Question

In the standardization of sodium hydroxide experiment, the following data were obtained: the mass of oxalic acid dihydrate used was 0.256 g; the volume of sodium hydroxide used was 19.45 mL. What is the molarity of the NaOH? (The molar mass of oxalic acid dihydrate is 126.07 g/mol) Helpful Equation:

H2C2O4 (aq) + 2 NaOH (aq) → Na2C2O4 (aq) + 2 H2O (l)

Answer 1

Standardization of sodium hydroxide is done by acid base titration. When oxalic acid is allowed to react with sodium hydroxide, sodium oxalate and water are obtained. In this titration, to locate the end point phenolphthalein indicator is used.

Answer 2

To calculate the molarity of the NaOH solution, determine the moles of oxalic acid used, which informs us of the moles of NaOH since they react in a 2:1 ratio. Then, use the volume of NaOH solution to find its molarity, which is calculated to be 0.2089 M.

Step-by-step explanation:

To find the molarity of the NaOH solution, we must first calculate the moles of oxalic acid dihydrate used.

Using the provided molar mass of oxalic acid dihydrate (126.07 g/mol), the moles of oxalic acid dihydrate can be calculated as follows:

Moles of oxalic acid dihydrate = Mass / Molar mass

= 0.256 g / 126.07 g/mol

= 0.002031 moles of oxalic acid dihydrate

The balanced chemical equation H2C2O4 (aq) + 2 NaOH (aq) → Na2C2O4 (aq) + 2 H2O (l) indicates that 2 moles of NaOH react with 1 mole of oxalic acid dihydrate.

Therefore, the moles of NaOH required are twice the moles of oxalic acid dihydrate:

Moles of NaOH = 2 * Moles of oxalic acid dihydrate

= 2 * 0.002031 moles

= 0.004062 moles of NaOH

Finally, the molarity of the NaOH can be calculated using the volume of NaOH solution:

Molarity (M) = Moles / Volume in liters

= 0.004062 moles / 0.01945 L

= 0.2089 M NaOH