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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?

2.9xx10^(-2) M

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 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.

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