What is the molar solubility in water of ag2cro4 ? (the ksp for ag2cro4 is 8 x 10-12. )
2 answers:
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 .
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Answer:
5.758 is the density of the metal ingot in grams per cubic centimeter.
Explanation:
1) Mass of pycnometer = M = 27.60 g
Mass of pycnometer with water ,m= 45.65 g
Density of water at 20 °C = d =
1 kg = 1000 g
Mass of water ,m'= m - M = 45.65 g - 27.60 g =18.05 g
Volume of pycnometer = Volume of water present in it = V
2) Mass of metal , water and pycnometer = 56.83 g
Mass of metal,M' = 9.5 g
Mass of water when metal and water are together ,m''= 56.83 g - M'- M
56.83 g - 9.5 g - 27.60 g = 19.7 g
Volume of water when metal and water are together = v
Density of metal = d'
Volume of metal = v' =
Difference in volume will give volume of metal ingot.
v' = v - V
Since volume cannot be in negative .
Density of the metal =d'
=