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