To determine the molarity of naoh solution, student took 3.5 g of khp (khp – potassium hydrogen phthalate; molar mass = 204.22 g/mol) and dissolved in 50 ml of water and titrated with the given unknown molarity naoh solution loaded in burette. his burette volume read 1.85 ml at the start of the experiment and 24.65 ml when the phenolphthalein indicator turned pink.

KHP molarity = (3.5/204.22) x (1000/50) = 0.343 M
NaOH volume = 24.65 - 1.85 = 22.8 ml
Molarity of NaOH = (weight/40) x (1000/22.8) = 1.096W
                                KH₂PO₄ + NaOH → NaKHPO₄ + H₂O
initial                         0.343       1.096W       0                 0
after reaction (0.343 - 1.096W)     0         1.096W        1.096W
KH₂PO₄ and NaKHPO₄ is considered as buffer and its (H₂PO₄⁻) pKa value is 7.21
phenolphthalein indicator turns pink means it is at the equivalence point
so pH = pKa
by using buffer equation
pH = pKa + log [salt] / [acid]
when pH = pKa so [salt] = [acid]
1.096 W = 0.343 - 1.096W
2.192 W = 0.343
W = 0.156 g
NaOH molarity = 1.096 W = 1.096 x 0.156 = 0.17 M
Molarity of NaOH = 0.17 M