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?

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?

Natalie Anton · Accepted Answer

#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 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.

Noah Aldrich · Answer

undefined To calculate the $K_{sp}$ for silver phosphate ($Ag_3PO_4$), use the formula $K_{sp} = [Ag^+]^3[PO_4^{3-}]$. Substitute the molar solubility of silver phosphate into the expression, assuming complete dissociation. The equilibrium concentrations of $Ag^+$ and $PO_4^{3-}$ are both equal to three times the molar solubility due to the stoichiometry of the balanced equation.