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?

1 Answer
Jul 29, 2017

Answer:

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#.........