Question

Oxalic acid, H2C2O4*2H2O (molar mass = 126.07 g/mol) is oftenused as a primary standard for the standardization of a NaOHsolution. If 0.147 g of oxalic acid dihydrate is nutralizedby 23.64mL of a NaOH solution, what is the molar concentration ofthe NAOH solution? Oxalic acid is a diprotic acid. (What is thebalanced equation?)

Answer 1

0.09865M of NaOH

Step-by-step explanation:

As a diprotic acid, 1 mole of oxalic acid reacts with 2 moles of NaOH as follows:

H2C2O4 + 2 NaOH → 2 H2O + Na2C2O4

With the mass of the acid and its molar mass we can obtain the moles of oxalic acid. Twice these moles are the moles of NaOH in the solution. As you required 23.64mL = 0.02364L of the NaOH, you can knkow molarity, thus:

Moles Oxalic acid:

0.147g * (1mol / 126.07g) = 1.166x10⁻³ moles Oxalic acid.

Moles NaOH:

1.166x10⁻³ moles Oxalic acid * (2 moles NaOH / 1 mole H2C2O4) = 2.332x10⁻³ moles NaOH

Molarity:

2.332x10⁻³ moles NaOH / 0.02364L =

0.09865M of NaOH