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Concentration/dilution problem

by Benzoate
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Benzoate
#1
Sep4-07, 04:21 PM
P: 569
1. The problem statement, all variables and given/known data

What is [H1+] in a solution prepared by mixing the following solutions

* 21.3 mL of 6.40 M HCl
* 30.8 mL of 5.20 M HCl
* 15.8 mL of 7.50 M HCl
* 34.6 mL of water


2. Relevant equations


C(final)= C(initial) * V(initial)/V(final). I think V(final) will be the volume of H2O for all 3 concentrations

3. The attempt at a solution
C(final,1)= 7.50 M*(15.8 mL/34.6mL) = 3.42 M
C(final,2) = (5.20 M) * (21.3 mL/34.6mL) = 3.20 M
C(final,3)= 6.40 M * (30.8 mL/34.6mL)= 5.70 M

C(final,total)= 5.70 M + 3.20 M + 3.42 M = 12.32

I don't think I calculated the concentration of [H+] correctly.
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chemisttree
#2
Sep4-07, 05:30 PM
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Quote Quote by Benzoate View Post
...I think V(final) will be the volume of H2O for all 3 concentrations.

You need to correct this before you go any further. There are four liquids given in your problem. Assume no change of volume due to mixing. For example, 54 mL and 65 mL of two solutions combined will produce 119 mL of solution.

Next you need to find the number of moles of acid added from each different solution.

Express the concentration in terms of M (moles/liter).
Benzoate
#3
Sep4-07, 06:29 PM
P: 569
Quote Quote by chemisttree View Post
You need to correct this before you go any further. There are four liquids given in your problem. Assume no change of volume due to mixing. For example, 54 mL and 65 mL of two solutions combined will produce 119 mL of solution.

Next you need to find the number of moles of acid added from each different solution.

Express the concentration in terms of M (moles/liter).
so the final volume is going to be the same for all 3 concentrations?

here is my attempted solution:

C(final ,1) =C(initial, 1) * (V(initial, 1)/V(final)

C(final ,2) =C(initial, 2) * (V(initial, 2)/V(final)

C(final ,3) =C(initial, 3) * (V(initial, 1)/V(final)

C(final , total) =C(final ,1) + C(final ,2) + C(final ,3)

Benzoate
#4
Sep4-07, 06:36 PM
P: 569
Concentration/dilution problem

Quote Quote by chemisttree View Post
You need to correct this before you go any further. There are four liquids given in your problem. Assume no change of volume due to mixing. For example, 54 mL and 65 mL of two solutions combined will produce 119 mL of solution.

Next you need to find the number of moles of acid added from each different solution.

Express the concentration in terms of M (moles/liter).
Quote Quote by Benzoate View Post
so the final volume is going to be the same for all 3 concentrations?

here is my attempted solution:

C(final ,1) =C(initial, 1) * (V(initial, 1)/V(final)

C(final ,2) =C(initial, 2) * (V(initial, 2)/V(final)

C(final ,3) =C(initial, 3) * (V(initial, 1)/V(final)

C(final , total) =C(final ,1) + C(final ,2) + C(final ,3)
for the concentration of H2O , what do you assumed about the initial concentration of H2O ?
chemisttree
#5
Sep4-07, 11:44 PM
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Another way to think of it would be to find the number of moles of HCl contributed by each of the four solutions.

#moles = concentration(moles/liter) X Volume (liters)

Simply do this math for each of the solutions and sum the moles of HCl. Add the volume for each of the solutions to determine Vfinal.

Remember, you are combining all of these solutions into one final solution. With that in mind, what significance does "C(final ,3)" or "C(final ,2)" or "C(final ,1)" have? Why are you performing this calculation? It doesn't make sense.

and this,
C(final , total) =C(final ,1) + C(final ,2) + C(final ,3)
is not correct. I have never seen any expression like this... anywhere. Think about it. If I were to take a 100 mL solution of 3.5M HCl and split it into 20, 5 mL solutions of 3.5M HCl, would recombining them give me 100 mL of (20 X 3.5M HCl)? Or would it only give me back the original 100 mL of 3.5M HCl?
atavistic
#6
Sep5-07, 09:52 AM
P: 106
Is the answer .13632g correct for part a)


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