Question

Table 1. The results of adding solid Mg(OH)2 to 10.0L of water. Total amount of Mg(OH)2 added (g) [Mg] in solution (M) [OH') in solution (M) Mass of Mg(OH)2 that does NOT dissolve (g) 0.00963 0.04815 0.09590 0.09630 0.09700 0.10000 0.15000 0.20000 Total amount of Mg(OH)2 added (mol) 1.65 x 104 8.26 x 10" 1.64 x 100 1.65 x 10 1.66 x 103 1.71 x 10 2.57 x 10 3.43 x 100 1.65x 10 8.26 x 10 1.64 x 104 1.65 x 104 1.65 x 104 1.65 x 104 1.65 x 104 1.65 x 104 3.30 x 10 1.65 x 104 3.29 x 104 3.30 x 104 3.30 x 104 3.30 x 104 3.30 x 104 3.30 x 10+ 0.0007 0.0037 0.0537 0.1037

Answer 1

a. We are given that 8.26 X 10-4 moles of Mg(OH)2 are added to 10 litres of water. So the concentraion of Mg(OH)2 in this solution is:

Moles of solute Molarity = Volume of solution in litres

8.26 x 10-4 moles Molarity = 10 L

Molarity = 8.26 x 10-5 M

So, the concentration of Mg(OH)2 is 8.26 X 10-5 M.

b. and c. Now, in order to answer parts b and c, we'll have to look at the dissociation of Mg(OH)2

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

This means that 1 mole of Mg(OH)2 dissociates into 1 mole of Mg2+ ions and 2 moles of OH- ions.

This means that after dissociation, the number of moles of Mg2+ are as follows:

1 mole of Mg2+ 8.26 X 10-moles of Mg(OH)2 * Imole of Mg(OH)2 - = 8.26 x 10-4 moles of Mg2+

and the concentraion of Mg2+ ions in the solution is:

Molarity = 8.26 x 10-4 moles -= 8.26 x 10-5M 10 L

So, the concentraion of Mg2+ is, [Mg2+ ] = 8.26 X 10-5 M

Now, coming to the concentration of OH- ions:

Since, 1 mole of Mg(OH)2 dissociates into 1 mole of Mg2+ ions and 2 moles of OH- ions.

after dissociation, the number of moles of OH- ions are as follows:

2 mole of OH- 8.26 x 10-4moles of Mg(OH)2 x 7, go)2 Imole of Mg(OH)2 - = 16.52 x 10-4 moles of OH-

and the concentraion of OH- ions in the solution is:

Molarity = 16.52 x 10-4 moles 10 L = 1.652 x 10-4 M or 16.52 x 10-5 M

So, the concentraion of OH- is, [OH- ] = 1.652 X 10-4 M