Question

Consider the reaction H2C2O4 + 2 NaOH -> Na2C2O4 + H2O You are standardizing (trying to find out molarity of NaOH solution) a NaOH solution with H2C2O4. You start with 0.4g of H2C2O4. At the end point, you have dispensed 17.75mL of NaOH. What is the molarity of the NaOH solution?

a. 4.5M

b. 5.0M

c. 0.5M

d. 0.4M

Answer 1

Answer : C ( 0.5 M)

Given, H2C2O4 + 2 NaOH -------> Na2C2O4 + H2O

volume of the solution = 17.75 mL = 0.01775 L

mass of H2C2O4 = 0.4 g

molar mass of H2C2O4 = 90 g / mol

no.of moles of H2C2O4 = mass of H2C2O4 / molar mass of H2C2O4

= 0.4 / 90

= 0.0044 moles

As per the stoichiometric equation,

1mole of H2C2O4 requires --------> 2 moles of NaOH

==> 0.0044 moles of NaOH requires -----> 2*0.0044 = 0.0088 moles of NaOH

( since, reaction reaches end point i.e., reaction completed )

Therefore, initial number of moles of NaOH = 0.0088 moles

Now,

Molarity of NaOH = no.of moles of the NaOH / Total volume in Liters

= 0.0088 / 0.01775 = 0.495 = 0.5 M

Therefore, Molarity of NaOH = 0.5 M