Question

I need help, I think the problem is easy, but I'm maybe too stressed to know what to do. Here it goes:

Because of the lack of reactives in a laboratory, you wish to prepare a solution of AgNO3 0.1 N, given two solutions, the first 0.0825N and the second 0.7280N. ¿What volume of each of them you'll have to mix to obtain 1lt of the solution you want?

Answer 1

this task has more math than chemistry but hey, you still have to know how to solve it, right?!
first you write down what you have:
c1=0,0825 M
c2=0,7280 M
c3=0,1 M
V3=1 dm3
now you set your equation:
c1V1+c2V2=c3V3
now you work a little magic:
V1+V2=V3 ---> V1=V3-V2
now you substitute that magic in your first equation:
c1(V3-V2) + c2V2=c3V3
and now you do a little math:
c1V3-c1V2 + c2V2=c3V3
now transfer parts with known volume on one side and unknown on other (in short some more math):
c1V3-c3V3=c1V2-c2V2
more magic:
V3(c1-c3)=V2(c1-c2)
so now V2 is:
V2= V3(c1-c3) / c1-c2
and you put in known values:
V2= 1 dm3 * (0,0825-0,1) / 0,0825 - 0,7280
and you get:
V2= 0,027 dm3
so V1 is:
V1=V3-V2
V1=0,973 dm2

ta-da! :D