Respuesta :
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
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
