Question

The concentration of a sample of milk of magnesia ,Mg(OH)2, was determined by titration with phosphorus acid (H3PO4) . 30ml of milk of magnesia .required 54.8ml of 0.5 phosphoric acid to neutralise it . The equation is shown below
q2H3P04 t+ 3Mg = 6H20 +Mg3(P04)2
What was the concentration for the milk of magnesia?²​

Answer 1

Answer:

1.37 M

Explanation:

moles = M * liters = 0.5 M * 0.0548 L = 0.0274 moles phosphoric acid

 q2H3P04 t+ 3Mg = 6H20 +Mg3(P04)2

As you can see, the mole conversion from phosphoric acid to magnesia is 2:3.

0.0274 moles H3P04 * 3 mole Mg/2 mole H3P04 = 0.0411 mol Mg

Molarity = moles/liters = 0.0411 mol/ 0.03 liters = 1.37 M Mg