Solving Molarity Problem: Ba(ClO4)2 from 0.800M HClO & 0.200M Ba(OH)2

[23scadoo]

Here is a question that i have on a test review and neither the book or the teachers examples are any help.

If 200mL of 0.800 molar HClO is added to 400mL of 0.200 molar Ba(OH)2 solution, the resulting solution will be ________M in Ba(ClO4)2.

I know that the answer is 0.133M

i know that the balanced equation is:
2HClO4 + Ba(OH)2 ----> Ba(ClO4)2 + 2H2O

The steps that she has given us to do are:
1. balance equation
2. reaction ration
3. Start
4. Change
5. After reaction
6. Final Volume of Solution
7. Molarity

If anyone could explain how this problem is worked i would be greatly appreciative.

 

[Hootenanny]

Start by working out the new concentrations of $HClO_{4}$ and $Ba(OH)_{2}$ in the mixed solution.

 

[23scadoo]

you mean like this:

200mL HClO4 x (1L/1000mL) x (0.800/1L) = 0.16M HClO4
&
400mL Ba(OH)2 x (1L/1000mL) x (0.200/1L) = 0.08M Ba(OH)2

then what?

 

[Borek]

No. What is the final volume of the solution?Borek

 

[23scadoo]

final volume of the solution is 600mL

 

[Hootenanny]

How many moles of $HClO_{4}$ and $Ba(OH)_{2}$ did you start with? HINT: This will be the same after the solutions are mixed.

-Hoot