You add an excess of Ca(OH)_2 to water maintained at a particular temperature, s
ID: 974479 • Letter: Y
Question
You add an excess of Ca(OH)_2 to water maintained at a particular temperature, stir until the solution is saturated, filter, then determine the [OH^-] in the solution by titration with acid. Titration of 10.0 mL of the calcium hydroxide solution to the endpoint requires 4.86 mL of 0.070 M HCl solution. What is the molar quantity of OH in the 10.0 mL of solution? What are the concentrations of Ca^2 and OH? What is the solubility of Ca(OH)_2 under these conditions? What is the K_sp for Ca(OH)_2 under these conditions? 2) You measure the solubility of a salt at four different temperatures and calculate the following K_SP values: DeltaGdegree for the solubility process is given by the relationship DeltaGdegree = -RTInK_sp, where R = 8.314 J/(K*mol) and T is the absolute temperature in degrees K. Calculatc DeltaGdegree (in kJ/mol) at each of these temperatures. As the temperature increase, how does, the solubility of the salt change? According Le Chatelier's principal, dissolving this salt should therefore be which type of process? In other w^ords, what is the sign of DeltaHdegree ?Explanation / Answer
I will answer (for now) question 1. The other question póst it in another question thread.
a) The reaction is as follow:
Ca(OH)2 + 2HCl -------> 2H2O + CaCl2
moles HCl = 2 moles Ca(OH)2
moles HCl = 0.07 * 0.00486 = 0.00034 moles
moles Ca(OH)2 = 0.00034 / 2 = 0.00017 moles
Cb = 0.07 * 4.86 / 2*10
Cb = 0.01701 M
b) Ca(OH)2 ------------> Ca2+ + 2OH-
[Ca2+] = [Ca(OH)2] = 0.01701 M
[OH-] = 0.01701 * 2 = 0.03402 M
c) Already answer in part a as concentration in mol/L, now in g/L is:
0.01701 mol/L * 74 g/mol = 2.2986x10-4 g/L or 0.22986 mg/L (ppm)
d) Ksp = [Ca][OH]2
Ksp = (0.01701)(0.03402)2
Ksp = 1.97x10-5
Hope this helps
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