A student titrates 0.139 g of an unknown monoprotic weak acid to the equivalence point with 44.6 ml of 0.100 m naoh (aq). what i

Question

AlexFokin [52]

4 years ago

9

A student titrates 0.139 g of an unknown monoprotic weak acid to the equivalence point with 44.6 ml of 0.100 m naoh (aq). what i

s the molar mass of the weak acid?

Chemistry

1 answer:

givi [52] · Answer 1 · 2021-12-20T20:08:17+03:00

The molar mass of the weak acid is calculated as follows

by use of an example of monoprotic weak such as CH3COOH reacting with NaOH

that is NaOH + CH3COOH = CH3COONa + H2O

calculate the moles of NaOH used
= molarity x volume/1000

= 44.6 ml x 0.100/ 1000 = 4.46 x10^-3 moles

by use of mole ratio between NaOH to monoprotic acid which is1:1 the moles of monoprotic acid is also 4.46 x10^-3 moles

molar mass is therefore = mass/ moles = 0.139 g/ (4.46 x10^-3) = 31.17 g/mol