Question

Unknown Acid Number Mass of weighing paper Mass of weighing paper and unknown acid Concentration of NaOH solution 615978 1.6610g 0.10M 24.5481 12. 2674 m2 4.91 Volume of NaOH to reach the equivalence point Volume of NaOH at half of the equivalence point pH at half of the equivalence point CALCULATIONS Show all calculations. Mass of unknown acid •5013g Moles of NaOH to reach equivalence point .602453 mol Moles of unknown acid Molar mass (equivalent mass) of unknown acid pKa of unknown acid Identity of unknown acid (from Table 1) Percent error of the molar mass of unknown acid

Answer 1

from Henderson - Hanelbalch equation we get, pH = pka + log CCA [CB where CB = conjug ate bare A- Acid At half eanrivalene point [A] = [ CB] then, pH=Pk a flog B = Pka to PH=pka . at half of the earrivalence point pH = 6.91 . Pka-4.91 unknown acid has one pka value. So unk nown acid har 1 disociable Proton. So, unknocon acid is mono- Protic ocide this unknown acid mill. react with Naoh at 1. ratio upto comivalluce point # Since moles of woolt required to reach eanävalence fount = 0.002453mol. moles of unknown seid also wan Present 0.002453 mol

#hans of unknown acid was present 0.50139 . According to calculation 0.002453 mo unknown aid = 0.50139 -0.2013 0.002453 = 110.66 g Molar man of unknown acid in 110.66 g/mol #pka of unknown aud=6.9) table 1 needed. But to # For rent here not answer given.