To what volume should 222 mL of a 12.47 M solution of nitric acid be diluted to, to prepare a 2.58 M solution?

Question

I just need to know the volume of this question, please help

Answer 1

V1M1 = V2M2

(222 ml)(12.47 M) = (x ml)(2.58 M)

x = 1073 ml final volume = 1070 ml (to 3 sig. figs.)

Answer 2

It's important to recognize right off the bat that the starting concentration (molarity) is about four times HIGHER than the ultimate concentration ... this will help in reviewing the answer at the very end ("does this make sense?").

Solution (no pun intended): 12.47 M nitric acid is the same as 12.47 moles/1 liter of water. It can also be depicted as follows:

12.47 moles

1000 mL water

1. First, let's calculate the amount of nitric acid in 222 mL of the above nitric acid solution:

12.47 moles nitric acid X 222 mL = 2.77 moles HNO₃ (aka nitric acid)

1000 mL water

1. Now, let's calculate what total volume of water would be required to convert the 2.77 moles nitric acid into a 2.58 M solution:

2.77 moles nitric acid = 2.58 moles nitric acid (Remember: molarity is simply moles/1000 mL H₂O)

X[total] mL water 1000 mL water

1. Next, let's solve for X(total) from the above equation:

(2.58)(X[total]) = (2.77)(1000)

X[total] = 2770 ÷ 258

X[total] = 1074 mL

1. Finally, let's answer the original question: "To what volume should 222 mL of .... be diluted to, to prepare a 2.58 M solution." Recognizing that X[total] is the sum of the original 222 mL of the starting solution plus some amount of water, we determine that:

X (added water) = 1074 mL - 222 mL, or 852 mL.

The final volume of this new, less concentrated solution would be 1074 mL, and it would require the addition of 852 mL of pure water to make up such a solution.