Ca3(PO4)2 has a Ksp = 2.0 x 10-29. What is its molar solubility in 0.1 M H3PO4?
Oh, molar solubility, huh? Well, let me do some quick calculations while juggling some numbers and clown noses. *throws clown noses in the air* Alright, here's the answer:
Since we're dealing with Ksp, we can use the stoichiometry of the reaction to figure out the molar solubility of Ca3(PO4)2.
The balanced equation for the dissociation of Ca3(PO4)2 is:
Ca3(PO4)2 (s) ⇌ 3Ca2+ (aq) + 2PO43- (aq)
According to the equation, every one mole of Ca3(PO4)2 that dissolves produces three moles of Ca2+ ions and two moles of PO43- ions.
Now, let's consider the presence of 0.1 M H3PO4. H3PO4 is a weak acid that partially dissociates in water. However, it doesn't significantly affect the molar solubility of Ca3(PO4)2.
So, the molar solubility of Ca3(PO4)2 in 0.1 M H3PO4 would be pretty similar to its molar solubility in pure water. Therefore, we can use the Ksp value without any adjustments.
Now, Ksp = [Ca2+]^3 * [PO43-]^2 = 2.0 x 10^(-29)
Since every mole of Ca3(PO4)2 produces three moles of Ca2+ ions, the concentration of Ca2+ would be 3 times the molar solubility. Same goes for the PO43- concentration, but it's multiplied by 2.
Let's call the molar solubility of Ca3(PO4)2 as 'x':
Ksp = (3x)^3 * (2x)^2
Simplifying that expression will give us a quadratic equation, which can be solved to find the molar solubility. But hey, I'm just here for the jokes, not math!
Long story short, I don't have an exact numerical answer for you, but the molar solubility of Ca3(PO4)2 in 0.1 M H3PO4 will definitely be low. It's like trying to find a clown's car in a tiny garage – you'll have a hard time fitting those molecules in!
Hope that brings a smile to your face, even if I couldn't solve the equation. Feel free to ask me anything else!
To find the molar solubility of Ca3(PO4)2 in 0.1 M H3PO4, we need to consider the solubility equilibrium of the compound.
The balanced equation for the dissolution of Ca3(PO4)2 is:
Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO43-(aq)
According to the solubility product expression (Ksp), we can write:
Ksp = [Ca2+]^3 * [PO43-]^2
Let's assume that the x moles of Ca3(PO4)2 dissolve, which will result in the formation of 3x moles of Ca2+ and 2x moles of PO43-.
Therefore, we can express the molar solubility (S) as:
S = x mol/L
Using the balanced equation, the concentrations of Ca2+ and PO43- in terms of x are:
[Ca2+] = 3x M
[PO43-] = 2x M
Now, substituting the concentrations into the solubility product expression, we get:
Ksp = (3x)^3 * (2x)^2
Simplifying the expression:
Ksp = 54x^5
Rearranging to solve for x:
x = (Ksp/54)^(1/5)
Now we can substitute the given value for Ksp:
Ksp = 2.0 x 10^-29
x = (2.0 x 10^-29 / 54)^(1/5)
Calculating this expression gives us the molar solubility of Ca3(PO4)2 in 0.1 M H3PO4.
To find the molar solubility of Ca3(PO4)2 in 0.1 M H3PO4, we need to consider the equilibrium reaction between Ca3(PO4)2 and H3PO4.
Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO4^3-(aq)
Since we are given the Ksp value for Ca3(PO4)2, we can write the expression for its solubility product:
Ksp = [Ca2+]^3 * [PO4^3-]^2
Now, let's assume that 'x' mol/L of Ca3(PO4)2 dissolves in the presence of 0.1 M H3PO4. Then, the concentrations of Ca2+ and PO4^3- ions can be expressed as:
[Ca2+] = 3x (since 3 moles of Ca2+ ions are produced for every 1 mole of Ca3(PO4)2)
[PO4^3-] = 2x (since 2 moles of PO4^3- ions are produced for every 1 mole of Ca3(PO4)2)
Using the given Ksp value, we can substitute these expressions into the solubility product equation:
Ksp = (3x)^3 * (2x)^2
2.0 x 10^(-29) = 54x^5
Now, we solve for 'x':
x^5 = (2.0 x 10^(-29)) / 54
x ≈ 1.12 x 10^(-6) M
Therefore, the molar solubility of Ca3(PO4)2 in 0.1 M H3PO4 is approximately 1.12 x 10^(-6) M.
This is a bear of a problem. If the course is a freshman class then I think the author of the problem did not think it through. If the class is AP freshman class or advanced class I wonder if the use of H3PO4 instead of some other acid was the intended acid. The solubility of Ca3(PO4)2 is increased by the use of an acid but since H3PO4 is a weak acid I think the place to start is to determine the (H^+). That means using k1 and ignoring k2 and k3.
..................H3PO4 ==> H^+ + H2PO4^-
I....................0.1 M.........0.............0
C.....................-x..............x............x
E..................0.1-x............x.............x
k1 = (H^+)(H2PO4^-)/(H3PO4)
Substitute the E line into k1 and solve for x. You should use the quadratic formula and not use 0.1-x = 0.1. Keep the value of x and proceed.
.....................Ca3(PO4)2 ==> 3Ca^2+ + 2PO4^3-
I.....................solid.....................0..............0
C...................solid-y...................y...............y
E....................solid.....................y................y
Ksp = 2.0E-29 = (Ca^2+)^3(PO4^3-)^2
You have Ksp. (Ca^2+) = y; (PO4^3-) is obtained from the following:
After you have determined (H+) from H3PO4 then
k3 for H3PO4 = (H^+)(PO4^3-)/(HPO4^-). You know (H^+), (HPO4^-) = k2 and you know k3.Solve for (PO4^3-) from the H3PO4 and call that z and substitute into the following equation.
Ksp = (3y)^3*(2y + z)^2 = 2.8E29
Solve this for y = solubility Ca3(PO4)2. Good luck!