Question

a. 0.2565 g of KHP was weighed and dissolved in 25 mL of distilled water. It was then titrated against an NaOH solution. The amount of NaOH needed to reach the end point was determined to be 1.54 mL. What is the concentration of the NaOH solution.

b. In the process of weighing out the KHP, you did not notice that some of the KHP that you weighed out spilled while you walked from the balance to your station. You continued your experiment and carried out the standardization process. The molarity of NaOH from you experiment will be higher or lower than the true value? Explain.

Answer 1

Number of moles of KHP, n= mass/molarmass

= 0.2565 g/204.2(g/mole)

= 0.0013 moles

1 mole of KHP reacts with NaOH

0.0013 moles of KHP reacts with 0.0013 moles of NaOH

So molarity of NaOH = number of moles/volume in L

= 0.0013 moles /(1.54mL*10^-3L/mL)

= 0.816 M

b. If some of KHP was expelled then the number of moles of KHP calculated was less than the expected.

So the number of moles of NaOH calculated was less so molarity as calculated were less.