Calculate the oxalic acid mass required for neutralization of 25 ML of 0.1 M NaOH solution.Dihydrate Organic Diacid (H2C2O4.2 H2O, molar mass = 126.04 g/mol)
H2ox = the oxalic acid dihydrate
H2ox + 2NaOH ==> Na2ox + 2H2O
millimols NaOH = mL x M = 25 x 0.1 = 2.5
You will need 1/2 that or 1.25 millimols H2ox = 0.00125 mols.
grams H2ox = mols H2ox x molar mass H2Ox = ?
To calculate the mass of oxalic acid required for neutralization, we can use the following formula:
Mass of oxalic acid = (volume of NaOH solution in L) × (Molarity of NaOH solution) × (Molar mass of oxalic acid)
Given:
Volume of NaOH solution = 25 mL = 0.025 L
Molarity of NaOH solution = 0.1 M
Molar mass of oxalic acid (H2C2O4.2H2O) = 126.04 g/mol
Plugging in the values, we get:
Mass of oxalic acid = (0.025 L) × (0.1 M) × (126.04 g/mol)
= 0.0313 g
Therefore, the mass of oxalic acid required for neutralization of 25 mL of 0.1 M NaOH solution is approximately 0.0313 grams.
To calculate the mass of oxalic acid required for neutralization, we need to use the concept of stoichiometry. In this reaction, the balanced chemical equation is:
H2C2O4 + 2NaOH → Na2C2O4 + 2H2O
From the equation, we can see that 1 mole of oxalic acid reacts with 2 moles of sodium hydroxide. Substituting the given concentration and volume of NaOH solution into the equation, we can calculate the number of moles of NaOH:
moles of NaOH = concentration of NaOH × volume of NaOH solution
= 0.1 mol/L × 0.025 L
= 0.0025 moles
Since 1 mole of oxalic acid reacts with 2 moles of NaOH, the number of moles of oxalic acid required is twice the number of moles of NaOH:
moles of H2C2O4 = 2 × moles of NaOH
= 2 × 0.0025 moles
= 0.005 moles
Now we can calculate the mass of oxalic acid using the molar mass of H2C2O4 dihydrate:
mass of H2C2O4 = moles of H2C2O4 × molar mass of H2C2O4
= 0.005 moles × 126.04 g/mol
= 0.6302 grams
Therefore, the mass of oxalic acid required for neutralization of 25 mL of 0.1 M NaOH solution is 0.6302 grams.