Question

Please help me with 6a calculations

Experimental Procedure, Part A.6 versus Part B.2. Would you expect more or less standard 0.05 M HCI to be used to reach the methyl orange endpoint in Part B.2? Explain. A determination of the molar solubility and the K_sp calcium hydroxide was completed according to the experimental procedure. The following data were collected for Trial 1. (See Report Sheet.) Complete the table for the determinations. Record the calculated values to the correct number of significant figures. Molar Solubility and Solubility Product of Calcium Hydroxide Calculation Zone Volume of saturated Ca(OH)_2 solution (mL) Part A.6 Molar concentration of standard HCI solution (mol/L) Buret reading, initial (mL) Part A.7 Buret reading, final (mL) Volume of HCI added (mL) Moles of HCI added (mol) Show calculation. Part A.8 Moles of OH^- in saturated solution (mol) Show calculation Part A.9 [OH^-], equilibrium (mol/L) Show calculation. [Ca^2+], equilibrium (mol/L) Show calculation. Part A.10 Molar solubility of Ca(OH)_2 (mol/L) Show calculation. Part A.11 K_sp of Ca(OH)_2 Show calculation. For Trials 2 and 3, the K_sp of Ca(OH)_2 was 5.2 times 10^-6 and 4.8 times 10^-6 respectively. What is the average K_sp of Ca(OH)_2? What are the standard deviation and the relative standard deviation (% RSD) for the K_sp of Ca(OH)_2?

Answer 1

Q5.

Vadded = Vf-Vi = 13.9-1.7 = 12.2 mL

Q6.

moles = MV

moles = (0.048)(12.2*10^-3) = 0.0005856 moles of HCl

Q7.

assume H+ = OH-

so

0.0005856 mol of H+ = 0.0005856 molf of OH-

For Q8:

[OH-]eq = [OH-]added - [OH-]reacted

[OH-]reacted = 0.0005856 mol / 25 mL = 0.023424 M

[OH-]added --> from your lab practice data

Q9.

[Ca+2] = from your lab data

Q10

Molar solubility

[Ca+2] = S

[Ca+2] = (from Q9)

S = [Ca+2]

Q11

Ksp = [Ca+2][OH-]^2

so

[Ca+2] = S

[OH-] = 2S

Ksp = S*(2S)^2) = 4*S^3

Ksp = 4*(S^3) S from Q9