Does Mixing CaCl2 and NaOH Solutions Form a Precipitate of Ca(OH)2?

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Mixing 100 mL of 0.10 mol/L CaCl2 with 50 mL of 0.070 mol/L NaOH will result in the formation of calcium hydroxide, Ca(OH)2, which is a precipitate. The solubility product (Ksp) of Ca(OH)2 is 7.9 x 10^-6 at 25°C. Calculations show that the ion product (Q) for the combined solutions is 1.4 x 10^-10, which is less than Ksp, indicating no precipitate will form. However, an alternative calculation suggests a K_trial of 3.5 x 10^-5, which is greater than Ksp, leading to conflicting conclusions. The discussion highlights discrepancies in calculations and emphasizes the need for accurate concentration assessments.
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70. The solubility product of calcium hydroxide, Ca(OH)2, is 7.9 x 10-6 at 25°C. Will a precipitate form when 100mL of 0.10 mol/L of CaCl2 solution and 50.0mL of 0.070 mol/L of NaOH solution are combined?

Solution

1. Determine the formula for the possible precipitate that might form, and write the Ksp expression of this insoluble salt. If the precipitate forms, it will be Ca(OH)2.

Ca(OH)2 <---> Ca2+ (aq) + 2OH- (aq)

Ksp = [Ca2+][OH-]2 = 7.9 x 10-6

2. Calculate the concentration of each of the Ca2+ and OH- ions available in solution:

[Ca2+] = number of mol of Ca2+ / total volume of solution

= 0.1 mol/L (1 x 10-1 L) / 0.05 L + 1 x 10-1 L

= 2.0 x 10-2 mol/L

[OH-]2 = 0.070 mol/L (0.05 x 10-3 L) / 0.1 L + 0.05 x 10-3 L

= 8.5 x 10-5 mol/L

3. Substitute the ion concentrations into an experimental ion product (Q) equation that is identical to the Ksp expression. This will allow you to compare two solubility product values and determine whether a precipitate will form.

Experimental ion product (Q)
= [Ca2+][OH-]2

= (2.0 x 10-2)(8.5 x 10-5)2

= 1.4 x 10-10

Since Q (1.4 x 10-10) is greater than Ksp (7.9 x 10-6), a precipitate will form.
 
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I am working on this now and have a different answer.
If anyone has any idea which is right could you please post.

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

Ca(OH)2 <---> Ca2+ (aq) + 2OH- (aq)

Ksp = [Ca2+][OH-]2 = 7.9 x 10-6

[Ca2+] = 0.1 mol/L x .100 L / 0.150 L
= .067 mol/L

[OH-] = 0.070 mol/L x 0.050 L / 0.150 L
= .023 mol/L

K_trial= [Ca2+][OH-]2
= (.067 mol/L)(.023 mol/L)^2
= 3.5 x 10^-5

Since K_trial (3.5 x 10-5) is greater than Ksp (7.9 x 10-6), a precipitate will form.
 
yellowduck: your concentrations are correct, as opposed to ScrubsFan.

ScrubsFan: check your math, you are doing some strange tricks when calculating volume. Besides, 1.4 x 10-10 is much LOWER then Ksp.
 
Thank you Borek for taking the time to check my work.
 
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