A solution of sodium hydroxide is standardized against potassium hydrogen phthalate. From the following data, calculate the molarity of NaOH solution? (See details below)
For a titration is necessary that the moles of NaOH are equivalent to the mole of KHP (that have a MM of 204,22g/mol).
V of NaOH used =(31,26-0,23) = 31,03 mL = 0,03103 L
but the mol of the two substance are the same therefore
Molarity of NaOH = (mol KHP)/ (V NaOH used for titration):
Since NaOH is monovalent its molarity is equal to its Normality =0,0688 (eq)/L.
For the oxalic acid (MM= 90,03 g/mol; ME = 45,01 g/eq) is necessary a double quantity of moles of NaOH, since oxalic acid is biprotic (1 eq = 2 mol), or (that is the same) the Equivalent Mass of the oxalic acid is the half of the Molecular Mass .
We need (i) a stoichiometrically balanced equation for our primary standard:
And this is a convienient 1:1 stoichiometry, so moles of phthalate is equivalent to moles of base:
And (ii) equivalent quantities of
Given the stoichiometry, this molar quantity was contained in the volume of titrant, i.e.
So we have
Again, we need a stoichiometric equation:
And thus, we require under
This is a long problem, so I certainly would check the arithmetic. It is all too easy to make an error.