Molar solubility?

Question

What is the molar solubility of silver oxalate? (Ksp for Ag2C2O4 is 5.4 * 10^-12)

Answer 1

The dissociation reaction is
Ag₂C₂O₄(s) → 2Ag⁺(aq) + C₂O₄⁻(aq)
the solubility product is
Ksp = [Ag⁺]² · [C₂O₄⁻]
where [...] ere the concentrations in mol/L

If you the dissolve x mol per liter silver oxalate the corresponding ion concentrations are:
[Ag⁺] = 2x
[C₂O₄⁻] = x

If x is the maximum soluble amount, the ion concentrations fulfill the solubility product.
Therefore
Ksp = (2x)² · x
=>
x = (Ksp/4)^(1/3) = 1.105 · 10^-4 mol/l

Answer 2

Start with the equation for Ksp:

Ksp=[Ag+]^2 [C2O42-] = 5.4 X 10^-12.

Let x equal the molar solubility. In the expression for Ksp, x will equal the concentration of C2O42- ions since you only get one oxalate ion for each mole of silver oxalate that dissolves. The concentration of silver ions will be 2x.

So, your equation for Ksp becomes:

(2x)^2 *x = 5.4 X 10^-12
4x^3 = 5.4 X 10^-12

You should be able to solve for x.

Answer 3

First of all, write the equation for the dissociation of the silver oxalate.

Ag2C2O4(s) --> 2Ag+ + C2O42-


Write the equilibrium expression, which is
Ksp = [Ag+]^2[c2042-]

The initial concentrations of silver and oxalate are zero, and define the change as 2s for silver and s for oxalate.

You then get Ksp = (2s)^2 (s)
or
Ksp = 4s^3

Solve for s and that is your molar solubility.

Answer 4

Remember molar solubility is another fancy name for the "K" constant in the equation: Ksp = k[A][B] of the reactants only.

If they give you the Ksp; 5.4 x 10^12, simply take this value and square root it.5.4 x 10^12 = k[Ag]^2[C2O4]
5.4 x 10^12 = k[[Ag]^2]^2
5.4 x 10^12 =k [C2O4]^2
You would then need to square root the Ksp value to find the K constant. You can treat C2O4 as a variable if you like. For ex: For the [[Ag]^2]^2, you can also write it out like [2x]^2. Then you solve for your variables like an algebra problem. :)

Answer 5

while a answer is declared to be saturated, it potential that the answer is protecting the utmost volume of solute it could carry. which you would be able to answer your question: definite, this is the kind you come across solubility in g/L. The Ksp is termed the solubility product consistent. it quite is a ratio of the multiplication of the respective concentrations of the products shaped from dissolving the solute to the concentration of the solute. If the solute is a organic sturdy, liquid, or gas (meaning its concentration won't be in a position to get replaced), then its concentration will take transport of as a million (or cohesion) interior the ratio. NiCl2 (s) --> Ni(2+) (aq) + 2 Cl(-) (aq) as a consequence, Ksp = [Ni(2+)] x [Cl(-a million)]^2. Moles NiCl2 = 2.a million g/129.596 g/mol = 0.0.5 mol Moles Ni(2+) = 0.0.5 mol NiCl2 x (a million mol Ni(2+)/a million mol NiCl2) = 0.0.5 mol Moles Cl(-) = 0.0.5 mol NiCl2 x (2 mol Cl(-)/a million mol NiCl2) = 0.030 mol as a consequence, at equilibrium, [Ni(2+)] = 0.0.5 mol/0.0.5 L = 3 M & [Cl(-)] = 0.030 mol/0.0.5 L = 6 M. consequently, Ksp = 3 M x (6 M)^2 = 108 M^3.