Answer:
The correct answer is 2.33g of , A
Explanation:
We can use basic stoich to get this question.
To start, lets convert our grams to moles in order to use our equation. 5.38g of moles.
We then can use our molar ratio of 13:10. Simply divide by our molar ratio in order to go from moles of to moles of
. This gets us 0.129335 moles of
.
Now lets convert to grams of by multiplying by the molar mass of
.
of
The amount of barium ions that must be present in order for the salt to precipitate is 0.245 M.
A solution's solubility product is the result of raising each ion's concentration to the power of its stoichiometric ratio. It is portrayed as
A combination of 1 barium ion and 2 fluoride ions results in the ionic compound known as barium fluoride.
The following equation describes the equilibrium reaction for barium fluoride ionization:
BaF₂ → Ba²⁺ + 2F⁻
Ksp = [Ba²⁺] · [F⁻]²
2.45*= [Ba²⁺] *
[Ba²⁺]=0.245 M
As a result, 0.245 M of barium ions must be present in order for the salt to precipitate.
<h3>Solubility </h3>Solubility in chemistry refers to a chemical's capacity to dissolve in another substance, the solvent, to produce a solution. Inability of the solute to create such a solution is the opposite quality, or insolubility. A substance's degree of solubility in a given solvent is often determined by the amount of the solute present in a saturated solution, which is a solution in which no additional solute can be dissolved. The solubility equilibrium between the two compounds is considered to have been reached at this time. If there is no such restriction for a given solute and solvent, the two are referred to as being "miscible in any amounts."
What concentration of the barium ion, ba2 , must be exceeded to precipitate baf2 from a solution that is 1. 00×10−2 m in the fluoride ion, f−? ksp for barium fluoride is 2. 45×10−5
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