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).

#mol KHP= (0,436g)/(204,22 (g/(mol))) = 0,00213 mol #.

V of NaOH used =(31,26-0,23) = 31,03 mL = 0,03103 L

#mol NaOH = M V =M 0,03103 L #
but the mol of the two substance are the same therefore
Molarity of NaOH = (mol KHP)/ (V NaOH used for titration): #M(NaOH)=(0,00213mol)/(0,03103 L)= 0,0688 (mol)/L #.
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 .
#eq H_2C_2O_4= (0,1453 g)/(45,01g/(eq)) = 0,003228 eq#
#V(NaOH) = (eq H_2C_2O_4) / (N(NaOH)) =(0,003228 eq)/(0,0688 (eq)/L)= 0,0469 L = 46,9 mL#

We need (i) a stoichiometrically balanced equation for our primary standard:

#C_6H_4(CO_2^(-)K^(+))(CO_2H) + NaOH rarr C_6H_4(CO_2^(-)K^(+))(CO_2^(-)Na^(+))+H_2O#

And this is a convienient 1:1 stoichiometry, so moles of phthalate is equivalent to moles of base:

And (ii) equivalent quantities of #"KHP"#,

i.e. #"Moles of KHP"=(0.4536*g)/(204.22*g*mol^-1)=2.221xx10^-3mol#.

Given the stoichiometry, this molar quantity was contained in the volume of titrant, i.e.

#[NaOH(aq)]="Moles of base"/"Volume of titrant"#

#((0.4536*g)/(204.22*g*mol^-1))/((31.26-0.23)xx10^-3L)=7.158xx10^-2*mol*L^-1#.

So we have #NaOH#, and we use to titrate a mass of oxalic acid. You must know this is a diacid, i.e. #HO_2C-CO_2H#, with a molecular mass of #90.03*g*mol^-1# (the dihydrate is much common, i.e. #HO_2C-CO_2H*(OH_2)_2#, so if your results are off, question the quality of your oxalic acid, and you can assess this easily by melting point).

Again, we need a stoichiometric equation:

#"HO"_2"CCO"_2"H"(aq) + "2NaOH"(aq) rarr "Na"^(+)""^(-)"O"_2"CCO"_2^(-)"Na"^+ +2"H"_2"O"(l)#

#"Moles of oxalic acid"# #=# #(0.1453*g)/(90.03*g*mol^-1)=1.614xx10^-3*mol#.

#"Volume of NaOH titrant required"# #=# #(2xx1.614xx10^-3*mol)/(7.158xx10^-2*mol*L^-1)xx1000*mL*L^-1=??*mL#.

And thus, we require under #50*mL# of titrant, and this is a good volume for a titration.

This is a long problem, so I certainly would check the arithmetic. It is all too easy to make an error.