Question

Titration problem? Which method of solving is the correct way?

100.0 mL OF 0.100 M NH₃ (BOH)        
             (NO HCl SOLUTION ADDED, weak base by itself!!)

             NH_{3 +} H₂O _---->  NH₄⁺+ OH^-
                                     <----

This what the notes say:

        K_b = [NH₄⁺][OH^-]/[ NH₃] = 1.8 x 10^-5 = x²/0.100-x

But isn't it supposed to be 1.8 x 10^-5 = x²/0.010-x because you have to take into account the volume and 100mL x .100M equals .01M?

Answer 1

In this relation

K_b = [NH₄⁺][OH^-]/[ NH₃]

[...] means concentration at equilibrium (mol/L= M).

Then use concentrations (mol/L), not quantities (mol). This relation

K_b = [NH₄⁺][OH^-]/[ NH₃] = 1.8 x 10^-5 = x²/0.100-x

is correct.

Your statement 100mL x .100M equals .01M is not correct. The result is not 0.01 M (concentration, i.e. mol/L), but mol (quantity).

If you need to calculate the mol number, the correct form is:

0.100 L x 0.1 mol/L = 0.010 mol    (pay attention to the measurement units)

So, :

- don