Question

A saturated solution of $Ca _{3}\left( PO _{4}\right)_{2}$ contains $2 \times 10^{-8} M\, Ca ^{2+}$ ions and $1.6 \times 10^{-5} \,M\, PO _{4}^{3-}$ ions at a particular temperature. The solubility product of $Ca _{3}\left( PO _{4}\right)_{2}$ at that temperature is $2.048 \times 10^{-x} .$ What is the value of ' $x$ '?

Answer 1

$Ca _{3}\left( PO _{4}\right)_{2} \rightleftharpoons 3 Ca ^{2+}+2 PO _{4}^{3-}$
$K _{ sp } =\left[ Ca ^{2+}\right]^{3} \times\left[ PO _{4}^{3-}\right]^{2}$
$=\left(2 \times 10^{-8}\right)^{3} \times\left(1.6 \times 10^{-5}\right)^{2}$
$=8 \times 10^{-24} \times 2.56 \times 10^{-10}$
$=20.48 \times 10^{-34}$
$=2.048 \times 10^{-33}$
$\therefore $ Value of $x$ is $33 .$