Given a solution containing 0.242 g of aqueous barium chloride, BaCl2, how many mL of 0.0581 M H2SO4 solution will completely precipitate the barium ions, Ba2+? The molar mass of BaCl2 is 208.3 g/mol.
BaCl2 + H2SO4 --> 2HCl + BaSO4
mols BaCl2 = grams/molar mass = 0.242/208 = 0.00116
From the equation, 1 mol BaCl2 = 1 mol H2SO4 so it will talke 0.00116 mols H2SO4. M H2SO4 = mols/L. You know M and mols, solve for L and convert to mL.
So it's A?
a) 20 mL
b) 18 mL
c) 24 mL
d) 30 mL
To determine the amount of H2SO4 solution needed to precipitate the barium ions from the BaCl2 solution, we can follow these steps:
Step 1: Calculate the number of moles of BaCl2.
Given mass of BaCl2 = 0.242 g
Molar mass of BaCl2 = 208.3 g/mol
Number of moles of BaCl2 = mass/molar mass = 0.242 g / 208.3 g/mol
Step 2: Calculate the number of moles of Ba2+ ions.
Barium chloride, BaCl2, dissociates into one mole of Ba2+ ions per mole of BaCl2. Therefore, the number of moles of Ba2+ ions is the same as the number of moles of BaCl2.
Step 3: Convert moles of Ba2+ ions to volume (mL) of H2SO4 solution using its molarity.
Molarity of H2SO4, M = 0.0581 M
Using the equation Moles = Volume (L) × Molarity, we can rearrange to solve for volume: Volume (L) = Moles / Molarity.
The volume will be in liters, so we need to convert to milliliters by multiplying by 1000.
Now let's calculate the final result:
Step 1: Calculate the number of moles of BaCl2.
Number of moles of BaCl2 = 0.242 g / 208.3 g/mol ≈ 0.0011625 mol
Step 2: Calculate the number of moles of Ba2+ ions.
Number of moles of Ba2+ ions ≈ 0.0011625 mol
Step 3: Convert moles of Ba2+ ions to volume (mL) of H2SO4 solution using its molarity.
Volume (L) of H2SO4 solution = 0.0011625 mol / 0.0581 M ≈ 0.02 L ≈ 20 mL
Therefore, approximately 20 mL of 0.0581 M H2SO4 solution is needed to completely precipitate the barium ions, Ba2+, from the given BaCl2 solution.