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

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

Nov 20, 2017

#### Explanation:

We use $\text{265 drops}$ to make up a volume of $\left(36.0 - 23.0\right) \cdot m L = 13.0 \cdot m L$.

And thus we take the quotient.....

$\frac{265 \cdot \text{drops}}{\left(36.0 - 23.0\right) \cdot m L}$

$= 20.4 \cdot \text{drops} \cdot m {L}^{-} 1$, i.e. $\text{1 drop} \equiv 0.05 \cdot m L$...

When we do a titration from a $50 \cdot m L$ or $25 \cdot m L$ burette one DROP of titrant is conceived to have a volume of $0.01 \cdot m L$... Of course a drop from a Pasteur pipette, or a drop from a tap, has a lesser or greater volume.