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

In summary, the conversation is about a question on a test review regarding the resulting solution when 200mL of 0.800 molar HClO is added to 400mL of 0.200 molar Ba(OH)2 solution. The answer is 0.133M Ba(ClO4)2 and the steps given are to balance the equation, calculate the reaction ratio, and determine the final volume and molarity of the solution. The final volume of the solution is 600mL and the moles of HClO4 and Ba(OH)2 are the same before and after mixing.
  • #1
23scadoo
4
0
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.
 
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  • #2
Start by working out the new concentrations of [itex]HClO_{4}[/itex] and [itex]Ba(OH)_{2}[/itex] in the mixed solution.
 
  • #3
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?
 
  • #4
No. What is the final volume of the solution?Borek
 
Last edited by a moderator:
  • #5
final volume of the solution is 600mL
 
  • #6
How many moles of [itex]HClO_{4}[/itex] and [itex]Ba(OH)_{2}[/itex] did you start with? HINT: This will be the same after the solutions are mixed.

-Hoot
 

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

1. How do I determine the molarity of Ba(ClO4)2 in this solution?

The molarity of Ba(ClO4)2 can be determined using the formula M1V1 = M2V2, where M1 and V1 are the molarity and volume of the original solution (0.800M HClO), and M2 and V2 are the molarity and volume of the diluted solution (after adding 0.200M Ba(OH)2). In this case, the molarity of Ba(ClO4)2 would be 0.800M x (0.800L/1.000L) = 0.640M.

2. What is the purpose of adding Ba(OH)2 to the solution?

Ba(OH)2 is added to the solution to neutralize the excess HClO and create Ba(ClO4)2, which is the desired compound for the molarity problem.

3. How do I calculate the volume of Ba(OH)2 needed to reach a final volume of 1.000L?

The volume of Ba(OH)2 needed can be calculated using the formula M1V1 = M2V2, where M1 and V1 are the molarity and volume of the Ba(OH)2 solution (0.200M and unknown volume), and M2 and V2 are the molarity and volume of the diluted solution (0.800M and 1.000L). In this case, the volume of Ba(OH)2 needed would be 0.800M x (1.000L/0.200M) = 4.000L.

4. Can I use any other compound instead of Ba(OH)2 to neutralize the solution?

Yes, you can use any strong base (such as NaOH or KOH) to neutralize the solution and create Ba(ClO4)2. However, make sure to adjust the calculations accordingly based on the molarity of the chosen base.

5. Is it necessary to use a volumetric flask to prepare this solution?

No, a volumetric flask is not necessary. You can use any glassware (such as a beaker or graduated cylinder) as long as you accurately measure the volumes of the solutions used in the calculation.

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