The Ksp of Ag2CrO4(s) is 1.12 x 10^-12. Calculate the molar solubility of AgCrO4(s)
a)in pure water
b) in a solution of 0.10 mol/L sodium chromate, Na2CrO4(s)
Ag2CrO4 ==> 2Ag^+ + CrO4^=
Ksp = (Ag^+)^2 (CrO4^=) = 1.12 x 10^-12
solubility Ag2CrO4 = S
(Ag^+) = 2S
(CrO4^=) = S
Plug into Ksp and solve for S.
For b.
Everything stays the same except (CrO4^=) = S + 0.1. Substitute into Ksp and solve for S. You can make the simplifying assumption that S + 0.1 = 0.1. If you do, to avoid solving a quadratic equation, then check at the end to make sure S + 0.10 = 0.1. If it isn't, then you will need to regroup and solve the quadratic or use successive approximations. I THINK S will be negligible when compared to 0.1 M.
Check my thinking. Check my work. Post your work if you get stuck.
yeah... its cool
Im not sure ...need more explanations please?
To calculate the molar solubility of AgCrO4(s), we'll use the concept of Ksp (the solubility product constant) and the stoichiometry of the chemical equation.
a) Molar Solubility in Pure Water:
In pure water, we assume that the dissolution of AgCrO4(s) happens completely, so the molar solubility is equal to the concentration of the dissolved ions.
The balanced chemical equation for the dissociation of AgCrO4(s) is:
AgCrO4(s) ⇌ Ag+(aq) + CrO4^2-(aq)
From the equation, we can see that one mole of AgCrO4 dissociates into one mole of Ag+ ions and one mole of CrO4^2- ions.
Let's assume that the molar solubility of AgCrO4(s) is "S" M.
So, the concentration of Ag+ ions and CrO4^2- ions in the saturated solution will be "S" M.
According to the Ksp expression for Ag2CrO4(s):
Ksp = [Ag+][CrO4^2-]
Substituting the values, we have:
1.12 x 10^-12 = S * S
Taking the square root of both sides of the equation, we find:
S = √(1.12 x 10^-12) M
Therefore, the molar solubility of AgCrO4 in pure water is √(1.12 x 10^-12) M or approximately 1.06 x 10^-6 M.
b) Molar Solubility in a Solution of Sodium Chromate:
In this case, we have a solution of 0.10 mol/L sodium chromate, Na2CrO4(s).
The concentration of CrO4^2- ions from the sodium chromate can react with the Ag+ ions from AgCrO4(s) to form a precipitate. This will affect the molar solubility of AgCrO4.
From the balanced chemical equation:
2 AgCrO4(s) + Na2CrO4(aq) ⇌ 2 AgCrO4(aq) + Na2CrO4(s)
We can see that two moles of AgCrO4 are formed from one mole of Na2CrO4(aq).
Let's assume that the molar solubility of AgCrO4(s) in the presence of 0.10 mol/L Na2CrO4 is "S" M.
Since the concentration of CrO4^2- is double that of AgCrO4, we can write:
[S]^2 * [Na2CrO4] = Ksp
Substituting the values, we have:
[S]^2 * [0.10] = 1.12 x 10^-12
Simplifying the equation, we get:
[S]^2 = (1.12 x 10^-12) / [0.10]
Taking the square root of both sides, we find:
S = √[(1.12 x 10^-12) / 0.10] M
Therefore, the molar solubility of AgCrO4 in a solution of 0.10 mol/L Na2CrO4 is √[(1.12 x 10^-12) / 0.10] M or approximately 3.35 x 10^-6 M.