An impure sample of (COOH)2 · 2 H2O that has a mass of 2.9 g was dissolved in water and titrated with standard NaOH solution. The titration required 32.4 mL of 0.144 mo- lar NaOH solution. Calculate the percent (COOH)2 · 2 H2O in the sample.
After dissolving the oxalic acid into water, the problem reduces to 1mole Oxalic Acid + 2moles Sodium Hydroxide => 1mole Sodium Oxalate + 2moles Water. In this, moles of NaOH used > moles of OxA neutralized by 2-times. To equate moles of OxA and NaOH => 2(moles OxA) = 1(moles NaOH).
Then, 2 x moles Oxalic Acid neutralized = Molarity of Sodium Hydroxide x Volume of NaOH(liters)
2(moles OxA) = (0.144M)(.0324L) =>
moles OxA neutralized = [(0.144M)(.0324L)/2]mole OxA = 0.0023 mole OxA.Then, grams of OxA = (0.0023 mole OxA)(90 gms OxA/mole) = 0.21gms Oxa.
%OxA neutralized = (0.21g/2.90g)100% = 7.2%
To calculate the percent (COOH)2 · 2 H2O in the sample, we need to determine the number of moles of (COOH)2 · 2 H2O in the sample and divide it by the total moles present.
1. Convert the mass of (COOH)2 · 2 H2O to moles:
Mass of (COOH)2 · 2 H2O = 2.9 g
Molar mass of (COOH)2 · 2 H2O = (12.01 g/mol + 2 * 16.00 g/mol) * 2 + 2 * 1.01 g/mol = 90.04 g/mol
Number of moles of (COOH)2 · 2 H2O = (2.9 g) / (90.04 g/mol)
2. Determine the number of moles of NaOH used in the titration:
Volume of NaOH solution = 32.4 mL = 0.0324 L
Molar concentration of NaOH = 0.144 mol/L
Number of moles of NaOH = (0.0324 L) * (0.144 mol/L)
3. Calculate the molar ratio between (COOH)2 · 2 H2O and NaOH:
From the balanced chemical equation:
2 moles of NaOH react with 1 mole of (COOH)2 · 2 H2O.
4. Determine the number of moles of (COOH)2 · 2 H2O in the sample:
Number of moles of (COOH)2 · 2 H2O = (Number of moles of NaOH) / 2
5. Calculate the percent (COOH)2 · 2 H2O in the sample:
Percent (COOH)2 · 2 H2O = [(Number of moles of (COOH)2 · 2 H2O) / Total moles] * 100
Total moles = Number of moles of (COOH)2 · 2 H2O
Now, let's substitute the values and calculate the percent (COOH)2 · 2 H2O in the sample.
To calculate the percentage of (COOH)2 · 2 H2O in the sample, you need to determine the number of moles of (COOH)2 · 2 H2O and the total mass of the sample.
Let's break down the problem step by step:
Step 1: Calculate the number of moles of NaOH used in the titration.
We have the volume and molarity of NaOH solution used, so we can calculate the number of moles of NaOH:
moles NaOH = volume (L) × molarity (mol/L)
moles NaOH = 32.4 mL × (1 L / 1000 mL) × 0.144 mol/L
Step 2: Convert the moles of NaOH to moles of (COOH)2 · 2 H2O.
From the balanced chemical equation:
1 mole of NaOH reacts with 1 mole of (COOH)2 · 2 H2O
Therefore, moles of (COOH)2 · 2 H2O = moles of NaOH
Step 3: Calculate the mass of (COOH)2 · 2 H2O in the sample.
Using the mass and moles of (COOH)2 · 2 H2O:
mass (COOH)2 · 2 H2O = moles of (COOH)2 · 2 H2O × molar mass (COOH)2 · 2 H2O
Step 4: Calculate the percent (COOH)2 · 2 H2O in the sample.
percent (COOH)2 · 2 H2O = (mass (COOH)2 · 2 H2O / mass of the sample) × 100%
Now, let's plug in the provided values into the equations:
Given:
Mass of the impure sample = 2.9 g
Volume of NaOH used in titration = 32.4 mL
Molarity of NaOH = 0.144 mol/L
Step 1: Calculate the number of moles of NaOH used in the titration.
moles NaOH = 32.4 mL × (1 L / 1000 mL) × 0.144 mol/L = 0.0046656 mol NaOH
Step 2: Convert the moles of NaOH to moles of (COOH)2 · 2 H2O.
moles (COOH)2 · 2 H2O = 0.0046656 mol NaOH
Step 3: Calculate the mass of (COOH)2 · 2 H2O in the sample.
The molar mass of (COOH)2 · 2 H2O = (12.01 g/mol + 2(16.00 g/mol) + 2(1.01 g/mol) + 2(2(1.01 g/mol))) + 2(16.00 g/mol) = 114.11 g/mol
mass (COOH)2 · 2 H2O = 0.0046656 mol × 114.11 g/mol = 0.532 g
Step 4: Calculate the percent (COOH)2 · 2 H2O in the sample.
percent (COOH)2 · 2 H2O = (0.532 g / 2.9 g) × 100% ≈ 18.34%
Therefore, the percent (COOH)2 · 2 H2O in the impure sample is approximately 18.34%.