Question

Double checking my calculations

Answer 1

I made 100 mL of a .03 M H3AsO3 solution. If I then take 5 mL of that solution and add it to 95 mL of another solution, what is the final concentration of the H3AsO3?

I said: (.03)(.005)=(.1)(?), which equals .0015M. Is that right?

@Kainui

Answer 2

That is, what is the final concentration of the H3AsO3 now that it is part of a new 100 mL soln?

Answer 3

Yeah perfectly correct.

Answer 4

Oh good. Thanks!

Answer 5

The main concept is that when you are diluting something, the total number of molecules of H3AsO3 remains the same.

So when you look at your equation:

\[M_1 V_1 = M_2 V_2\]

since Molarity is particles per volume, really what we're saying is:
\[\frac{n}{V_1}= M_1\]and
\[\frac{n}{V_2}= M_2\]
So you can derive it from the definition of Molarity and keeping total number of particles the same... Hopefully this helps your doubts a little bit in the future.

Answer 6

I wish I would have had this a couple days ago! Thanks!!!