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Q: The solubility product of calcium hydroxide, Ca(OH)2, is 7.9 x 10^-6 at 25 deg. celcius. 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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# Precipitate question

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