Will a Precipitate Form When Combining CaCl2 & NaOH Solutions?

[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. 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

 

[GCT]

Hi there

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

 

[ravenprp]

for asking for clarification! Your calculations are correct and your conclusion is also correct. According to the solubility product (Ksp) and the concentrations of the two solutions, a precipitate will not form. This is because the Ksp value for Ca(OH)2 is much smaller than the Ktrial value, indicating that the solution is not saturated and therefore no precipitate will form. Keep in mind that this is assuming ideal conditions and that other factors, such as temperature and pH, could also affect the formation of a precipitate.