November 28, 2021, 07:12:32 AM
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Topic: How do I prepare solution that is mixture of 5% NaCl and 99.9% Acetic Acid?  (Read 619 times)

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Offline allbright1747

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How do I prepare a solution that is a mixture of 5% NaCl and 99.9% acetic acid so that the final volume of mixture is 2.5 liters and final pH is 3.2?

Offline Babcock_Hall

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It is a forum rule that you must show us your work before we can help you.  Let me ask for some clarification.  I would interpret 5% NaCl to mean its final strength.  I would also interpret this to be an aqueous solution that was prepared from some volume of concentrated acetic acid and some volume of water.  Is that your understanding?

Offline allbright1747

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Yes, final strength is 5% NaCl and it is an aqueous solution prepared from concentrated acetic acid, water and NaCl. So far I have converted percent into molar. I calculated that the 99.9% concentrated acetic acid is 17.34 M and the 5% NaCl solution to be 0.855 M. Right now I'm having difficulty figuring out how to calculate the volume I would need of both the 17.34 M acetic acid and 0.855 M NaCl solution so that if I mixed them together I would end up with 2.5 L of a 0.04 M solution. My plan was to make out the NaCl solution first, which I know how to do I just don't know how much of it to make up, and then add the concentrated acetic acid to this until mixture reached 2.5 L.

Offline Borek

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I think you are misreading the question. You can make solution that is either 99.9% acetic acid or a solution that has pH of 3.2, it can't be both at the same time.

Not to mention the fact that solution can't be both 5% in NaCl and 99.9% in acetic acid, fractions must sum up to 100%.
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Offline Babcock_Hall

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@OP, I interpret the question to mean that the final concentration of NaCl is 5%.  I also interpret the question to mean that 99.9% acetic acid is something that one starts with.  One must use the final pH to decide how much acetic acid to use.

Offline allbright1747

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Yes, you're right I did misread it. It is asking to make up a 5% NaCl solution then adding a 0.04 M acetic acid solution to it.

Offline allbright1747

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alright thanks guys that clarifies everything :)

Offline Babcock_Hall

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There is one complication, namely that the apparent pKa of acetic acid is a function of ionic strength.  That may not be a concern, however.

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