Question

Ca3(po4)2(s)↽−−⇀3ca2 (aq) 2po3−4(aq) if ksp=2.1×10−33, what is the molar solubility of ca3(po4)2?

Answer 1

The molar solubility of
`Ca_3(PO_4)_2`, given that the solubility product, Ksp is
`2.1*10^(-33)`, is
`1.14*10^(-7)\ M`

How to calculate the molar solubility of
`Ca_3(PO_4)_2`?

First, we shall analyze the question to determine the useful information as regarding the question. This is shown below:

• Equation given:
  `Ca_3(PO_4)_2(s)\ \rightleftharpoons\ 3Ca^(2+)(aq)\ +\ 2PO_4^(3-)(aq)`
• Solubility product (Ksp) =
  `2.1*10^(-33)`
• Molar solutibility =?

The molar solubility can be calculated as follow:

`Ksp = [Ca^(2+)]^3[PO_4^(3-)]^2`

Let the molar solubility by y, thus, at we will have the following concentration for each species at quilibrium:

• 
  `[Ca^(2+)] =` 3y
• 
  `[PO_4^(3-)] =` 2y

Thus, the Ksp will be written as follow:

`Ksp = (3y)^3\ *\ (2y)^2\\\\Ksp = 27y^3\ *\ 4y^2\\\\Ksp = 108y^5\\\\But,\\\\Ksp\ = 2.1*10^(-33)\\\\Therefore, \\\\2.1*10^(-33) = 108y^5\\\\y^5 = (2.1*10^(-33))/(108) \\\\Take\ the\ fifth\ root\\\\y =\sqrt[5]{(2.1*10^(-33))/(108)} \\\\y = 1.14*10^(-7)\ M`

Thus, the molar solubility is
`1.14*10^(-7)\ M`