# Given that for Ca(OH)_2, K_"sp"=4.8xx10^-6, will precipitation occur if a 125*mL volume of calcium chloride at 2.9xx10^-2*mol*L^-1 concentration is mixed with a 175*mL volume of sodium hydroxide at 2.9xx10^-2*mol*L^-1 concentration?

Mar 20, 2017

We interrogate the equilibrium:

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

FOR which ${K}_{\text{sp}} = 4.8 \times {10}^{-} 6$

#### Explanation:

Now we are given that ${K}_{\text{sp}} = 4.8 \times {10}^{-} 6 = \left[C {a}^{2 +}\right] {\left[H {O}^{-}\right]}^{2}$.

And thus we must work out the concentrations with respect to $C {a}^{2 +}$ and $H {O}^{-}$, and determine the initial ion product........$Q$

$C {a}^{2 +} =$

$\text{Moles of calcium ion"/"volume of solution} = \frac{0.125 \cdot L \times 2.9 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1}{0.125 \cdot L + 0.175 \cdot L}$

$= 0.0121 \cdot m o l \cdot {L}^{-} 1$

$H {O}^{-} =$

$\text{Moles of hydroxide ion"/"volume of solution} = \frac{0.175 \cdot L \times 2.9 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1}{0.125 \cdot L + 0.175 \cdot L}$

$= 0.0169 \cdot m o l \cdot {L}^{-} 1$

And the ion product $Q = \left[C {a}^{2 +}\right] {\left[H {O}^{-}\right]}^{2} = 3.46 \times {10}^{-} 6 \cdot m o l \cdot {L}^{-} 1$

Since ${K}_{\text{sp}} = 4.8 \times {10}^{-} 6$ $>$ ${Q}_{\text{ion product}}$, precipitation SHOULD NOT occur..........