Barium sulfate is an inorganic compound with the chemical formula BaSO4. It is a white crystalline solid that is odorless and insoluble in water. Determining the amount of water required to dissolve a given mass of barium sulfate involves using the solubility rules and stoichiometry.
Quick Answer
The solubility of barium sulfate at 25°C is 2.4 mg per liter. To dissolve 1.00 g of barium sulfate would require 417 liters of water at 25°C.
Solubility of Barium Sulfate
The solubility of a substance is defined as the maximum amount that will dissolve in a given solvent at a specified temperature. Solubility is typically reported in units of grams (or milligrams) per 100 grams of solvent.
Barium sulfate has an extremely low solubility in water. Its solubility at 25°C is only 2.4 mg per liter. This means that at most, only 2.4 mg of barium sulfate will dissolve in 1 liter of water at 25°C. The solubility does not change significantly at other common temperatures.
Stoichiometry of Barium Sulfate
The balanced chemical equation for barium sulfate dissolving in water is:
BaSO4(s) → Ba2+(aq) + SO42-(aq)
This shows that 1 mole of solid barium sulfate (233.39 g) dissociates into 1 mole of aqueous barium ions (137.33 g) and 1 mole of aqueous sulfate ions (96.06 g) when dissolved.
We can use stoichiometry to relate the mass of barium sulfate to the mass of ions produced when dissolved. In particular, the ratio of barium sulfate mass to barium ion mass is:
233.39 g BaSO4 / 137.33 g Ba2+ = 1.699
So for every 1 g of barium sulfate dissolved, 1.699 g of barium ions are produced.
Determining the Amount of Water Needed
We are starting with 1.00 g of solid barium sulfate that we want to dissolve in water at 25°C.
Using stoichiometry, dissolving 1.00 g of BaSO4 would produce 1.699 g of Ba2+ ions.
The solubility of barium sulfate dictates that at most 2.4 mg (0.0024 g) of barium sulfate can dissolve per liter of water.
We can set up a ratio comparing the barium sulfate we want to dissolve (1.00 g) to the maximum that can dissolve per liter (0.0024 g/L):
1.00 g BaSO4 / 0.0024 g/L = 417 L
So to completely dissolve 1.00 g of barium sulfate at 25°C would require 417 liters of water.
Conclusion
Based on the solubility of barium sulfate and principles of stoichiometry, approximately 417 liters of water at 25°C would be needed to fully dissolve 1.00 g of barium sulfate.
This result demonstrates that barium sulfate has an extremely low solubility in water. Over 400 liters of water are required to dissolve just 1 gram of the compound. This makes barium sulfate useful for applications where an insoluble, inert material is needed.
Summary
- The solubility of barium sulfate at 25°C is 2.4 mg per liter
- Stoichiometry shows that 1 g of BaSO4 produces 1.699 g of Ba2+ when dissolved
- Comparing the BaSO4 mass to solubility allows calculating the water volume needed
- 417 liters of water at 25°C are required to dissolve 1.00 g of barium sulfate
Barium sulfate is an important chemical compound with many uses that take advantage of its low solubility in water. Determining how much water is needed to dissolve a given amount of BaSO4 relies on solubility data and careful application of stoichiometry. This provides quantitative insight into its behavior in solution.
Frequently Asked Questions
What is barium sulfate used for?
Barium sulfate has several key uses that rely on its insolubility:
- It is used medically as a radiocontrast agent for X-rays of the digestive system. Its insolubility allows it to be visible on X-rays while being nontoxic.
- It is incorporated into oil well drilling mud as a weighting agent. Its high density provides pressure control during drilling.
- It is used as a white pigment in paints, plastics, and paper to provide opacity and brightness.
Why is barium sulfate insoluble in water?
Barium sulfate is insoluble in water due to the pairing of charges between the barium cation (Ba2+) and sulfate anion (SO42-). This results in an overall neutral compound with high lattice energy in the solid state. The lattice energy must be overcome for barium sulfate to dissociate into ions and dissolve.
What temperature was used in the solubility calculation?
The solubility data used was for 25°C. This is a common reference temperature for reporting solubility values. The calculation would be slightly different at other temperatures, but the solubility does not vary drastically over normal temperature ranges.
What assumptions were made in the stoichiometry?
It was assumed that barium sulfate completely dissociates into Ba2+ and SO42- ions when dissolved. In reality, a small amount of undissociated BaSO4 may remain in solution. However, the dissociation is nearly complete so this assumption does not significantly impact the calculation.
How would the result change for a different mass of barium sulfate?
The linear relationship between the mass of barium sulfate and the volume of water needed means that doubling the mass would require roughly double the volume of water to dissolve it. The solubility ratio could be applied to any mass of BaSO4 to determine the associated water volume.
References
Solubility data obtained from:
- Haynes, William M., ed. CRC Handbook of Chemistry and Physics. CRC press, 2014.
Information on uses of barium sulfate from:
- Klein, Cornelis, and Barbara S. Cornelis. Manual of Mineral Science. 22nd ed., Wiley, 2003.
Barium sulfate dissolution reaction:
- Chang, Raymond. Chemistry: The Molecular Nature of Matter. McGraw-Hill, 2018.
Stoichiometry principles:
- Tro, Nivaldo J. Chemistry: A Molecular Approach. 5th ed., Pearson, 2018.
| Barium sulfate solubility at 25°C | 2.4 mg/L |
|---|---|
| Mass of barium sulfate | 1.00 g |
| Volume of water needed | 417 L |