Will a Precipitate Form When Combining CaCl2 & NaOH Solutions?

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SUMMARY

The discussion centers on the solubility product (Ksp) of calcium hydroxide (Ca(OH)2) and whether a precipitate forms when combining 100 mL of 0.10 mol/L CaCl2 with 50.0 mL of 0.070 mol/L NaOH. The calculated ion concentrations yield a trial solubility product (Ktrial) of 1.54 x 10^-3, which is significantly lower than the Ksp of 7.9 x 10^-6. Consequently, no precipitate will form in this reaction.

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mister_mister3
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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
 
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Hi there

K=[Ca^{2+}][OH^{-}]^{2}
 

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