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[The Lamp]

Sorry that this is kind of a specific thread, but I really wanted help with this kind of problem and our professor never showed us how to do it, and this other girl who thought she knew how to do it got it wrong, and it's guaranteed to be on the test in some way/shape/form.

It's a buffer calculation.

Calculate the pH of a buffer solution composed of 1.0L of 0.50M formic acid (HCO2H) and 0.60M sodium formate (NaHCO2) before and after adding 0.01 mol NaOH (solid). (Ka = 1.0 x 10^-5).

pH before?

pH after?

I know how to calculate pH but he never explained what to do if we change concentrations of things or add stuff with buffer solutions.

If anyone could help me, that would be wonderful! Thank you!

 

[B-Dex]

Do your own homework!

 

[Boogiepop]

I've often contemplated asking GAF to do my particularly frustrating hw, only to realize it probably wouldn't end well. Good luck to you.

 

[jmdajr]

Gaf doing your homework?

For real though I wouldn't know....

 

[Lebron]

I'm zoning out right now, but I figure you have to:

First half plug into ice table and find your concentration and pH without factoring in the NaOH. Then take the concentrations and convert them into moles(using the 1L) and plug it into a BRA table. You'll subtract the NaOH from the moles you find of your acid and conjugate base, then reconvert back into Molarity(your buffer). Then redo the problem in a ice table to get the pH after.


Correct me if I'm wrong obviously.

 

[Al-ibn Kermit]

I forgot all of that general chem stuff but I do remember answers.yahoo.com being very helpful for learning how to use those formulas.

Is this specific question assigned out of a textbook? If it is then you can go to cramster.com and get a step-by-step explanation.

 

[The Lamp]

The class doesn't really use a textbook, there is no homework, and the professor didn't cover this. No one else I've talked to got this problem correct so I'm sorry, lol, I was just really wanting to know how to do this kind of problem.

There's a Math help thread...there should be a physics/chemistry/biology one too I think...haha.

At the above post, thanks, let me see how that works. I don't know what a BRA table is, though...And don't you mean add instead of subtract the NaOH, since I'm trying to find the pH with NaOH added?

EDIT: Actually when do I use ICE tables? I thought I just use the Henderson-Hasselbach equation, pH = -log(Ka) + log(conj. base/acid)?

I forgot all of that general chem stuff but I do remember answers.yahoo.com being very helpful for learning how to use those formulas.

Is this specific question assigned out of a textbook? If it is then you can go to cramster.com and get a step-by-step explanation.

I think it's hand-written by the professor. But I didn't know about cramster, I'll try to look at that.

 

[Deleted member 81567]

Did you try yahoo answers? They basically answer everything like gangstas.

 

[Lebron]

Hmm, a BRA table is basically an ICE table only you're using moles instead of Molarity. Also, I you subtract the NaOH since you're adding it and then using it up(think Titration). It should be the lowest number in regards to moles, also, you can't use your buffer calculation method if you're CB or A are all used up. Your pH should increase though, obviously.

Does he give you the answer?

 

[The Lamp]

Hmm, a BRA table is basically an ICE table only you're using moles instead of Molarity. Also, I you subtract the NaOH since you're adding it and then using it up(think Titration). It should be the lowest number in regards to moles, also, you can't use your buffer calculation method if you're CB or A are all used up. Your pH should increase though, obviously.

Does he give you the answer?

No he doesn't give an answer, it's just counted wrong.
Tried googling this question for yahoo! answers and nothing comes up.

 

[Lebron]

Is this on Mastering Chem?

 

[Deleted member 81567]

Tried googling this question for yahoo! answers and nothing comes up.

I meant ask it manually. You can get a reply in a few minutes.

 

[dvolovets]

I can help you, just give me a sec to work out the problem...

For the pH before you add the solid, use the Henderson-Hasselbalch equation because you have a buffer solution.

pH = pKa + log([base]/[acid])
pH = -log (1*10^-5) + log(0.6 / 0.5)
pH = 5 + 0.079181246
pH = 5.08

For the pH after you add the solid, make an ICE table with the following reaction.

HCO2H + NaOH --> NaHCO2 + H2O
I: 0.5 mols 0.01 mols 0.6 mols
C: -0.01 mols -0.01 mols +0.01 mols
E: 0.49 mols 0 0.61 mols

You do not need to solve for x because NaOH dissociates completely and is thus used up completely assuming the other reactant is in excess. Now do the Henderson-Hasselbalch equation again with the new concentrations (which are the same numbers as the mols because you have 1 liter of solution).

pH = pKa + log([base]/[acid])
pH = -log (1*10^-5) + log(0.61 / 0.49)
pH = 5 + 0.095133755
pH = 5.10

I am less sure on the second problem than on the first, but the answer makes sense because pH goes up with the addition of base. If I actually do sig figs correctly, then both answers are 5.1. Haha. That also makes sense because you're not adding a lot of NaOH, and the whole point of a buffer solution is to keep pH relatively constant.

