We were titrating 15 mL of a saturated Ca(OH)2 solution with .050M HCl solution.

Question

We were titrating 15 mL of a saturated Ca(OH)2 solution with .050M HCl solution. The equivalence point for trial 1 is 11.75 mL Determining the K, of Calcium Hydro (mL) 2 DATA ANALYSIS 1. Calculate the [OH] from the results of your titrations. Explain your calculations. 2. Calculate the [Ca?-]. Explain your calculations. 3. Calculate the Kyp for calcium hydroxide. Explain your calculations. 4. Find the accepted value of the K, for calciun hydroxide and compare it with your value. Discuss the discrepancy and suggest possible sources of experimental error. 26

Explanation / Answer

Ksp for Ca(OH)2

1. average volume of HCl used = 11.935 ml

moles HCl used = 0.05 M x 0.011935 L = 6 x 10^-4 moles

[OH-] present = 6 x 10^-4 moles/0.015 L = 0.04 M

2. [Ca2+] present = 0.04/2 = 0.02 M

3. Ksp = [Ca2+][OH-]^2

           = (0.02)(0.04)^2

           = 3.2 x 10^-5

4. Ksp of Ca(OH)2 (literature) = 6 x 10^-6

the values calculated in the experiment is much greater than literature Ksp value.

The error could be due to incorrect measurement of HCl volume added to the solution for titration. The intiial mass volume of Ca(OH)2 might be higher than the value reported here.

1.08 x 10^-10

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.