.007696-x) = (0.0037+x)(x)

x = 1.6*10^-11

[H3O+] = (0.0037 + 1.6*10^-11) / (0.026 + 0.0337)= 0.062

pH = -log(0.062) = 1.21

Hope this helps!