What is the molarity of a NaOH solution if 21.93 mL of base is required to titrate 0.243g of oxalic acid?

Answer 1

To find the molarity of the NaOH solution, you need to use the equation for molarity:

Molarity (M) = moles of solute / volume of solution in liters.

First, you need to find the moles of oxalic acid. The molar mass of oxalic acid is 90.03 g/mol.

moles of oxalic acid = mass / molar mass
= 0.243 g / 90.03 g/mol
= 0.00271 mol

Since the balanced equation for the reaction between NaOH and oxalic acid is 2 NaOH + H2C2O4 -> Na2C2O4 + 2 H2O, the number of moles of NaOH is twice the number of moles of oxalic acid.

moles of NaOH = 2 * moles of oxalic acid
= 2 * 0.00271 mol
= 0.00542 mol

Next, you need to find the volume of the NaOH solution in liters. The volume of the NaOH solution is given as 21.93 mL.

volume of NaOH solution = 21.93 mL / 1000 mL/L
= 0.02193 L

Finally, you can calculate the molarity of the NaOH solution:

Molarity = 0.00542 mol / 0.02193 L
= 0.247 M

Therefore, the molarity of the NaOH solution is 0.247 M.