Question

An analytical chemist weighs out 0.144 g of an unknown monoprotic acid into a 250 mL volumetric flask and dilutes to the mark with distilled water. She then titrates this solution with 0.0900 M NaOH solution. When the titration reaches the equivalence point, the chemist finds she has added 22.2 mL of NaOH solution. Calculate the molar mass of the unknown acid. Be sure your answer has the correct number of significant digits. mol x 6 ?

Answer 1

Balanced chemical equation is:

NaOH + HA ---> NaA + H2O

lets calculate the mol of NaOH

volume , V = 22.2 mL

= 2.22*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 9*10^-2*2.22*10^-2

= 1.998*10^-3 mol

According to balanced equation

mol of HA reacted = (1/1)* moles of NaOH

= (1/1)*1.998*10^-3

= 1.998*10^-3 mol

This is number of moles of HA

mass(HA)= 0.144 g

use:

number of mol = mass / molar mass

1.998*10^-3 mol = (0.144 g)/molar mass

molar mass = 72.07 g/mol

Answer: 72.1 g/mol