I'm not quite sure what i'm doing wrong here:
The Ksp for Ga(OH)3 = 7.28 X 10-36. What is the molar solubility of Ga(OH)3 in a 3.80M solution of gallium sulfate?
And I can't tell you since you didn't show your work? Post your work and I'll find the error. I would rather you spend you time typing and me analyzing than the other way around.
To determine the molar solubility of Ga(OH)3 in a 3.80M solution of gallium sulfate, we need to understand the concept of solubility product constant (Ksp) and the stoichiometry of the reaction.
The Ksp represents the equilibrium constant for the dissolution of an ionic compound in water. It indicates the degree of solubility of the compound. For Ga(OH)3, the Ksp is given as 7.28 x 10^-36, which means it has an extremely low solubility in water.
The chemical equation for the dissociation of Ga(OH)3 can be written as:
Ga(OH)3(s) ⇌ Ga3+(aq) + 3OH-(aq)
Since we have a 3.80M solution of gallium sulfate (Ga2(SO4)3), we can assume that Ga3+ ions will be present in excess, as they are derived from a strong electrolyte. Thus, we can consider Ga3+ concentration as 3.80M.
To find the molar solubility of Ga(OH)3, we need to determine the concentration of OH- ions, which is related to the solubility of Ga(OH)3 since it forms three OH- ions per Ga(OH)3 molecule.
Let's assume the molar solubility of Ga(OH)3 is 'x'.
So, the concentration of OH- ions can be expressed as 3x, as there are three OH- ions produced for every molecule of Ga(OH)3 that dissolves.
Using the stoichiometry from the balanced equation, we can write an expression for the equilibrium concentration of Ga3+ as [Ga3+] = 3.80M (concentration of Ga2(SO4)3).
Similarly, the expression for the equilibrium concentration of OH- can be written as [OH-] = (3x).
Since Ga(OH)3 dissolves completely and dissociates to Ga3+ and 3OH-, at equilibrium, the product of their concentrations must be equal to the Ksp:
Ksp = [Ga3+][OH-]^3
Therefore, we can substitute the above expressions to obtain the following relationship:
Ksp = (3.80M)(3x)^3 = 7.28 x 10^-36
Now we can solve the equation for 'x', which represents the molar solubility of Ga(OH)3.
1. Cubing the expression:
(3.80M)(27x^3) = 7.28 x 10^-36
2. Rearranging the equation:
(3.80M)(27x^3) = 7.28 x 10^-36
98.28Mx^3 = 7.28 x 10^-36
3. Dividing both sides by 98.28M:
x^3 = (7.28 x 10^-36) / (98.28M)
4. Taking the cube root of both sides:
x = (7.28 x 10^-36)^(1/3) / (98.28M)^(1/3)
Calculating this value will give you the molar solubility of Ga(OH)3 in the given 3.80M solution of gallium sulfate.