difficult buffer problem

[matt46]

We add a certain volume of .10M NH3 to 50mL of .1M HCl. Which one of these NH3 volumes will produce a buffer solution?
a. 25mL
b. 50mL
c. 75 mL
d. none of the above

I don't understand this because we don't have NH4's concentration
Any help would be greatly appreciated. THank you.

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[blastula]

Find moles of HCl = 50ml / 1000 x 0.1 M HCl = 0.05 moles HCl
Buffered sol will be near the half eq point. So 0.05 / 2 = 0.025 moles
Find vol NH3 = 0.025 moles / 0.1 M NH3 x 1000 = 25mL NH3

Think about the titration curve when doing buffer problems. Hope that helps!

 

[RSAgator]

I have a question. A buffer is an equal mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. How does that work with a strong acid like HCl mixed with a weak base? My intuition was to think 100mL of NH3 because the first 50mL would be protonated by the strong acid into NH4+ and the next 50mL would remain as NH3 leaving you with a solution of equal NH3 and NH4+ and the inert Cl- ions. Do you have the answer matt?

 

[blastula]

Yeah now that I think about it HCl would not create a buffered solution since it is a strong acid and the reaction would go to completion. Answer should be D.

My calculations above would only work for weak acids and their conjugate base.

 

[bozz]

HCL would create a buffer solution!

At the point where NH3 and its conjugate are equal! NH3 and NH4+

Think about the titration curve (half equivalence pt)

Half of NH3 would have to be converted to NH4+ ... to get get NH3 = NH4+
think about it

do the math now... where do the H+ ions come from? HCl

If you're having trouble with this stuff, I strongly suggest you take a look at Berkeley Review's chem books... I never understood acid/base chem... within 2 days, I totally understood it... try these passages:

http://www.berkeley-review.com/TBR/bookdetails/passages/genchem1frame.html

 

[RSAgator]

HCL would create a buffer solution!

At the point where NH3 and its conjugate are equal!

So it would be at 100mL like I said?

 

[blastula]

HCL would create a buffer solution!

At the point where NH3 and its conjugate are equal! NH3 and NH4+

Think about the titration curve (half equivalence pt)

Half of NH3 would have to be converted to NH4+ ... to get get NH3 = NH4+
think about it

do the math now... where do the H+ ions come from? HCl

If you're having trouble with this stuff, I strongly suggest you take a look at Berkeley Review's chem books... I never understood acid/base chem... within 2 days, I totally understood it... try these passages:

http://www.berkeley-review.com/TBR/bookdetails/passages/genchem1frame.html

HCl will not create a buffer. Buffer solutions consist of a weak acid and its conjugate base (more common) or a weak base and its conjugate acid (less common). The resistive action is the result of the equilibrium between the weak acid and its conjugate base. The reaction of NH3 with HCl will not create an equilibrium. It will go to completion therefore no buffer. No matter how much NH3 you add HCl will not change.

 

[Kaustikos]

HCl will not create a buffer. Buffer solutions consist of a weak acid and its conjugate base (more common) or a weak base and its conjugate acid (less common). The resistive action is the result of the equilibrium between the weak acid and its conjugate base. The reaction of NH3 with HCl will not create an equilibrium. It will go to completion therefore no buffer. No matter how much NH3 you add HCl will not change.

You're saying you can't make a buffer solution with NH3? That makes absolutely no sense at all. The question asks how much NH3 to make a buffer solution in HCl. You're implying that if you add, say 10 gallons of NH3, you won't come close to buffering the solution? That would imply taht NH3 fails as a buffering solution, which goes against everything NH3 stands for (poor guy).

 

[blastula]

You're saying you can't make a buffer solution with NH3? That makes absolutely no sense at all. The question asks how much NH3 to make a buffer solution in HCl. You're implying that if you add, say 10 gallons of NH3, you won't come close to buffering the solution? That would imply taht NH3 fails as a buffering solution, which goes against everything NH3 stands for (poor guy).

If you have to add 10 gallons it would not be a buffer. How many drops of 0.1 NH3 would that be? Look up the definition of a buffer. That should end your confusion.

 

[iA-MD2013]

If you have to add 10 gallons it would not be a buffer. How many drops of 0.1 NH3 would that be? Look up the definition of a buffer. That should end your confusion.

i think you should look it up...you can make a buffer: it would be NH3 and the cojugate acid NH4+...

 

[fileserver]

Answer will either be c or d. I am leaning toward c bc a buffer does not need their acid and cb to be equal.

 

[RSAgator]

i think his point is if you add a lot of NH3 (10 gallons) you wouldn't have a buffer, you'd simply have a very basic solution and one that is not necessarily very resistant to change. Can someone actually solve the initial problem, i'm curious myself.

