How many millilitres of 0.250 m naoh are required to neutralize 0.150 g of oxalic acid dihydrate h2c2o4 · 2h2o?

Given:

Concentration of NaOH = 0.250 M

Mass of oxalic acid = 0.150 g

To determine:

Volume of NaOH required to neutralize the given oxalic acid

Explanation:

The chemical reaction is:

H2C2O4 + 2NaOH → Na2C2O4 + 2H2O

Therefore: 1 mole of H2C2O4 requires 2 moles of NaOH

i.e. moles of NaOH = 2(moles of H2C2O4)

# moles of H2C2O4 = mass of oxalic acid/molar mass

 Mass of oxalic acid = 0.150 g

molar mass of H2C2O4.2H2O = 126 g/mol

# moles of oxalic acid = 0.150/126 = 0.00119 moles

Thus, moles of NaOH = 2(0.00119) = 0.00238 moles

Volume of NaOH = Moles NaOH/ Molarity NaOH

V(NaOH) = 0.00238/0.250 mole.L-1 = 0.00952 L = 9.52 ml

Ans: Volume of NaOH required = 9.52 ml

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