What volume of stock hydrochloric acid (12 mol/L) is required to make 500 mL of 0.075 mol/L acid solution?

Answer 1

Previous answer is absolutely correct, but an easier way to do these types of dilution problems is by using

M1V1 = M2V2

M1 = initial molarity = 12 M (12 mol/L)

V1 = initial volume = ?

M2 = final molarity = 0.075 (0.075 mol/L)

V2 = final volume = 0.500 L (500 ml x 1 L/1000 ml = 0.500 L)

(12M)(V1) = (0.075 m)(0.5 L)

V1 = 0.003125 L = 3.125 ml = 3.1 ml (to 2 significant figures)

Answer 2

The desired solution is 500 mL of 0.075 mol/L. From this we can get the number of moles of acid in the solution using: 0.075 * 0.5 which gives 0.0375 mol.

We need enough of the 12 mol/L solution to get 0.0375 mol which can be solved by diving 0.0375 by 12 giving 0.003125 L or 3.125 mL of the stock hydrocholoric acid.