Question

If Compound A (256 mg) was hydrolyzed using 12.5 mL of 1.0 M KOH and the amount of remaining KOH was back titrated and required 44.9 mL of 0.25 M HCL to neutralize the excess NaOH, what were the saponification equivalent and neutralization equivalent, and what is compound A?

Answer 1

KOH    +    HCL --------> KCL + H2O

1 mole     1 mole

no of moles of HCl = molarity * volume in L

                              = 0.25*0.0449 =0.011225 moles of HCl

no of moles of KOH = no of moles of HCl

no of moles of KOH = 0.011225 moles

Remaining no of moles of KOH = 0.011225 moles

Actual no of moles of KOH = molarity * volume in L

                                          = 1*0.0125 = 0.0125 moles

used no of moles of KOH   = 0.0125-0.011225 = 0.001275moles

compound A was hydrolyzed use KOH = 0.001275 moles

no of moles of A compound   = no of moles of hydrolyzed use KOH

no of moles of A compound = 0.001275 moles

no of moles = weight of substance/Gram molar mass

0.001275   = 256*10^-3 gm/Gram molar mass

Gram molar mass = 256*10^-3/0.001275 = 200.78gm/mole