As a result of mineral erosion and biological activity, phosphate ion is common in natural waters, where it often precipitates as insoluble salts, such as Ca3(PO4)2. Calculate Qsp when [Ca2+]init = [PO43-]init = 2.161 10-6 M in a given river.
Ksp of Ca3(PO4)2 = 1.2 ✕ 10-29.
I don't understand the trouble you're having. Qsp is just the ion product as if it were Ksp; then you can compare with Ksp and see if the Ksp is exceeded. If so a ppt forms, if not there is no ppt.
To calculate Qsp, which is the reaction quotient for the precipitation reaction of Ca3(PO4)2, we need to write the balanced equation for the reaction and determine the concentrations of the ions involved.
The balanced equation for the precipitation reaction of Ca3(PO4)2 (calcium phosphate) can be written as:
3Ca2+ + 2PO43- → Ca3(PO4)2
From the given information, we know that [Ca2+]init = [PO43-]init = 2.161 × 10^-6 M. Therefore, the initial concentration of calcium ions ([Ca2+]) and phosphate ions ([PO43-]) in the river is 2.161 × 10^-6 M.
Now we can calculate Qsp by substituting the initial concentrations of the ions into the expression for the reaction quotient for the precipitation reaction:
Qsp = [Ca2+]init^3 [PO43-]init^2
Substituting the given values, we have:
Qsp = (2.161 × 10^-6)^3 (2.161 × 10^-6)^2
Calculating this expression gives us the value of Qsp.