Question

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?

Answer 1

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

Explanation:

`Ca(OH)_2(aq) +2HCl(aq) rarr CaCl_2(aq) + 2H_2O(l)`

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 `Ca(OH)_2` was equal to the concentration as specified by the following equilibrium......

`Ca(OH)_2(s)rightleftharpoonsCa^(2+) + 2HO^-`

We had `8.13xx10^-3*Lxx0.102*mol*L^-1=8.29xx10^-4*mol` with respect to `HCl`, and thus there were HALF this molar quantity present in solution with respect to the saturated `Ca(OH)_2`; you with me.....?

This `Ca(OH)_2` was dissolved in a `25xx10^-3*L` volume, so `[Ca(OH)_2]=1/2xx(8.29xx10^-4*mol)/(25xx10^-3*L)=1.659xx10^-2*mol*L^-1`

And thus............................. `S_(Ca(OH)_2)=1.659xx10^-2*mol*L^-1xx74.09*g*mol^-1`

`=1.229*g*L^-1`.........