A 0.2638 g impure sodium carbonate sample is analyzied by titrating the sodium carbonate with the 0.1288 M hydrochloride solution requiring 38.27 mL.
The reaction is CO3^2- + 2H^+ --> H2O + CO2
Calculate the percent sodium carbonate in the sample.
mols HCl = M x L = ?
From the equation, mols Na2CO3 = 1/2 that.
grams Na2CO3 = mols Na2CO3 x molar mass Na2CO3.
%Na2CO3 = (grams Na2CO3/0.2638)*100 = ?
To calculate the percent sodium carbonate in the sample, we need to determine the number of moles of sodium carbonate used in the titration.
1. Calculate the number of moles of HCl used in the titration:
Moles of HCl = Molarity of HCl * Volume of HCl used
= 0.1288 M * 0.03827 L
2. Since the balanced equation shows that there is a 1:2 stoichiometric ratio between HCl and Na2CO3, we can deduce that the moles of HCl used are equal to half the moles of Na2CO3 used.
Moles of Na2CO3 = 0.5 * Moles of HCl
3. Calculate the number of moles of Na2CO3 in the impure sample:
Moles of Na2CO3 in sample = Moles of Na2CO3 / Volume of sample used in titration
= Moles of Na2CO3 / (0.2638 g / Molar mass of Na2CO3)
4. Calculate the percent sodium carbonate in the sample:
Percent Na2CO3 = (Moles of Na2CO3 in sample / Total moles in sample) * 100
Now let's plug in the values and calculate:
Molarity of HCl = 0.1288 M
Volume of HCl used = 0.03827 L
Molar mass of Na2CO3 = 106 g/mol
Mass of the impure sample = 0.2638 g
1. Moles of HCl = 0.1288 M * 0.03827 L = 0.004933 moles of HCl
2. Moles of Na2CO3 = 0.5 * 0.004933 moles = 0.002467 moles of Na2CO3
3. Moles of Na2CO3 in sample = 0.002467 moles / (0.2638 g / 106 g/mol) = 0.9931 moles of Na2CO3
4. Percent Na2CO3 = (0.9931 moles / Total moles in sample) * 100
Total moles in sample = Mass of the impure sample / Molar mass of Na2CO3
= 0.2638 g / 106 g/mol
= 0.002486 moles
Percent Na2CO3 = (0.9931 moles / 0.002486 moles) * 100
= 39.95%
Therefore, the percent sodium carbonate in the impure sample is approximately 39.95%.
To calculate the percent of sodium carbonate in the sample, we first need to determine the number of moles of sodium carbonate that reacted with the hydrochloric acid solution. Since we know the volume and concentration of the hydrochloric acid solution used, we can use the equation:
Molarity (M) = moles (mol) / volume (L)
First, let's calculate the number of moles of hydrochloric acid solution used:
Volume of hydrochloric acid solution = 38.27 mL = 0.03827 L
Molarity of hydrochloric acid solution = 0.1288 M
Using the equation, we can rearrange it to solve for moles:
moles of hydrochloric acid solution = Molarity × volume
= 0.1288 M × 0.03827 L
≈ 0.004938756 mol
According to the balanced chemical equation, we need 2 moles of hydrochloric acid solution to react with 1 mole of sodium carbonate. Hence, 0.004938756 mol of hydrochloric acid solution reacted with (0.004938756 / 2) ≈ 0.002469378 mol of sodium carbonate.
Next, let's calculate the molecular mass of sodium carbonate (Na2CO3):
(Na: 22.99 g/mol, C: 12.01 g/mol, O: 16.00 g/mol)
Molecular mass of Na2CO3 = (2 × 22.99 g/mol) + (1 × 12.01 g/mol) + (3 × 16.00 g/mol)
= 106 g/mol
Now that we know the number of moles and the molecular mass of sodium carbonate, we can calculate the mass of sodium carbonate in the sample:
Mass of sodium carbonate = moles × molecular mass
= 0.002469378 mol × 106 g/mol
≈ 0.2617 g
Finally, to calculate the percent of sodium carbonate in the sample, we divide the mass of sodium carbonate by the mass of the impure sample and multiply by 100:
Percent sodium carbonate in the sample = (Mass of sodium carbonate / Mass of impure sample) × 100
= (0.2617 g / 0.2638 g) × 100
≈ 99.20%
Therefore, the percent of sodium carbonate in the impure sample is approximately 99.20%.