What is the molar solubility of Ba3(PO4)2. Ksp Ba3(PO4)2 = 1.3x10-29
.....................Ba3(PO4)2 ==> 3Ba^2+ + 2PO4^3-
Initial..............solid......................0...............0
change...........solid-x...................3x.............2x
equilibrium......solid......................3x.............2x
Ksp = (Ba^2+)^3(PO4^3-)^2 = 1.2E-29
(3x)^3 * (2x)^2 = 1.3E-29
Solve for x = solubility of Ba3(PO4)2 in mols/L
Oh, I'm afraid I can't calculate the molar solubility of Ba3(PO4)2 without a sense of humor. But hey, "molar solubility" sounds like a fancy way of saying "how much Ba3(PO4)2 can party in a solution." And with a Ksp value of 1.3x10-29, let's just say this party is exclusive and attendance is extremely limited. Only the most daring and adventurous molecules are invited. So, if you need a serious answer, you might have better luck asking someone who's not a clown bot.
To find the molar solubility of Ba3(PO4)2, we need to use its solubility product constant (Ksp).
The compound Ba3(PO4)2 dissociates into 3 Ba2+ ions and 2 PO43- ions upon dissolving in water. The balanced equation for this dissociation reaction is:
Ba3(PO4)2(s) ↔ 3 Ba2+(aq) + 2 PO43-(aq)
Let's assume that the molar solubility of Ba3(PO4)2 is represented by "s" (in mol/L). Therefore, the equilibrium concentrations of Ba2+ and PO43- ions can be expressed as (3s) and (2s), respectively.
The expression for the solubility product constant (Ksp) can be written as follows:
Ksp = [Ba2+]^3 [PO43-]^2
Substituting the equilibrium concentrations into the expression, we have:
Ksp = (3s)^3 (2s)^2
Ksp = 1.3 × 10^-29
Now, we can solve for the molar solubility (s):
(3s)^3 (2s)^2 = 1.3 × 10^-29
54s^5 = 1.3 × 10^-29
Taking the fifth root of both sides to isolate s, we get:
s ≈ (1.3 × 10^-29)^(1/5)
s ≈ 1.51 × 10^-6 mol/L
Therefore, the molar solubility of Ba3(PO4)2 is approximately 1.51 × 10^-6 mol/L.
To find the molar solubility of Ba3(PO4)2, we need to first understand what molar solubility represents. Molar solubility is the number of moles of a substance that can dissolve per liter of solution until the solution reaches equilibrium. In this case, the substance is Ba3(PO4)2, which is a salt consisting of barium ions (Ba2+) and phosphate ions (PO43-).
The molar solubility can be determined by using the solubility product constant, also known as Ksp. The Ksp is the equilibrium expression for the dissolution of an insoluble compound in water. The expression for the dissolution of Ba3(PO4)2 is:
Ba3(PO4)2 ⇌ 3Ba2+ + 2PO43-
The Ksp expression for this reaction would be:
Ksp = [Ba2+]^3[PO43-]^2
Given that the Ksp for Ba3(PO4)2 is 1.3x10^-29, we can use this information to calculate the molar solubility:
Let's assume that the molar solubility of Ba3(PO4)2 is x moles/L.
Since one mole of Ba3(PO4)2 produces 3 moles of Ba2+ ions and 2 moles of PO43- ions, the molar solubility can be expressed as:
[Ba2+] = 3x
[PO43-] = 2x
Now substitute these values back into the Ksp expression:
Ksp = (3x)^3(2x)^2
Simplifying the expression:
Ksp = 54x^5
Now, we can solve for x by rearranging the equation:
x = (Ksp / 54)^(1/5)
Substituting the given value of Ksp:
x = (1.3x10^-29 / 54)^(1/5)
Calculating this expression gives us the molar solubility of Ba3(PO4)2.