Question

1. what minimum number of grams of oxalic acid monohydrate, h₂c₂o₄·h₂o, would you specify for a titration of no fewer than 15.0 ml of 0.100 m naoh? both of the hydrogens from oxalic acid are replaceable in this reaction.

Explanation:

Step1: Calculate moles of NaOH

Moles = Molarity × Volume (L)
$n_{\text{NaOH}} = 0.100\ \text{M} × 0.0150\ \text{L} = 0.00150\ \text{mol}$

Step2: Find mole ratio of acid to NaOH

Reaction: $\text{H}_2\text{C}_2\text{O}_4·\text{H}_2\text{O} + 2\text{NaOH} → \text{Na}_2\text{C}_2\text{O}_4 + 3\text{H}_2\text{O}$
Ratio: $1:2$, so $n_{\text{acid}} = \frac{0.00150}{2} = 0.000750\ \text{mol}$

Step3: Calculate molar mass of acid

Molar mass = $2×1 + 2×12 + 4×16 + 2×1 + 16 = 126\ \text{g/mol}$

Step4: Compute mass of acid

Mass = Moles × Molar mass
$m = 0.000750\ \text{mol} × 126\ \text{g/mol} = 0.0945\ \text{g}$

Answer:

0.0945 g