Question

Suppose 3.50 g of solid Mg(OH)2 is added to 30.0 mL of 0.500 M H2SO4 solution. What
will the concentration of Mg2+ be when all of the acid has been neutralized? How many
grams of Mg(OH)2 will not have dissolved?

Please explain work!

Answer 1

Mg(OH)₂ + H₂SO₄ --> Mg(SO₄) + H₂O

(a) since 1 mol H2SO4 --> 1 mol Mg+2 (MgSO4)
[Mg+2] = [H2SO4] (Molarity)

So, the concentration when all the acid has been neutralized is: 0.500 M

(b) How many grams of Mg(OH)2 will not have dissolved?

Let's find the moles of H2SO4, remember convert ml to liters:
0.0300 L * 0.500 M = 0.015 moles H2SO4

Now, we know the moles of H2SO4 are equal to the moles of Mg(OH)2
0.015 moles H2SO4 reacts with 0.015 moles Mg(OH)₂

Now we have to find the grams of Mg(OH)2 that are consumed in the reaction using the molar mass:
0.015 moles Mg(OH)2 * 58.33 g/mol = 0.875 grams Mg(OH)2 (this is what reacts)

3.50 g Mg(OH)2 - 0.875 g = 2.63 grams Mg(OH)2 (this is what remains)

That the amount of Mg(OH)2 that will not have dissolved because it didn't react.