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Topic: Acid concentration after salt  (Read 3676 times)

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Offline Balancegenerally

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Acid concentration after salt
« on: April 02, 2017, 06:29:53 AM »
So I have a quick question about how concentration is calculated. I have a H2SO4 solution that needs to have a final concentration of .5M. Before the concentration measurement it is being used to devolve various metals and so it will have the salt from those reactions still in solution. When calculating how much H2SO4 I need to add at the start would I need to add extra to account for the SO4 2- used up in the reaction?

Offline Arkcon

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Re: Acid concentration after salt
« Reply #1 on: April 02, 2017, 09:36:06 AM »
OK.  This one could be simple, or it could be tough to follow.  Let's work together, point by point.

So I have a quick question about how concentration is calculated. I have a H2SO4 solution that needs to have a final concentration of .5M.

This is easy to follow, and easy to work with.  Its not your real question, but I just wanted to point out -- it makes the problem look easy.

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Before the concentration measurement it is being used to devolve various metals and so it will have the salt from those reactions still in solution.

OK.  This easy to understand, but hard to calculate.  You have "some" acid, you dissolve "some" metal, and want to know what's left.

Here's two problems for you:

You have 20 dimes, and 15 nickels.  How much money do you have?  You take away all the nickels, how much money do you now have.

or

You have a bunch of coins, you replace some with some others. Take some away, I dunno, put some others in there. There's some pennies in there too, I saw 'em.  How much money do you have?

The second question is a bad one to ask.  So lets avoid questions like that one.

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When calculating how much H2SO4 I need to add at the start would I need to add extra to account for the SO4 2- used up in the reaction?

That is one way to know what's left, but not the best way I think.  Consider:  how much SO4 did you start with?  How much is left after the metal reacts?  Draw the chemical equation.  This is a trick question.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Balancegenerally

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Re: Acid concentration after salt
« Reply #2 on: April 02, 2017, 09:52:34 AM »
Ok, so I have the lab data for the 45 elements being dissolved by the acid. In order to form the salts it will take ~4.5 mol of H2SO4. Specifically I was told to have an exiting acid concentration of .5M. the amount of SO4 ions in solution after the reaction would include the salt but if I only go based of that I would end up with ~3.9 mols h2so4... not enough to form all the salts.

Offline Arkcon

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Re: Acid concentration after salt
« Reply #3 on: April 02, 2017, 10:18:51 AM »
Please write a generic balanced chemical reaction for dissolving metal in sulfuric acid.  Use M as the elemental symbol for all metals.  See how much sulfate is on both sides of the reaction.  You will see why it appears not to work.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Balancegenerally

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Re: Acid concentration after salt
« Reply #4 on: April 02, 2017, 10:24:54 AM »
Please write a generic balanced chemical reaction for dissolving metal in sulfuric acid.  Use M as the elemental symbol for all metals.  See how much sulfate is on both sides of the reaction.  You will see why it appears not to work.

2M+3(H2SO4) :rarrow: 3H2(g)+2M3-(aq)+3SO42+(aq)

so if they are specifying the acid concentration could they actually be specifying the concentration of H

Offline Arkcon

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Re: Acid concentration after salt
« Reply #5 on: April 02, 2017, 04:12:32 PM »
so if they are specifying the acid concentration could they actually be specifying the concentration of H

Very good, so you realize now that sulfate won't change, its either "attached" to H+ or to M+ - well, not really "attached", sort of floating free.

Now, to your question:how much acid did you start with?  How much metal did you add?  How much acid is left over?
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Balancegenerally

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Re: Acid concentration after salt
« Reply #6 on: April 09, 2017, 12:40:27 PM »
Now, to your question:how much acid did you start with?  How much metal did you add?  How much acid is left over?

Sorry for the delayed response. This is for a sr. design project so I am trying to solve for how much acid I need to add. I added enough metal to require 4.152750919 mol H2SO4 It is my understanding that when the acid reacts with the metal it will produce H2 (g) and so if I need an exiting solution concentration of .5M I will need to add 793.4018622 lbs H2SO4 and 7662.86816 lbs H2O as 407.2976574 lbs of SO42+ will be 'consumed' by the various metals. what remains should give me my .5M solution.

Edit: Im also planning on using selective precipitation of the salts by using the common Ion effect later, how would this additional SO4 influence that?
« Last Edit: April 09, 2017, 12:58:54 PM by Balancegenerally »

Offline Arkcon

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Re: Acid concentration after salt
« Reply #7 on: April 12, 2017, 06:18:40 AM »
And I'm sorry to take so long too, but I have work of my own to do, and you've dumped a simple, but highly involved problem into my lap.  You have a highly advanced engineering problem, but you lack the very basics of high school chemistry, and you need to use some of that, to solve this problem. 

Some points to consider:

Don't abuse significant figures.  You are using Google calculate, or some scientific calculator, robotically, and that is wrong.  And it makes you look bad -- lazy, or uninformed.  And that's bad for the people reviewing your work to see.  The weight -- 407.2976574 lbs, what does that mean in real weights?  What does 0.00006 lbs mean?  That is 0.02 grams, a raisin weighs less than 1 gram -- you're going to put hundreds of pounds on a balance, plus or minus the shavings of a raisin.  That's impossible to do.  So fix that.

You switch from moles to lbs too seamlessly.  I don't have time to convert.  Show the entire thing, theoretically, in moles first, then convert to lbs at the end.

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Edit: Im also planning on using selective precipitation of the salts by using the common Ion effect later, how would this additional SO4 influence that?

I don't know what  you mean here.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

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