Titration

[bri1ash2]

50.0 mL of 0.0018 M aniline (a weak base) is titrated with 0.0048 M HNO3. How many mL of the acid are required to reach the equivalence point?

[Paul Robbins]

Originally Posted by bri1ash2

50.0 mL of 0.0018 M aniline (a weak base) is titrated with 0.0048 M HNO3. How many mL of the acid are required to reach the equivalence point?

Each molecule of aniline will react with one molecule of HNO₃.

We need to calculate how many moles of aniline there are. This will be the same number as the moles of HNO₃ we need. Using moles of acid and the molarity of the acid we will be able to calculate the volume of acid we need.


1. Moles of Aniline

Moles = Volume x Molarity

= 0.050 l x 0.0018 mol/l

= 0.00009 mol

2. Hence we require 0.00009 moles of HNO₃.

3. Volume = Moles / Molarity

= 0.00009 / 0.0048

= 0.01875 l

= 18.75 ml of HNO₃ needed.

[robin]

You'll already know this Paul, but for anyone who doesn't, this is a 1:1 reaction, so all you need to do is look at the concentrations of the two reactants. The ratio is 0.0018 M : 0.0048 M.

There's 50 ml of the aniline so the amount of nitric acid needed is

(0.0018/0.0048) x 50 ml = 18.75 ml

If you're not confident about doing this sort of thing, it's safer to follow the method Paul showed above.

[Paul Robbins]

I agree with everything you said robin, thanks.