[ArbuzToWoda]

I'm really confused now. I've run 4 distinct titrations with 1g of my presumably phtalic/3-nitrophtalic crystals and the results are at least weird. Here are the gram amounts of sodium hydroxide each titration needed to get to the faint pink:
1. 0.519g
2. 0.511375g
3. 0.4438g
4. 0.5185g
Not sure what happened with the third titration. but the results from others are pretty similar. The problem is, when substituted to the system of equations:
n1+n2=2*(m NaOH/ M NaOH)
n1 * M1 + n2 * M2 = 1g

where n1 and n2 (x and y) are amounts of moles of 3/4-nitrophtalic acid and phtalic acid.
The results pop up with either x or y negative. Look at this one from wolfram in the attachement.

[AWK]

1 g of phthalic acid needs 2*NaOH*(1/C8H6O4) = 0.4815 g of NaOH
1 g of nitrophthalic acid needs 2*NaOH*(1/C8H5NO6) = 0.3780 g of NaOH
How did you prepare the NaOH solution for titration?

[ArbuzToWoda]

Yeah, that's why I was surprised. Depends on the run, I did 1%, 2.5% and 3% by mass solutions I believe. I dissolved the corresponding amounts of NaOH in distilled water, didn't do it molar. Is that wrong?

[AWK]

Solid NaOH always contains Na₂CO₃.

[rolnor]

Sooner or later you will need better analysis method and TLC is the cheapest, its just a fact.

[ArbuzToWoda]

So how should I conduct a proper titration with solid NaOH? Maybe precipitate out carbonates with barium nitrate and then somehow go further? Not sure though, that would disrupt the concentration of solution.

[AWK]

You must prepare a standard NaOH solution. For standardization samples of potassium acid phthalate (KHP) or oxalic acid dihydrate may be used.
cerritos.edu/chemistry/_includes/docs/Chem_111/Lab/Exp%2012%20-A%20Standardization%20of%20NaOH%20F%2008.pdf