# 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?

Jun 10, 2016

$2.9 \times {10}^{- 2} M$

#### Explanation:

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

$A {g}_{3} P {O}_{4} \left(s\right) r i g h t \le f t h a r p \infty n s 3 A {g}^{+} \left(a q\right) + P {O}_{4}^{- 3} \left(a q\right)$

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.

$K s p = 1.8 \times {10}^{- 5} M = {\left[A {g}^{+}\right]}^{3} \left(a q\right) + \left[P {O}_{4}^{3 -}\right] \left(a q\right)$

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.

$1.8 \times {10}^{- 5} M = {\left(3 X\right)}^{3} \times \left(X\right)$ (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.

$\frac{1.8 \times {10}^{- 5} M}{27}$ = $6.67 \times {10}^{- 7} M$

Now you're left with ${X}^{3} \times X$, so multiply the X's to get ${X}^{4}$. Take the fourth root of the $6.67 \times {10}^{- 7} M$ 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.