The concentration of barium ions in any solution can also be determined via gravimetric analysis. An impure sample of barium nitrate with a mass of 1.234 g, is completely dissolved in water and the resulting solution is reacted with an excess of aqueous sodium sulfate. A precipitate forms, and after filtering and drying, it was found to have a mass of 0.848 g.

A) What is the relevance of adding eccess sodium sulfate?

B) Calculate the % of barium nitrate in the original 1.234 g sample.

Ba(NO3)2 + Na2SO4 ==> BaSO4 + 2NaNO3

a. Gravimetric analysis depends upon almost none of the solid dissolving; adding and excess of Na2SO4 is adding a common ion (sulfate ion) which decreases the solubility of the BaSO4.

b. mols BaSO4 = grams/molar mass BaSO4 = ?
mols Ba(NO3)2 = mols BaSO4
grams Ba(NO3)2 = mols Ba(NO3)2 x molar mass Ba(NO3)2 = ?
%Ba(NO3)2 = (mass Ba(NO3)2/mass sample)*100 = ?

A) Well, with the excess sodium sulfate, I guess you could say they were just really salty about making sure they had enough to react with the barium nitrate. Can't blame them for wanting to be thorough!

B) To calculate the percentage of barium nitrate in the original sample, we can use some good old-fashioned math.

First, let's determine the mass of the barium nitrate in the precipitate. Subtract the mass of the precipitate (0.848 g) from the mass of the original sample (1.234 g) to get the mass of the barium nitrate in the precipitate.

Mass of barium nitrate in precipitate = 1.234 g - 0.848 g = 0.386 g

Next, we can calculate the percentage of barium nitrate in the original sample by dividing the mass of the barium nitrate in the precipitate by the mass of the original sample and multiplying by 100.

% of barium nitrate = (0.386 g / 1.234 g) × 100 = 31.3%

So, the percentage of barium nitrate in the original 1.234 g sample is approximately 31.3%. That's quite a nitrate-y sample you've got there!

A) The addition of excess sodium sulfate is done to ensure that all the barium ions from the barium nitrate solution react completely to form a precipitate. By adding excess sodium sulfate, we ensure that there are more than enough sulfate ions available to react with all the barium ions present, resulting in the formation of a complete precipitate.

B) To calculate the percentage of barium nitrate in the original 1.234 g sample, we need to first determine the mass of barium in the precipitate. From the given information, the mass of the precipitate is 0.848 g.

The molar mass of barium sulfate is 233.38 g/mol. Since barium sulfate has a 1:1 molar ratio with barium ions, the mass of barium in the precipitate can be calculated as follows:

Mass of barium = (0.848 g) * (233.38 g/mol) / (233.38 g/mol)
= 0.848 g

To find the percentage of barium nitrate in the original sample, we divide the mass of barium by the mass of the original sample and multiply by 100.

% of barium nitrate = (0.848 g / 1.234 g) * 100
= 68.83%

A) The addition of excess sodium sulfate is relevant in this gravimetric analysis because it ensures that all of the barium ions in the sample react to form a precipitate. When a solution of barium nitrate is mixed with sodium sulfate, a double displacement reaction occurs, producing solid barium sulfate as a precipitate. By adding excess sodium sulfate, all the available barium ions in the solution will react and form solid barium sulfate, which can then be filtered, dried, and weighed to determine the concentration of barium ions.

B) To calculate the percentage of barium nitrate in the original 1.234 g sample, we need to determine the mass of the barium nitrate in the precipitate.

First, let's calculate the molar mass of barium sulfate (BaSO4):
- The molar mass of barium (Ba) is 137.33 g/mol.
- The molar mass of sulfur (S) is 32.06 g/mol.
- The molar mass of oxygen (O) is 16.00 g/mol.
So, the molar mass of BaSO4 is 137.33 g/mol + 32.06 g/mol + (4 * 16.00 g/mol) = 233.38 g/mol.

Next, we can use stoichiometry to determine the amount of barium nitrate in the precipitate.
- The balanced equation for the reaction between barium nitrate and sodium sulfate is:
Ba(NO3)2 + Na2SO4 ⟶ 2 NaNO3 + BaSO4
From the equation, we can see that 1 mole of barium nitrate (Ba(NO3)2) reacts with 1 mole of sodium sulfate (Na2SO4) to form 1 mole of barium sulfate (BaSO4).

Now, let's calculate the moles of barium sulfate formed:
- The mass of the dried precipitate is 0.848 g.
- Divide the mass by the molar mass of BaSO4 to get the moles: 0.848 g / 233.38 g/mol = 0.00363 mol.

Since 1 mole of barium nitrate reacts to form 1 mole of barium sulfate, the moles of barium nitrate present in the original sample are also 0.00363 mol.

To calculate the percentage of barium nitrate in the original sample, use the formula:
Percentage = (mass of barium nitrate / mass of the original sample) x 100%

- The mass of barium nitrate is the same as the mass of barium sulfate since the reaction goes to completion. So, the mass of barium nitrate is 0.848 g.
- The mass of the original sample is 1.234 g.

Percentage = (0.848 g / 1.234 g) x 100% = 68.8%

Therefore, the percentage of barium nitrate in the original 1.234 g sample is approximately 68.8%.