A solution contains an unknown mass of dissolved barium ion. When sodium sulfate is added to the solution, a white precipitate forms. The precipitate is filtered and dried and then found to have a mass of 210 mg. What mass of barium was in the original solution? (Assume that all of the barium was precipitated out of the reaction.)
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To find the mass of barium in the original solution, we need to use stoichiometry and the information provided. Here's how we can solve the problem step by step:
Step 1: Write the balanced chemical equation for the reaction between barium ions and sulfate ions:
Ba^2+ + SO4^2- -> BaSO4
Step 2: Determine the molar mass of BaSO4:
BaSO4 = (1 * 137.33 g/mol) + (1 * 32.07 g/mol) + (4 * 16.00 g/mol) = 233.38 g/mol
Step 3: Convert the mass of the precipitate (210 mg) to grams:
210 mg = 210/1000 = 0.21 g
Step 4: Use the molar mass of BaSO4 to convert the mass of the precipitate to moles:
moles of BaSO4 = mass of precipitate (in grams) / molar mass of BaSO4
moles of BaSO4 = 0.21 g / 233.38 g/mol = 0.0009 mol
Step 5: Use stoichiometry to find the moles of barium in the reaction. From the balanced chemical equation, we can see that 1 mole of BaSO4 is equivalent to 1 mole of Ba^2+:
moles of Ba^2+ = moles of BaSO4
Step 6: Convert moles of Ba^2+ to grams of barium:
mass of barium = moles of Ba^2+ * molar mass of Ba
mass of barium = 0.0009 mol * 137.33 g/mol = 0.124 g
Therefore, the mass of barium in the original solution is 0.124 grams.