Question

If you were to overshoot the endpoint by 1 drop while you were standardizing the NaOH solution, what would be your % error? Assume the actual volume is 30.00 mL and there are exactly 20 drops in 1.00 ml for the sake of this calculation.

Answer 1

Answer:

$e(\%)=0.17\%$

Explanation:

Volume of a drop:

$V_{drop}=\frac{1 mL}{20 drop}$

$V_{drop}=0.05 mL$

To estimate the error:

$e(\%)=\frac{V_{real}-V{theorerical}}{V{theoretical}}*100\%$

$e(\%)=\frac{(30 mL+0.05mL)-30mL}{30mL}*100\%$

$e(\%)=0.17\%$