Question

The following data were collected during the titration of a solid sample of potassium acid phthalate ("KHP") with a solution of sodium hydroxide. The NaOH solution was added to the KHP from a buret.

From the experiment data below, calculate the molarity of the NaOH solution.

Data:

Molar Mass of potassium acid phthalate= 204.23g/mol

Mass of weighing bottle plus KHP= 23.4061 mL

Mass of weighing bottle= 22.0515 g

Initial buret reading= 0.20 mL

Final buret reading= 39.13 mL

Answer 1

The molarity of NaOH is found by the law of equivalence which equates the volume of KHP of a certain concentration required to fully neutralize the volume of NaOH taken, in an unknown molarity. Mathematically, it is given as V_KHP x M_KHP = V_NaOH x M_NaOH and M_NaOH = (V_KHP x M_KHP) / V_NaOH.

So, for this the quantities required first is the molarity of KHP. The mass of KHP in the bottle is mass of bottle with KHP - mass of empty bottle = 1.3546g. This gives the molarity of KHP as 1.3546/204.23 = 6.6327mmol. (1mmol = 0.001mol).

Now that its given that an actual volume of 39.13 - 0.2 = 38.93mL of NaOH was required to neutralize the KHP completely, we can find molarity of NaOH by taking that the same molarity of KHP in 1.3546g was in the 38.93mL of NaOH used, again by the law of equivalence. This gives 6.6327mmol of NaOH in 38.93mL. To find molarity, we use the formula (weight/molecular weight) x (1000/volume). Here, we already know the numerator value so we get 0.0066327 x 1000/38.93 = 0.17037M of NaOH in the unknown solution.