Answer to Question #98877 in Physical Chemistry for nehemiah

Question

the solubility of calcium phosphate is 7.7x10\'-4 g/dm\'3 at 25\'C calcutate the solubility product at this temparature

Answer 1

Solution.

The solubility product is calculated by the following formula:

$SP = n^n \times m^m \times p^{n+m}$

n, m are the coefficients before the substance formula.

Ca3(PO4)2

n = 3

m = 2

$SP = 3^3 \times 2^2 \times (7.7 \times 10^{-4})^{3+2} = 2.92 \times 10^{-14}$

$SP(theor.) = 2.00 \times 10^{-29}$

Since the theoretical value of SP is less than the experimental SP, the precipitate will fall.

Answer:

$SP = 2.92 \times 10^{-14}$