 

[djkimothy]

Lol. I have an M.Sc. In chemistry and I don't even remember how to do that.

 

[The Lamp]

Is this on Mastering Chem?

If this is an online homework thing, no, because the class doesn't have homework and this professor does everything completely different from the other chemistry profs.

 

[Escape Goat]

Remember kids. The pH of water is 0.

 

[djkimothy]

Remember kids. The pH of water is 0.

Lol <avatar quote>.

I think we'd be dead.

 

[Lebron]

If this is an online homework thing, no, because the class doesn't have homework and this professor does everything completely different from the other chemistry profs.

That sucks. Kind of stupid not to assign homework problems


Well if dvolovets isn't able to answer your question then I'll take a whack at it.

 

[dvolovets]

Just did both problems. Check out my post above, OP. I hope I actually did them right, haha. This is good practice for my chem final!

 

[The Lamp]

I just looked over it. I think that's right (though I'll never be sure) but it looks logical.

I just want to clarify with your ICE table....

Initially I have 0.5 mols of HCO2H and I add 0.01 mols of NaOH, which then -----> NaHCO2, of which there is 0.6 mols, and H20 which is not important knowing the concentration of?

Then as the reaction takes place, the only thing that takes place is the NaOH dissociating (I'm not sure why that's the only thing that changes, though)? So -0.01 mols with everything on each side of the equation, and that gives me my final concentrations?

I suppose I just don't understand the behavior of what's happening with the NaOH. You're just adding it in and nothing is happening to the other products/reactants, it's just dissociating (and therefore -0.01)?

Other than that, looks great

 

[Lebron]

Think titration. You're adding NaOH then using it up. Your buffer remains(in this case A and CB), if it's before the EQ point, which this is(moles aren't equal). That's why you have to make sure you subtract it out of your calculations and make sure your buffer remains.

 

[The Lamp]

Think titration. You're adding NaOH then using it up. Your buffer remains(in this case A and CB), if it's before the EQ point, which this is(moles aren't equal).

I see. So when I'm at equilibrium, the moles will be the same on both sides?

But for the H20, why don't I add 0.01 moles of NaOH to it on the right side of the ICE table as well? I suppose H20 is completely ignored in this problem but I'm not sure why.

 

[dvolovets]

I just looked over it. I think that's right (though I'll never be sure) but it looks logical.

I just want to clarify with your ICE table....

Initially I have 0.5 mols of HCO2H and I add 0.01 mols of NaOH, which then -----> NaHCO2, of which there is 0.6 mols, and H20 which is not important knowing the concentration of?

Then as the reaction takes place, the only thing that takes place is the NaOH dissociating (I'm not sure why that's the only thing that changes, though)? So -0.01 mols with everything on each side of the equation, and that gives me my final concentrations?

I suppose I just don't understand the behavior of what's happening with the NaOH. You're just adding it in and nothing is happening to the other products/reactants, it's just dissociating (and therefore -0.01)?

Other than that, looks great

I'm defining "initially" as the conditions immediately after adding NaOH. So, you have 0.5 mols of HCO2H, 0.01 mols of NaOH, and 0.6 mols of NaHCO2. You're right, H2O is not important in ICE tables.

Normally with ICE tables, you're dealing with weak acids/bases where the dissociation is small. In this case, your "change" line is going to be "-nx" for reactants, where n is the stoichiometric coefficint", and "+nx" for products. Looking at this problem, you know that NaOH is a strong acid. You also know that strong acids dissociate completely. Thus, it is going to be completely used up in the reaction. The key point here is that your other reactant is in excess. In other words, when the reaction is done, you still have some HCO2H that has not reacted. If you had, say, 0.6 mols of NaOH, all of the HCO2H would be used up and you'd have some amount of base and 0.1 mols of NaOH because your limiting reagent is HCO2H (i.e., it can't proceed past 0.5 mols because that's all there is in the first place).

In those kind of ICE tables (where you have a strong base/acid and a weak base/acid left over), you then add the STRONG base to water and do another ICE table because it will be the primary determinant of pH. In this ICE table, all the added base is used up, leaving you with less acid (because that's what reacted with the base) and more base (because that's what formed).