Powdered MgO was added to a beaker filled with deionized water. The mixture was

Question

Powdered MgO was added to a beaker filled with deionized water. The mixture was covered with parafilm, and allowed to stir overnight. The mixture was filtered, and the filtrate was collected. A 100 mL sample of the solution was titrated using 0.0010 M HCI. The endpoint of the titration was reached after the addition of 22.5 mL HCI. Calculate the number of moles of HCI needed to reach the endpoint. Determine the molar solubility, and the solubility of Mg(OH)_2 in grams/100 mL. Calculate the Ksp of Mg(OH)_2.

Explanation / Answer

The overall reaction is as follow:

MgO(s) + H2O(l) --------> Mg(OH)2(s)

Now, in the titration with HCl:

Mg(OH)2 + 2HCl <----------> MgCl2 + 2H2O

Now, the moles of HCl are:

moles = 0.0010 mol/L * 0.0225 L = 2.25x10-5 moles

Thje molar solubility of Mg(OH)2 would be:

MbVb = MaVa / 2

100Mb = 0.0010 * 22.5 / 2

Mb = 0.0010 * (22.5/2*100)

Mb = 1.125x10-4 M

Now, if we take the soolubility in g/100 mL then:

S = 1.125x10-4 mol/L * 58.32 g/mol = 6.561x10-3 g/L * 1 L / 1000 mL * 1000 mL / 100 mL = 6.561x10-5 g/100 mL

The Ksp finally would be:

Mg(OH)2(s) ---------< Mg2+ + 2OH-

Ksp = [Mg2+] [OH-]2

Ksp = s * (2s)2 = 4s3

Ksp = 4*(1.125x10-4)3

Ksp = 5.69x10-12

Hope this helps

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