Recently I solved for how many mL of 5% BaCl2.2H20 solution was required to precipitate sodium sulfate. Now the question asks what if the sodium sulfate was potassium sulfate instead. Would the volume required be higher or lowered than the sodium sulfate and why?
I got that the potassium sulfate required less volume though I don't understand the reason why.
The volume of 5% BaCl2.2H2O solution required to precipitate a particular amount of sulfate compound depends on the stoichiometry of the reaction between the barium chloride and the sulfate ion. Since BaCl2.2H2O is a salt composed of barium ions (Ba2+) and chloride ions (Cl-), its reaction with a sulfate compound, such as sodium sulfate (Na2SO4) or potassium sulfate (K2SO4), can be represented by the following balanced chemical equation:
BaCl2 + Na2SO4 -> BaSO4 + 2NaCl
BaCl2 + K2SO4 -> BaSO4 + 2KCl
As you can see, when barium chloride reacts with either sodium sulfate or potassium sulfate, the products formed are barium sulfate (BaSO4) and either sodium chloride (NaCl) or potassium chloride (KCl). The stoichiometry of the reaction indicates that one mole of barium chloride is required to react with one mole of sulfate ion to form one mole of barium sulfate.
Now, let's consider the molar masses of the sulfate compounds involved:
- Sodium sulfate (Na2SO4): Molar mass = 22.99 g/mol (sodium) + 2x 32.06 g/mol (sulfur) + 4x 16.00 g/mol (oxygen) = 142.04 g/mol
- Potassium sulfate (K2SO4): Molar mass = 39.10 g/mol (potassium) + 2x 32.06 g/mol (sulfur) + 4x 16.00 g/mol (oxygen) = 174.26 g/mol
Now, let's assume that you are given a certain mass or amount of the sulfate compound (either sodium sulfate or potassium sulfate) and you want to calculate the volume of the 5% BaCl2.2H2O solution required to precipitate it.
To determine which sulfate compound requires a higher or lower volume of the BaCl2.2H2O solution, you need to consider their molar masses. Since potassium sulfate has a higher molar mass than sodium sulfate, it means that for the same mass or amount of either sulfate compound, there will be fewer moles of potassium sulfate compared to sodium sulfate.
Since the stoichiometry of the reaction indicates that one mole of barium chloride reacts with one mole of sulfate ion, it means that for a given mass or amount of sodium sulfate, there will be more moles of sodium sulfate compared to the same mass or amount of potassium sulfate. Therefore, the volume of the BaCl2.2H2O solution required to react with sodium sulfate will be higher than the volume required to react with potassium sulfate.
In summary, the volume of a 5% BaCl2.2H2O solution required to precipitate potassium sulfate would be lower compared to the volume required for sodium sulfate because potassium sulfate has a higher molar mass, resulting in fewer moles of potassium sulfate for the same mass or amount.
The volume of 5% BaCl2.2H2O solution required to precipitate a certain amount of sulfate ions depends on the stoichiometry of the precipitation reaction.
First, let's consider the balanced chemical equation for the precipitation reaction of sodium sulfate with barium chloride:
Na2SO4 + BaCl2 -> 2 NaCl + BaSO4
From the equation, we can see that for every 1 mole of sodium sulfate, we need 1 mole of barium chloride to fully react and precipitate all the sulfate ions.
Now, let's consider the balanced chemical equation for the precipitation reaction of potassium sulfate with barium chloride:
K2SO4 + BaCl2 -> 2 KCl + BaSO4
Similarly, for every 1 mole of potassium sulfate, we need 1 mole of barium chloride to precipitate all the sulfate ions.
Since the stoichiometry is the same for both reactions, the amount of barium chloride required to precipitate the sulfate ions will be the same regardless of whether it is sodium sulfate or potassium sulfate.
However, since potassium sulfate has a lower molar mass than sodium sulfate (174.259 g/mol for K2SO4 compared to 142.044 g/mol for Na2SO4), the mass of potassium sulfate required to reach the same number of moles will be lower than that of sodium sulfate.
Because the mass of sulfate ions in the potassium sulfate solution is lower than that in the sodium sulfate solution, less volume of the barium chloride solution will be needed to precipitate all the sulfate ions when potassium sulfate is used compared to when sodium sulfate is used.
Therefore, the volume of the 5% BaCl2.2H2O solution required will be lower when precipitating potassium sulfate compared to precipitating sodium sulfate.