A sample of pure oxalic acid (H2C2O4.2H2O) weighs 0.2000 g and requires 30.12 ml of KOH solution for

complete neutralization. Calculate the molarity of the KOH solution. (0.105 M)

The molarity of KOH is 0.1055 M

calculation

Step 1: write the equation for reaction between H₂C₂O₄.2H₂O and KOH

H₂C₂O₄.2H₂O + 2 KOH → K₂C₂O₄ + 4 H₂O

step 2: find the moles of H₂C₂O₄.2H₂O

moles = mass: molar mass

from periodic table the molar mass H₂C₂O₄.2H₂O = (1 x2) + (12 x2) + (16 x4) + 2 (18) = 126 g/mol

= 0.2000 g : 126 g/mol = 0.00159 moles

step 3: use the mole ratio to calculate the moles of KOH

H₂C₂O₄.2H₂O : KOH is 1:2

therefore the moles of KOH = 0.00159 x 2 = 0.00318 moles

step 4: find molarity of KOH

molarity = moles/volume in liters

volume in liters = 30.12/1000=0.03012 L

molarity is therefore = 0.00318/0.03012 = 0.1055 M