Question

A 25.5 ml aliquot of HCl of unknown concentration was titrared with .113 M NaOH. It took 51.2 ml of the base to reach the endpoint of the titration. The concentration (M) of the acid was ?

Answer 1

Answer;

0.227 M

Explanation;

Volume of HCl = 25.5 mL or 25.5 / 1000 => 0.0255 L

Molarity or concentration of NaOH = 0.113 M

Volume of NaOH = 51.2 mL / 1000 = 0.0512 L

Number of moles NaOH:

Moles = Molarity x Volume

= 0.113 x 0.0512

= 0.0057856 moles of NaOH

From the equation we can obtain the mole ratio:

HCl + NaOH = NaCl + H2O

1 mole HCl : 1 mole NaOH

Therefore;

Moles of HCl : 0.0057856 moles NaOH

Hence; moles of HCl = 0.0057856 x 1 / 1

= 0.0057856 moles of HCl

Molarity of HCl = moles of HCL / Volume of HCl

M = 0.0057856 / 0.0255

= 0.227 M