 

[RSAgator]

Answer will either be c or d. I am leaning toward c bc a buffer does not need their acid and cb to be equal.

possible. the whole point of a buffer is to be resistant to change. 1:1 is the point at which a buffer is most resistant to change (most resistant on either side). After that it may be a matter of semantics. I suppose a 3:2 ratio would still technically be a buffer. I think c or d

 

[iA-MD2013]

I'm like 95% sure its A....but not completely...
So, the equivalence pt (where moles NH3=mols H+) would occur when 50 mL NH3 is added. At this point, the pH of the solution would be slightly below 7. But, we're not looking for that, we're looking for the half-equivalence pt (in which conc of NH3= conc of NH4+). The half equivalence pt is half the volume of NH3 needed in the equivalent pt (that seems like a stupid sentence ). That would be 25 mL. So, a really good buffer would form if 50 mL HCl were added to 25 mL NH3.
Somebody please correct me if I'm wrong. It looks like Blastula also got 25 mL using the same calculations...so, I hope its right.

 

[fileserver]

When you add 25 mL of weak base into 50 mL of strong acid, all of the base will convert to its c.a. There is no base left to act as a buffer.

Btw, the ratio is 2 to 1,;p .

 

[RSAgator]

When you add 25 mL of weak base into 50 mL of strong acid, all of the base will convert to its c.a. There is no base left to act as a buffer.

Btw, the ratio is 2 to 1,;p .

you're right =(

 

[iA-MD2013]

ohhh wow...i read it wrong
so would it be C? its annoying cuz it would produce a buffer solution, but not the best one (100 ml).

 

[Kaustikos]

When you add 25 mL of weak base into 50 mL of strong acid, all of the base will convert to its c.a. There is no base left to act as a buffer.

Btw, the ratio is 2 to 1,;p .

It depends on the concentration as well. Buffers resist great changes to pH, so what I said earlier could exist but it's far too dramatic, as blastula mentioned. The idea is that we would have a lot of the NH3 left over after reacting with the HCl, or that's what we want, in order to create a "buffer".

I'm saying C.

 

[bozz]

HCl will not create a buffer. Buffer solutions consist of a weak acid and its conjugate base (more common) or a weak base and its conjugate acid (less common). The resistive action is the result of the equilibrium between the weak acid and its conjugate base. The reaction of NH3 with HCl will not create an equilibrium. It will go to completion therefore no buffer. No matter how much NH3 you add HCl will not change.

Kaplan and other books tend to not be so helpful and sometimes make you think incorrectly... and only teach you to regurgitate, "A buffer solution is a combination of a weak acid and its conjugate base" without actually thinking about how the process works. Buffers are ALMOST ALWAYS made by using a strong/weak combo

It's absolutely necessary to do so in order to have a 50:50 mixture of a compound and its conjugate

Why?

You need a strong acid (HCl) in order to readily donate H+ ions to NH3... NH3/NH4+ won't magically appear out of nowhere
Think about titration curves. Combination of strong/weak. Weak + Weak = useless.

In fact, most of the buffer problems I have done involved HCl reacting with a weak base (NH3 for example) in order to create a buffer solution.

2 ways to create buffers:

1) already have a 50:50 solution of something and its conjugate

2) React a weak base with a STRONG ACID OR react a weak acid with a STRONG BASE

 

[bozz]

Now OP... for practice, answer this question

If you get this question, you understand buffers completely.

Use this passage: http://www.berkeley-review.com/TBR/bookdetails/passages/genchem1frame.html

To make a buffer at pH = 10.83, which of the following should be mixed?

1. One-half equivalent of NaOH with one equivalent of H2CO3.
   [*]One and one-half equivalents of NaOH with one equivalent of H2CO3.
   [*]One and one-half equivalents of NaOH with one equivalent of H3PO4.
   [*]Two and one-half equivalents of NaOH with one equivalent of H3PO4.

Now before you start going, "there's something wrong with the question! wtf is NaOH doing? It can't be! THIS IS MADNESS"... think about what is happening

 

[fileserver]

I would guess 2.

Anyway, you add NH3 to the 50 mL of acid. Not acid to 50mL NH3. (I think this is why you gave the other answer, bozz)

 

[RSAgator]

You need a strong acid (HCl) in order to readily donate H+ ions to NH3... NH3/NH4+ won't magically appear out of nowhere
Think about titration curves. Combination of strong/weak. Weak + Weak = useless.

....


