Given that for Ca(OH)_2Ca(OH)2, K_"sp"=4.8xx10^-6Ksp=4.8×10−6, will precipitation occur if a 125*mL125⋅mL volume of calcium chloride at 2.9xx10^-2*mol*L^-12.9×10−2⋅mol⋅L−1 concentration is mixed with a 175*mL175⋅mL volume of sodium hydroxide at 2.9xx10^-2*mol*L^-12.9×10−2⋅mol⋅L−1 concentration?

1 Answer
anor277
Mar 20, 2017

We interrogate the equilibrium:

Ca^(2+) + 2HO^(-) rightleftharpoons Ca(OH)_2(s)darrCa2++2HO−⇌Ca(OH)2(s)⏐⏐↓

FOR which K_"sp"=4.8xx10^-6Ksp=4.8×10−6

Explanation:

Now we are given that K_"sp"=4.8xx10^-6=[Ca^(2+)][HO^-]^2Ksp=4.8×10−6=[Ca2+][HO−]2.

And thus we must work out the concentrations with respect to Ca^(2+)Ca2+ and HO^-HO−, and determine the initial ion product........QQ

Ca^(2+)=Ca2+=

"Moles of calcium ion"/"volume of solution"=(0.125*Lxx2.9xx10^-2*mol*L^-1)/(0.125*L+0.175*L)Moles of calcium ionvolume of solution=0.125⋅L×2.9×10−2⋅mol⋅L−10.125⋅L+0.175⋅L

=0.0121*mol*L^-1=0.0121⋅mol⋅L−1

HO^(-)=HO−=

"Moles of hydroxide ion"/"volume of solution"=(0.175*Lxx2.9xx10^-2*mol*L^-1)/(0.125*L+0.175*L)Moles of hydroxide ionvolume of solution=0.175⋅L×2.9×10−2⋅mol⋅L−10.125⋅L+0.175⋅L

=0.0169*mol*L^-1=0.0169⋅mol⋅L−1

And the ion product Q=[Ca^(2+)][HO^-]^2=3.46xx10^-6*mol*L^-1Q=[Ca2+][HO−]2=3.46×10−6⋅mol⋅L−1

Since K_"sp"=4.8xx10^-6Ksp=4.8×10−6 >> Q_"ion product"Qion product, precipitation SHOULD NOT occur..........

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