Question

1. A sample of 0.2140 g of an unknown monoprotic weak acid was dissolved in 25.0 mL of water and titrated with 0.0950 M NaOH.  The acid required 15.50 mL of NaOH to reach the equivalence point.  What is the molar mass of the unknown acid?

Answer 1

HA + NaOH -----> NaA+ H2O

moles of NaOH = moles of HA

moles of NaOH = M*V

                         = 0.0950*(15.50/1000)

                         = 0.00147 moles

moles = mass/molarmass

molarmass = mass/moles

                  = 0.2140/0.00147

                  = 145.578

                  = 145.6 g/mol