Given that for #Ca(OH)_2#, #K_"sp"=4.8xx10^-6#, will precipitation occur if a #125*mL# volume of calcium chloride at #2.9xx10^-2*mol*L^-1# concentration is mixed with a #175*mL# volume of sodium hydroxide at #2.9xx10^-2*mol*L^-1# concentration?

1 Answer
Mar 20, 2017

Answer:

We interrogate the equilibrium:

#Ca^(2+) + 2HO^(-) rightleftharpoons Ca(OH)_2(s)darr#

FOR which #K_"sp"=4.8xx10^-6#

Explanation:

Now we are given that #K_"sp"=4.8xx10^-6=[Ca^(2+)][HO^-]^2#.

And thus we must work out the concentrations with respect to #Ca^(2+)# and #HO^-#, and determine the initial ion product........#Q#

#Ca^(2+)=#

#"Moles of calcium ion"/"volume of solution"=(0.125*Lxx2.9xx10^-2*mol*L^-1)/(0.125*L+0.175*L)#

#=0.0121*mol*L^-1#

#HO^(-)=#

#"Moles of hydroxide ion"/"volume of solution"=(0.175*Lxx2.9xx10^-2*mol*L^-1)/(0.125*L+0.175*L)#

#=0.0169*mol*L^-1#

And the ion product #Q=[Ca^(2+)][HO^-]^2=3.46xx10^-6*mol*L^-1#

Since #K_"sp"=4.8xx10^-6# #># #Q_"ion product"#, precipitation SHOULD NOT occur..........