# What is the molar solubility of Cd(OH)_2 given K_"sp"=5.3xx10^-15?

Mar 20, 2017

#### Answer:

$\left[C d {\left(O H\right)}_{2}\right] = 1.10 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

For the reaction:

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

We write ${K}_{\text{sp}} = \left[C {d}^{2 +}\right] {\left[H {O}^{-}\right]}^{2}$

And if we call the solubility of $C d {\left(O H\right)}_{2}$, $S$, then.........

${K}_{\text{sp}} = S \times {\left(2 S\right)}^{2} = 4 {S}^{3}$

And so $S = {\text{^3sqrt(K_"sp"/4)=}}^{3} \sqrt{\frac{5.3 \times {10}^{-} 15}{4}} = 1.10 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$, with respect to $C d {\left(O H\right)}_{2}$..........

If the $p H$ of the solution were buffered to HIGH or LOW $p H$ how would the solubility evolve?