Question

A sample of 0.2140 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0950MNaOH. The acid required 27.4 mL of base to reach the equivalence point. What is the molar mass of the acid?

Answer 1

moles of NaOH = Molarity of naOH volume of NaOH in Liters = 0.095 M x 0.0274 L = 0.002603 mol

he said acid given is mono protic acid

let mono protic acid = HA

write the balanced equation

HA + NaOH ---> NaA + H2O

from the balanced equation

1 mol of NaOH nuetralize the 1 mol of HA accordingly

0.002603 mol of NaOH nuetralize the 0.002603 mol of HA

so we got th emol of mono protic acid = 0.002603 mol

molar mass of HA = mass of HA / moles of HA

= 0.214 g / 0.002603 mol

= 80.67 g/mol