# Calcium carbonate, CaCO_3 has a Ksp value of 1.4 x 10^-8, what is the solubility of CaCO_3?

Aug 24, 2016

$C a C {O}_{3} \left(s\right) r i g h t \le f t h a r p \infty n s C {a}^{2 +} + C {O}_{3}^{2 -}$

#### Explanation:

${K}_{s p} = \left[C {a}^{2 +}\right] \left[C {O}_{3}^{2 -}\right]$ $=$ $1.4 \times {10}^{-} 8$.

If we call the solubility $S$, then $S = \left[C {a}^{2 +}\right] = \left[C {O}_{3}^{2 -}\right]$, and,

${K}_{s p} = \left[C {a}^{2 +}\right] \left[C {O}_{3}^{2 -}\right]$ $=$ $1.4 \times {10}^{-} 8$ $=$ ${S}^{2}$

So $C {a}^{2 +}$ $=$ $\sqrt{{K}_{\text{sp}}}$ $=$ $\sqrt{1.4 \times {10}^{-} 8}$ $=$ $1.18 \times {10}^{- 4}$ $m o l \cdot {L}^{-} 1$.

$\text{Solubility}$ $=$ $1.18 \times {10}^{-} 4 \times 100.09 \cdot g \cdot m o {l}^{-} 1$ $=$ ??g*L^-1