Question

A determination the molar solubility and the K_sp calcium hydroxide was completed according to the experimental procedure. The following data were collccicd for Trial 1 (Sec Repon sheet. minations. Record the calculated values to the correct number of significant ligures. A. Molar Solubility and Solubility Product of Calcium Hydroxide 1.Volume of saturated Ca(OH)_2 solution (mL) 2.Molar concentration of standard HCI solution (mol/L) 3.Buret reading, initial (mL) 4.Buret reading, final (mL) 5.Volume of HCI added (mL) 6.Moles of HCI added (mol) Show calculation. 7.Moles of OH in saturated solution (mol) Show calculation. 8.[OH^-],equilibrium (mol/L) Show calculation. 9.[Ca^2+], equilibrium (mol/L) Show calculation 10.Molar solubility of Ca(OH)_2 (mol/L) Show calculation. 11. K_sq of Ca(OH)_2

Answer 1

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Answers:

5. Vol HCl added: 12.2 mL

6. Mol of HCl added

n(HCl) = Vol(HCl) x M = 12.2 mL x 0.0480 mmol/mL = 0.5856 mmol

7. Moles of OH^- in saturated solution. The number of moles of OH^- in the saturated solution should be exactly the same as the number of moles of HCl used in the titration since the reaction has a relation one to one:

HCl + OH (aq)H20 Cl (aq)

So n(OH^-) = 0.5856 mmol

8. [OH^-] = 0.5856 mmol / 25 mL = 0.0234 mmol/mL (which is the same as mol/L)

9. [Ca⁺²]

We can derive the concentration of calcium ions from the solubility reaction:

In the solubility reaction we can see that for each mol of Calcium ion dissolved there are 2 moles of OH^- ion. So calcium concentration should be half of the [OH^-]

[Ca⁺²] = [OH^-]/2 = 0.0234 M / 2 = 0.0117 M

10. Molar solubility is the same [Ca⁺²] concentration, this is 0.0117 M

11. Ksp

We apply the equation:

Ksp = (0.0117) x (0.0234)² = 6.4 x 10^-6

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