When we use a burette in an acid base titration, how many drops are equivalent to a 13.0*mL13.0⋅mL volume?

2 Answers
Kathryn Bell
Nov 20, 2017

20.4 drops per 1 mL

Explanation:

First we need to calculate how much the volume of the solution has increased. We do this by finding the difference between the end volume and the start volume:

36.0 mL - 23.0 mL = 13.0 mL

265 drops increases the volume of water by 13.0 mL, so to calculate the number of drops per mL we just divide the number of drops by 13.0 mL.

265 drops / 13.0 mL = 20.4 drops per mL

anor277
Nov 20, 2017

About 20 drops....

Explanation:

We use "265 drops"265 drops to make up a volume of (36.0-23.0)*mL=13.0*mL(36.0−23.0)⋅mL=13.0⋅mL.

And thus we take the quotient.....

(265*"drops")/((36.0-23.0)*mL)265⋅drops(36.0−23.0)⋅mL

=20.4*"drops"*mL^-1=20.4⋅drops⋅mL−1, i.e. "1 drop"-=0.05*mL1 drop≡0.05⋅mL...

When we do a titration from a 50*mL50⋅mL or 25*mL25⋅mL burette one DROP of titrant is conceived to have a volume of 0.01*mL0.01⋅mL... Of course a drop from a Pasteur pipette, or a drop from a tap, has a lesser or greater volume.

Related questions
Impact of this question
4633 views around the world
You can reuse this answer
Creative Commons License