Question

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

Answer 1

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

Step-by-step 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

Answer 2

9.52 ml The balanced equation is H2C2O4(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(i) So for every mole of oxalic acid dihydrate, you need 2 moles of NaOH to neutralize. Let's determine the number of moles of oxalic acid dihydrate we have. Start with the atomic weights of the elements involved. Atomic weight hydrogen = 1.00794 Atomic weight carbon = 12.0107 Atomic weight oxygen = 15.999 Molar mass oxalic acid dihydrate = 6 * 1.00794 + 2 * 12.0107 + 6 * 15.999 = 126.06304 g/mol Moles oxalic acid dihydrate = 0.150 g / 126.06304 g/mol = 0.001189881 mol So we need 2 * 0.001189881 mol = 0.002379762 mol of NaOH for the reaction. Since 1 L of 0.250 m NaOH solution has 0.250 moles of NaOH, divide the number of moles we need by the number of moles in a liter of the solution to get the number of liters needed. The convert from liters to milliliters. 0.002379762 mol / 0.250 mol/l = 0.009519047 l = 9.52 ml