Question

In standardizing the solution of aqueous sodium hydroxide, a chemist overshoots the end point and adds too much naoh(aq). would this error result in a calculated concentration of naoh that was overestimated or underestimated? explain your reasoning.

Answer 1

Answer: Concentration of NaOH calculated will be underestimated.

Explanation:

End point is an observational point , which tells us about the completion of reaction between the titrant (solution in burette) and titre(solution in conical flask) in titration experiment.

In this case , NaOH is titrant whose concentration is unknown.

$M_1=\text{molarity of titre}$ , $M_2=\text{molarity of NaOH}$

$V_1=\text{volume of titre}$ , $V_2=\text{volume of NaOH}$

$M_1V_1=M_2V_2$

$M_2=\frac{M_1V_1}{V_2}$....(1)

According to question a chemist overshoots the end point and adds to much of NaOH solution, which means increase in the value of $V_2$.

Then the value of $M_2$ in equation (1), will get lowered , which means that the concentration of NaOH was lower than that of the actual value. Hence underestimated concentration of NaOH.