To determine how many mL of 5% BaCl2.2H20 solution would be required to precipitate the sulfate, we need to consider the reaction between sodium sulfate (Na2SO4) and barium chloride (BaCl2). The balanced chemical equation for this reaction is:
Na2SO4 + BaCl2 → BaSO4 + 2NaCl
This equation shows that one mole of sodium sulfate reacts with one mole of barium chloride to produce one mole of barium sulfate and two moles of sodium chloride.
To calculate the volume of the BaCl2.2H20 solution needed, we follow these steps:
1. Determine the molar mass of Na2SO4:
- Na (sodium) = 22.99 g/mol
- S (sulfur) = 32.07 g/mol
- O (oxygen) = 16.00 g/mol (x 4) = 64.00 g/mol
- Total molar mass = 22.99 g/mol + 2(32.07 g/mol) + 64.00 g/mol = 142.06 g/mol
2. Calculate the moles of Na2SO4:
- Given that the sample is pure sodium sulfate, let's assume we have "x" moles.
3. From the balanced equation, we can see that the ratio between Na2SO4 and BaCl2 is 1:1, meaning that the moles of BaCl2 needed to react fully will also be "x".
4. Determine the molar mass of BaCl2.2H20:
- Ba (barium) = 137.33 g/mol
- Cl (chloride) = 35.45 g/mol (x 2) = 70.90 g/mol
- H (hydrogen) = 1.01 g/mol (x 2) = 2.02 g/mol
- O (oxygen) = 16.00 g/mol
- Total molar mass = 137.33 g/mol + 70.90 g/mol + 2.02 g/mol + 16.00 g/mol = 226.25 g/mol
5. Calculate the mass of BaCl2.2H20 needed:
- Given that BaCl2.2H20 is present in a 5% solution, this means there are 5g of BaCl2.2H20 in every 100 mL of the solution.
- The density of the solution is given as 1.00 g/mL, so the mass of BaCl2.2H20 in 100 mL is 100g.
- Therefore, we have 5g of BaCl2.2H20 in 100g of the solution.
- Using the equation: mass = (percentage/100) x total mass
mass of BaCl2.2H20 = (5/100) x 100g = 5g
6. Calculate the moles of BaCl2.2H20:
- Using the molar mass of BaCl2.2H20, we find:
moles of BaCl2.2H20 = mass of BaCl2.2H20 / molar mass
= 5g / 226.25 g/mol
= 0.0221 moles
7. Since the ratio between Na2SO4 and BaCl2 is 1:1, we know that:
- 1 mole of Na2SO4 reacts with 1 mole of BaCl2
- 0.0221 moles of Na2SO4 will require 0.0221 moles of BaCl2
8. Convert the moles of BaCl2 to volume using the density given:
- volume (in mL) = moles of BaCl2 x molar mass of BaCl2 / density of the solution
= 0.0221 moles x 226.25 g/mol / 1.00 g/mL
= 4.99 mL
Therefore, approximately 4.99 mL of the 5% BaCl2.2H20 solution would be required to precipitate the sulfate in the sample.