What volume of 12.0 m hcl (aq) is required to make 75.0 ml of 3.50m hcl (aq) ?

We need an equation that would relate the concentration of the original solution to that of the desired solution. To solve this we use the equation expressed as follows,

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

M1V1 = M2V2

12.0 M x V1 = 3.50 M x 75.0 mL

V1 = 21.88 mL

Therefore, we need about 21.88 mL of the 12.0 M of hydrochloric acid solution to make 75.0 mL of the 3.50 M hydrochloric acid solution.