a. A determination of the molar solubility and the K_sp calcium hydroxide was co

Question

a. 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. b. 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. a. What is the average K_sp of Ca(OH)_2? b. What are the standard deviation and the relative standard deviation (%RSD) for the K_sp of Ca(OH)_2?

Explanation / Answer

Q5.

Volume added = Vfinal - Vinitial = 13.90-1.79 = 12.11 mL added

Q6.

mol of HCl = MV = (0.048 mol/L ) ( 12.1 mL ) = 0.5808 mmol = 0.5808*10^-3 mol

Q7.

moles of OH- in saturated solution, due to stoichiometry, must be equal to mol of HCl

so

OH- = 0.5808*10^-3 mol of OH

Q8

OH- in equilibiriumn

[OH-] = mol of OH- /t otal V = (0.5808*10^-3) / (25*10^-3) = 0.023232 M of OH-

Q9

Ca+2 in equilibrium is equal to the HALF of those of OH- due to 1:2 ratio

[Ca+2] = 1/2*[OH-] = 1/2*0.023232 = 0.011616 M

Q10

Molar solubility:

S = Ca(OH)2 = [Ca+2] = 0.011616 M

so..

Q11

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

Ksp = (0.011616)(0.023232^2) = 0.00000626945

Ksp = 6.269 *10^-6

B)

average Ksp = ((5.2*10^-6) + (4.8*10^-6) + (6.269 *10^-6) ) / 3= 5.43*10^-6

C)

std. dev --> 0.76*10^-6 (from data sheet)

% dev --> 0.76/5.43 *100 = 14%

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