3. The solubility of BaSO4 at 25C is 9.04 mg/L. Based on this, how many grams and moles of BaSO4 remained dissolved in the 100 mL of solution? What percent of your yield does this represent?
2. The precipitate is washed with warm water not cold, why is this? What might happen if cold water was used? How would this affect the measured yield of BaSO4?
3. Solubility is 9.04 mg/L; therefore, in 100 mL, the solubility is 9.04 mg x 100/1000 = ?
I don't know the percent of the yield since you didn't give a quantity you collected.
2. You want to avoid peptization; i.e., colloid formation. Of course, BaSO4 is more soluble in hot water than cold water so there will be some extra loss due to increased solubility in warm water; however, most procedures call for adding an excess of the precipitating agent and that decreases the solubility due to Le Chatelier's Principle.
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To determine how many grams and moles of BaSO4 remained dissolved in the 100 mL of solution, we can use the solubility value given.
First, we need to convert the solubility from mg/L to g/L by dividing it by 1000:
9.04 mg/L ÷ 1000 = 0.00904 g/L
Next, we can calculate the grams of BaSO4 dissolved in 100 mL (0.1 L) of solution by multiplying the solubility by the volume:
0.00904 g/L × 0.1 L = 0.000904 g
To determine the moles of BaSO4, we need to use the molar mass of BaSO4, which is 233.38 g/mol. So, we can divide the mass (in grams) by the molar mass:
0.000904 g ÷ 233.38 g/mol = 3.872 × 10^-6 mol
Now, let's calculate the percent yield by comparing the amount of dissolved BaSO4 to the expected yield. If we assume that the entire 100 mL of solution contained the dissolved BaSO4, we can find the expected mass and moles.
The molar mass of BaSO4 is still 233.38 g/mol, so the expected mass in grams is:
0.1 L × 9.04 mg/L ÷ 1000 = 0.000904 g
And the expected moles:
0.000904 g ÷ 233.38 g/mol = 3.872 × 10^-6 mol
To calculate the percent yield, we divide the actual moles by the expected moles and multiply by 100:
(3.872 × 10^-6 mol ÷ 3.872 × 10^-6 mol) × 100 = 100%
Therefore, the percent yield is 100%, indicating that all of the dissolved BaSO4 remained in the solution.
Moving on to the second question, the precipitate is washed with warm water instead of cold water due to a few reasons.
Firstly, warm water helps to improve the solubility of any remaining impurities. By using warm water, the impurities are more likely to dissolve, leaving behind a purer precipitate.
Secondly, warm water helps to speed up the washing process by enhancing the rate of diffusion. The particles in the precipitate move faster due to increased kinetic energy, allowing for a quicker removal of impurities.
If cold water was used instead, several things might happen. Firstly, the solubility of BaSO4 would decrease, which means that more of the precipitate may redissolve into the cold water, resulting in a lower yield of BaSO4.
Additionally, using cold water may not effectively remove impurities since their solubility may not be significantly affected by lower temperatures. This could lead to a less pure product.
In summary, using warm water for washing the precipitate improves solubility and enhances impurity removal, ensuring a higher yield and a purer product.