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Answer :
Final answer:
To find out how many mL of 0.25 M NaOH are needed to neutralize 0.15 g of oxalic acid dihydrate, calculate the moles of oxalic acid, account for it being diprotic, and then use the molarity equation to find the volume of NaOH required. The calculated value is 9.52 mL, which does not match any of the provided options.
Explanation:
To determine how many milliliters of 0.25 M NaOH are required to neutralize 0.15 g of oxalic acid dihydrate (H2C2O4 · 2H2O), we must first calculate the number of moles of oxalic acid dihydrate present. Oxalic acid dihydrate has a molar mass of approximately 126.07 g/mol.
The number of moles of oxalic acid dihydrate is calculated using the formula: moles = mass (g) / molar mass (g/mol).
moles of oxalic acid = 0.15 g / 126.07 g/mol = 0.00119 mol.
Oxalic acid is a diprotic acid, which means it can donate two protons per molecule during a reaction. Therefore, the mole ratio of oxalic acid to NaOH is 1:2. To neutralize 0.00119 mol of oxalic acid, we need twice that amount of NaOH:
moles of NaOH needed = 2 * 0.00119 mol = 0.00238 mol.
Next, we can find the volume of NaOH solution using the molarity formula: Molarity (M) = moles / Volume (L). Therefore, the volume (L) is equal to moles / Molarity (M).
Volume of NaOH = 0.00238 mol / 0.25 M = 0.00952 L.
To convert liters to milliliters, multiply by 1,000:
Volume of NaOH = 0.00952 L * 1,000 mL/L = 9.52 mL.
Since none of the options match our calculated value, and we are confident in our methodology and calculations, we refuse to answer from the provided options.
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