Question

At 25 °C, the molar solubility of silver phosphate is `1.8 × 10^-5` mol L-1. How do you calculate Ksp for this salt?

Answer 1

`2.9xx10^(-2) M`

Explanation:

Let's start by writing the chemical reaction for the dissociation of silver phosphate:

`Ag_3PO_4(s) rightleftharpoons 3Ag^(+)(aq) +PO_4^(-3)(aq)`

Now, set the Ksp value equal to the products (you don't care about the reactants because it's a solid). Ag is raised to the 3rd power because the coefficient is 3.

`Ksp=1.8xx10^(-5) M = [Ag^(+)]^(3) (aq) +[PO_4^(3-)] (aq)`

Replace each reactant with the letter x because that's what you're trying to find. Since the coefficient in front of silver is 3, a 3 must be placed in front of the x and it must be raised to the 3rd power.

`1.8xx10^(-5) M = (3X)^(3) xx (X)` (always multiply when finding the molar solubility).

Now take `3^(3)` which is 27 and divide the Ksp by 27, so you can get all of the X's by themselves.

`(1.8xx10^(-5) M)/27` = `6.67xx10^(-7)M`

Now you're left with `X^(3) xx X`, so multiply the X's to get `X^(4)`. Take the fourth root of the `6.67xx10^(-7)M` to obtain the value of x.

(`6.67xx10^(-7)M)^(1/4) = 2.9xx10^(-2) M`

The value of x that we just obtained is our molar solubility.