Excess Ca(OH)2 is shaken with water to produce a saturated solution. The solution is filtered, and a 60.00 mL sample titrated with HCl requires 11.13 mL of 9.83×10−2 M HCl to reach the endpoint. calculate ksp for Ca(OH)2?

1 Answer
anor277
Feb 24, 2018

#K_"sp"=3.03xx10^-6#

Explanation:

We work out the moles of hydroxide....

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

#"Moles of HCl"=11.13*mLxx10^-3*L*mL^-1xx9.83xx10^-2*mol*L^-1=0.001094*mol#...and given this molar quantity there were HALF this molar quantity with respect to #Ca(OH)_2# in the starting #60.00*mL# aliquot....

And so #[Ca(OH)_2(aq)]=(1/2xx0.001094*mol)/(60.00*mLxx10^-3*L*mL^-1)# #=9.12xx10^-3*mol*L^-1#...

But this concentration derived from SOLUBLE #Ca(OH)_2#, the which had participated in the solubility equilibrium...

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

for which we write the solubility expression....

#K_"sp"=[Ca^(2+)][HO^-]^2#..and we plug in the concentrations...

#=(9.12xx10^-3)xx(2xx9.12xx10^-3)^2#

#=4xx(9.12xx10^-3)^3=3.03xx10^-6#

Please check my arithmetic....no money-back guarantees...

Related questions
Impact of this question
4205 views around the world
You can reuse this answer
Creative Commons License