Question

Mr. Whiteman has two stock solutions of sulfuric acid (H2SO4) in the supply closet: 12.1 M and 6.0 M. If he wants to make 10.0 mL of a 1.0 M solution, which stock solution makes more sense to use? Justify with two calculations.

Answer 1

12.1-M requires less volume of stock solution.

Step-by-step explanation:

Hello there!

In this case, according to the foundations of dilution processes, it turns out convenient to remember that the moles of the solute before and after the dilution remains the same; for that reason we use the following equation:

`M_1V_1=M_2V_2`

Relating the initial and final molarities and volumes. Now, it is seen that V2 is 10.0 mL and M2 is 1.0 M, which means that if we solve for V1 for 12.1 M and 6.0 M, we obtain the following volumes:

`V_1=(M_2V_2)/(M_1) \\\\V_1^(12.1M)=(1.0M*10.0 mL)/(12.1M)=0.826mL\\\\V_1^(6.0M)=(1.0M*10.0 mL)/(6.0M)=1.67mL`

Therefore, since the 12.1-M solution requires less volume of stock solution, it makes sense to use this one as the initial solution.

Best regards!