# To determine the solubility of calcium hydroxide, a chemist took 25mL of a saturated calcium hydroxide solution and found that it reacted completely with 8.13mL of 0.102mol L^-1 hydrochloric acid?

Jul 29, 2017

We assess the following reaction.........

#### Explanation:

$C a {\left(O H\right)}_{2} \left(a q\right) + 2 H C l \left(a q\right) \rightarrow C a C {l}_{2} \left(a q\right) + 2 {H}_{2} O \left(l\right)$

Now the calcium hydroxide present in solution derived from a saturated solution; and by saturation we specify an equilibrium quantity. That is the concentration of $C a {\left(O H\right)}_{2}$ was equal to the concentration as specified by the following equilibrium......

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

We had $8.13 \times {10}^{-} 3 \cdot L \times 0.102 \cdot m o l \cdot {L}^{-} 1 = 8.29 \times {10}^{-} 4 \cdot m o l$ with respect to $H C l$, and thus there were HALF this molar quantity present in solution with respect to the saturated $C a {\left(O H\right)}_{2}$; you with me.....?

This $C a {\left(O H\right)}_{2}$ was dissolved in a $25 \times {10}^{-} 3 \cdot L$ volume, so $\left[C a {\left(O H\right)}_{2}\right] = \frac{1}{2} \times \frac{8.29 \times {10}^{-} 4 \cdot m o l}{25 \times {10}^{-} 3 \cdot L} = 1.659 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$

And thus............................. ${S}_{C a {\left(O H\right)}_{2}} = 1.659 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1 \times 74.09 \cdot g \cdot m o {l}^{-} 1$

$= 1.229 \cdot g \cdot {L}^{-} 1$.........