Laboratory 10 Determination of the K of a Slightly Soluble Compound NAME DATE: T

Question

Laboratory 10 Determination of the K of a Slightly Soluble Compound NAME DATE: TA: SECTION: Post-Lab Report (Use the In-lab observations to complete the laboratory report. Turn in to your Instructor when you have completed the report.) Temperature of the Lab Concentration of HCI used Volume saturated Ca(OH), used 20 oC mL Trial 1 Trial 2 Trial 3 Volume HCI used mL Average Volume HCI:SmL Average conc. of [OH] Average conc. of [Ca2 K, of Ca(OH) Molar solubility Ca(OH), Solubility of Ca(OH)_ Show calculation of K, M Show calculation of [OH] mol/L Show calculation of [Ca] g/L Show calculation of molar solubility and solubility in g/L of Ca(OH),

Explanation / Answer

1.The chemical equation=

Ca(OH)2 = Ca2+ + 2OH-

The Ksp equation=

Ca(OH)2 = [Ca2+] [OH-]2

Use titration data to determine moles of OH- in the 20 ml of sample,

Molarity of Hcl = 0.05mol/L

Volume of Hcl used in Liter = 0.00156 L (Here, taking average of three trials)

Molarity = Moles / Volume (L)

0.05 = x / 0.00156 (L)

x = 0.05 X 0.00156

x = 0.000078 mol of (OH-)

2. Determine [OH-]

Use moles of OH- and sample volume,

0.000078 mol / 0.020 (L) = 0.039 mol/L

Concentration of [OH-] is 0.0039 M

3. Determine (Ca2+)=

[Ca2+] is exactly half of [OH-] because 1:2 molar ration from equation.

so, [Ca2+] = 0.0039 / 2 = 0.0019 mol/L

Concentration of [Ca2+] is 0.0019 M

4. Ksp of Ca(OH)2 =

Ksp = [Ca2+] [OH-]2

Ksp = 0.0019 X (0.0039)2

Ksp = 0.0019 X 0.00001521

Ksp = 0.0000000288

Ksp of Ca(OH)2 is 2.88X10-8

5. Molar solubility of Ca(OH)2=

Ksp = [Ca2+] [OH-]2

2.88X10-8 = (x) (2X)2

2.88X10-8 = 4X3

2.88X10-8 / 4 = 4X3 / 4

7.22X10-9 = X3

By calculating X we get 0.00193 mol/L

Molar solubility of Ca(OH)2 is 0.00193mol/L

6. Solubility of Ca(OH)2=

Determine moles of HCL used,

Moles of HCL = 0.05 mol X 0.00156 (Average volume of HCL in L)

Moles of HCL = 0.000078 mol

Determine moles of Ca(OH)2 titrated, (every one Ca(OH)2 titrated requires 2H+)

0.000078 mol / 2 = 0.000039 mol

Convert moles of Ca(OH)2 to gram Ca(OH)2

0.000039 mol times 74.0918 g/mol

0.000039 X 74.0918 = 0.00288 g. This is gram per 20 ml

Convert to grams/Liter

0.00288(g) / 0.020(L) = 0.144 g/Liter

Solubility of Ca(OH)2 is 0.144g/L.

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