In a saturated solution of silver phosphate, the concentration of silver is 4.5 x 10^-4 mol/L. What is the Ksp of silver phosphate?

Question

In a saturated solution of silver phosphate, the concentration of silver is 4.5 x 10^-4 mol/L. What is the Ksp of silver phosphate?

Answer 1

Ag3PO4(s) <=> 3Ag^+(aq) + PO4^3+(aq)

Ksp = [Ag^+]^3•[PO4^3+]
Let [PO4^3+] = s
then [Ag^+]= 3s = 4.5 x 10^-4 mol/L.
s = 1.5x10^-4
Ksp = (3s)^3•(s) = 27s^4
substitute the value of s and solve

Answer 2

GK made a a couple of typos. PO4^3+ should have been PO4^3-. That doesn't change anything in the calculation.

Answer 3

To find the solubility product constant (Ksp) of silver phosphate, we need to use the concentration of silver ions.

The balanced equation for the dissolution of silver phosphate (Ag3PO4) is:

Ag3PO4(s) ↔ 3Ag+(aq) + PO4-3(aq)

From the equation, we can see that one mole of silver phosphate produces three moles of silver ions.

Given that the concentration of silver ions is 4.5 x 10^-4 mol/L, we can calculate the concentration of silver phosphate using the stoichiometric ratio:

[Ag3PO4] = [Ag+] / 3

[Ag3PO4] = 4.5 x 10^-4 mol/L / 3

[Ag3PO4] = 1.5 x 10^-4 mol/L

Now, the solubility product constant (Ksp) expression for silver phosphate can be written:

Ksp = [Ag+]^3 * [PO4-3]

Since [Ag+] = 4.5 x 10^-4 mol/L and [PO4-3] = [Ag3PO4] = 1.5 x 10^-4 mol/L, let's substitute these values into the expression:

Ksp = (4.5 x 10^-4)^3 * (1.5 x 10^-4)

Ksp = 91.125 x 10^-12

Ksp = 9.1125 x 10^-10

Therefore, the solubility product constant (Ksp) of silver phosphate (Ag3PO4) is 9.1125 x 10^-10.

Answer 4

To find the solubility product constant (Ksp) of silver phosphate, you need to know the molar concentrations of the ions in the saturated solution. In the case of silver phosphate (Ag3PO4), it dissociates into three silver ions (Ag+) and one phosphate ion (PO43-).

Let's assume the solubility of silver phosphate is 's' mol/L. In a saturated solution, the concentration of the silver ions will be equal to 3s (since there are three Ag+ ions in one formula unit of silver phosphate).

According to the given information, the concentration of silver ions in the saturated solution is 4.5 x 10^-4 mol/L. Therefore, we can write:

3s = 4.5 x 10^-4

Now, we can solve for the value of 's':

s = (4.5 x 10^-4) / 3

Once you have the value of 's', you can substitute it into the Ksp expression for silver phosphate:

Ksp = [Ag+]^3[PO43-] = (3s)^3 × (s)

Using the value of 's' calculated above, you can find the Ksp of silver phosphate.