A 0.5216 ?g sample of an unknown monoprotic acid was titrated with 9.94×10?2 M NaOH. The...
A 0.5216 ?g sample of an unknown monoprotic acid was titrated with 9.94×10?2 M NaOH. The equivalence point of the titration occurs at 23.76 mL . Part A Determine the molar mass of the unknown acid. Express your answer using three significant figures
Homework Answers
HA + NaOH -----------------> NaA + H2O
no of moles of NaOH = molarity * volume in L
= 9.94*10^-2 *0.02376 = 0.00236moles of NaOH
1 mole of NaOH react with 1 moles of HA
0.00236 moles of NaOH react with 0.00236 moles of HA
molar mass of HA = mass of HA/no of moles of HA
= 0.5216/0.00236 = 221g/mole
molar mass of Acid = 221g/mole
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