Question

A solution of an unknown acid is prepared by dissolving 5.573 g of the solid acid in sufficient DI water to make 300.00 mL of solution 19.92 mL of a 0.1253 M NaOH solution are required to neutralize 10.00 mL of this acid solution? What is the equivalent molar mass of the acid?

Answer 1

moles of NaOH required to neutralize 10.00 mL of the acid = MxV

= 0.1253 mol/L x 19.92 mL x (1L / 1000 mL) = 0.00250 mol

Hence moles of acid in 10.00 mL of the acid solution = moles of NaOH = 0.00250 mol acid

Hence moles of acid in 300.00 mL of the acid solution = (0.00250 mol acid / 10 mL solution) x 300 mL solution

= 0.075 mol

=> total moles of acid = 0.075 mol

=> mass of acid / molar mass = 0.075 mol

=> 5.573 g / molar mass = 0.075 mol

=> molar mass = 5.573 g / 0.075 mol = 74.3 g/mol (answer)