Will Ca(OH)2 precipitate form when mixing CaCl2 and NaOH?

[mister_mister3]

Question as posed in my text:

Q: The solubility product of calcium hydroxide, Ca(OH)2, is 7.9 x 10^-6 at 25 deg. celsius. Will a precipitate form when 100mL of 0.10 mol/L of CaCl2 solution and 50.0 mL of 0.070 mol/L of NaOH solution are combined?

My answer:

CaCl2 = 0.10 mol/L x 100mL/150mL = 0.067 mol/L
NaOH = 0.070 mol/L x 50mL/150mL = .023 mol/L

Ktrial = 0.067 x .023 = 1.54 x 10^-3

Therefore, since the Ktrial result is smaller than Ksp, no precipitate will form.

So this looks right to me, but something is nagging me about it...Anyone? Have I done this properly?

Thanks

 

[GCT]

Hi there

$$K=[Ca^{2+}][OH^{-}]^{2}$$