How many mL of 1.555 M NaOH are required to completely neutralize 3.587 g of H2C2O4 (MM = 90.0) ?

51.45 mL

Explanation:

We are given;

Molarity of NaOH as 1.555 M Mass of the acid, H₂C₂O₄ as 3.587 g

We are required to determine the volume of NaOH required.

Step 1: Write the balanced equation for the reaction.

The balanced equation for the reaction between NaOH and H₂C₂O₄ will be;

2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O

Step 2: Determine the number of moles of the acid, H₂C₂O₄

We know that, moles are calculated by dividing the mass of the compound by its molar mass.

Moles = Mass : Molar mass

Therefore;

Moles of H₂C₂O₄ = 3.587 g : 90.0 g/mol

= 0.040 moles

Step 3: Determine the number of moles of NaOH that reacted

From the equation, the mole ratio of NaOH to H₂C₂O₄ is 2 : 1

Therefore; Moles of NaOH = Moles of H₂C₂O₄ * 2

Thus;

Moles of NaOH = 0.040 moles * 2

= 0.080 moles

Step 4: Determine the volume of NaOH that reacted

Molarity of NaOH = 1.555 M

But; Molarity = Moles : Volume

Rearranging the formula;

Volume = Moles : Molarity

Therefore;

Volume of NaOH = 0.080 moles : 1.555 M

= 0.05145 L

But, 1 L = 1000 mL

Therefore;

Volume = 51.45 mL

Hence, the volume of HCl required is 51.45 mL