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

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Discussion Overview

The discussion revolves around whether mixing solutions of calcium chloride (CaCl2) and sodium hydroxide (NaOH) will result in the formation of a precipitate of calcium hydroxide (Ca(OH)2). Participants explore the solubility product (Ksp) of Ca(OH)2 and perform calculations to determine the concentrations of ions in solution, assessing the conditions under which a precipitate might form.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the concentrations of Ca2+ and OH- ions and concludes that a precipitate will form since the experimental ion product (Q) is greater than Ksp.
  • Another participant presents a different calculation for the concentrations of Ca2+ and OH- ions, arriving at a different value for K_trial, which also suggests that a precipitate will form.
  • A third participant challenges the calculations of the second participant, asserting that their concentrations are correct and questioning the mathematical approach taken.
  • There is an acknowledgment of the need to verify calculations, with one participant thanking another for checking their work.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there are competing calculations regarding the concentrations of ions and the resulting K_trial values. Disagreement exists on the correctness of the calculations and the implications for precipitate formation.

Contextual Notes

Participants' calculations depend on the accuracy of their volume and concentration determinations, which are under scrutiny. There are unresolved aspects regarding the methods used to calculate the ion concentrations and the subsequent K_trial values.

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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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