# 2xx10^-3*L of NaOH of 0.200*mol*L^-1 concentration is added to 1.00*L of 1.00*mol*L^-1 Ca(NO_3)_2(aq)...will calcium hydroxide precipitate?

Mar 10, 2017

Calcium hydroxide has a solubility of $1.73 \cdot g \cdot {L}^{-} 1$ under standard conditions. This should have been quoted in the question. Calcium hydroxide should not precipitate.

#### Explanation:

We work out the equivalent quantity of calcium hydroxide, for which we (i) need a stoichiometric equation:

$C {a}^{2 +} + 2 H {O}^{-} \rightarrow C a {\left(O H\right)}_{2} \left(s\right) \downarrow$

And (ii) equivalent quantities of the ions in solution:

$\text{Moles of sodium hydroxide,}$ $=$ $2.00 \times {10}^{-} 3 L \times 0.200 \cdot m o l \cdot {L}^{-} 1 = 4.0 \times {10}^{-} 4 \cdot m o l$.

And thus when this is added to the bulk solution, we have $\left[H {O}^{-}\right]$ $=$ $\frac{4.0 \times {10}^{-} 4 \cdot m o l}{1.002 \cdot L} = 4.0 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1$.

Given the stoichiometry, we thus have a solution that is NOMINALLY $2.0 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1$ with respect to calcium hydroxide.

And this is a mass solubility of,

$74.09 \cdot g \cdot m o {l}^{-} 1 \times 2.0 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1 = 14.8 \cdot m g \cdot {L}^{-} 1$ or $\text{15 ppm}$.

And so calcium hydroxide should NOT precipitate.