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The molar solubility in water of Ag₂CrO₄ is 2.52 * 10⁻⁴ M .
<h3>How do you define Solubility ?</h3>
The measure of the degree to which a substance gets dissolved in a solvent to become a solution.
The formula for determination of molar solubity is
= solubility product constant
= cation in an aqueous solution
= anion in an aqueous solution
a, b = relative concentrations of a and b
We have the data given as
for Ag₂CrO₄ is 8 x 10⁻¹²
For Silver chromate , the solubility would be only one-half of the Ag⁺ concentration.
We denote the solubility of Ag₂CrO₄ as S mol L⁻¹.
Then for a saturated solution, we have
[Ag⁺]=2S
[CrO₂⁴⁻]=S
the relation between the solubility and the solubility product constant depends on the stoichiometry of the dissolution reaction
(2S)²(S)=4S³ = 8 * 10⁻¹²
S=∛ (2*10⁻¹²)
S= 1.26 * 10⁻⁴ M
2S = 2.52 * 10⁻⁴ M
Therefore the molar solubility in water of Ag₂CrO₄ is 2.52 * 10⁻⁴ M .
To know more about Solubility
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Steps To Solve:
~Substitute
3(2) - 2(3)
~Subtract
6 - 6
~Simplify
0
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Answer:
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