Btw, A) is the answer.. you got the question right as shown in your calculations.. I don't understand what the confusion is about

well lets say you put 25mol NH3 in a soln of 50mol HCl. Wouldn't this just leave you with 25 mol HCl and 25 mol NH4+ (no NH3). Is this still a buffer? If so please explain.

 

[iA-MD2013]

well lets say you put 25mol NH3 in a soln of 50mol HCl. Wouldn't this just leave you with 25 mol HCl and 25 mol NH4+ (no NH3). Is this still a buffer? If so please explain.

The annoying thing about this question is that when you usually do a titration, you add the strong acid/base to the weak acid/base. Otherwise the sol'n will bubble and spill all over the place. Because of that, it's easy to misread the question. I think bozz may have read it the same way i did at first...adding HCl to 50 mL NH3. In that case, you would add 25 mL of HCl to 50 mL NH3 and you have an awesome buffer.
But I think youre right, you would completely protonate the NH3 and end up with a really acidic sol'n if you only add 25 mL NH3.
Buffers are so confusing

 

[bozz]

well lets say you put 25mol NH3 in a soln of 50mol HCl. Wouldn't this just leave you with 25 mol HCl and 25 mol NH4+ (no NH3). Is this still a buffer? If so please explain.

True.. in this case, you wouldn't get one. You're right... but it seemed like there was confusion as to how HCl along with something else could possibly give you a buffer.... in buffer problems that actually require you to calculate a number, it's possible

When you add 25 mL of weak base into 50 mL of strong acid, all of the base will convert to its c.a. There is no base left to act as a buffer.

Btw, the ratio is 2 to 1,;p .

EDIT: oops misread the question haha.. that's exactly what I read haha

Answer is D) ... I believe... you'd need a 100 ml... (for MCAT purposes) you'd need a 50:50 ratio for it to be classified as a buffer... to eliminate any subjectivity

 

[wannabedocta]

We add a certain volume of .10M NH3 to 50mL of .1M HCl. Which one of these NH3 volumes will produce a buffer solution?
a. 25mL
b. 50mL
c. 75 mL
d. none of the above

I don't understand this because we don't have NH4's concentration
Any help would be greatly appreciated. THank you.

Well, a buffer can be a weak base and its conjugate acid. Equal molars of each will neutralize each other (50 mls of NH3) and will give you the equivalence point. We have to figure out whether a buffer would be created above or below 50 mL.

Since the initial reaction is HCl + H20 --> H3O+ + Cl-, adding 25 ml of NH3 will give NH4+ and some HCl which is strong acid and conjugate base. This isn't a buffer. At 50 mL, all the HCl has donated it's protons to NH4+, so NH4+ is the only species. Above 50 mL, presumably we'd have excess NH3, NH4+, and Cl- in the reaction mixture which contains a weak base and conjugate acid. So 75 mL should do it, I would go with C.

 

[wannabedocta]

Now OP... for practice, answer this question

If you get this question, you understand buffers completely.

Use this passage: http://www.berkeley-review.com/TBR/bookdetails/passages/genchem1frame.html

To make a buffer at pH = 10.83, which of the following should be mixed?

1. One-half equivalent of NaOH with one equivalent of H2CO3.
   [*]One and one-half equivalents of NaOH with one equivalent of H2CO3.
   [*]One and one-half equivalents of NaOH with one equivalent of H3PO4.
   [*]Two and one-half equivalents of NaOH with one equivalent of H3PO4.

Now before you start going, "there's something wrong with the question! wtf is NaOH doing? It can't be! THIS IS MADNESS"... think about what is happening

I'd guess 2. If the buffer pH is relatively high, i'd guess it would be a weak acid, and a strong conjugate base mixture. Both H3PO4 and H2CO3 are weak but based on the Ka values, H2CO3 is weaker. I'd want more of the conjugate base in the buffer so I'd want more NaOH to deprotonate the weak acid to form the conjugate base and choice two seems to do the job. This question looks as if it requires calculation but I have no clue how to do it.

 

[Sam212]

Now OP... for practice, answer this question

If you get this question, you understand buffers completely.

Use this passage: http://www.berkeley-review.com/TBR/bookdetails/passages/genchem1frame.html

To make a buffer at pH = 10.83, which of the following should be mixed?

1. One-half equivalent of NaOH with one equivalent of H2CO3.
   [*]One and one-half equivalents of NaOH with one equivalent of H2CO3.
   [*]One and one-half equivalents of NaOH with one equivalent of H3PO4.
   [*]Two and one-half equivalents of NaOH with one equivalent of H3PO4.

Now before you start going, "there's something wrong with the question! wtf is NaOH doing? It can't be! THIS IS MADNESS"... think about what is happening

I think it'll be 1 because NaOH will completely dissociate being a strong base. So it will provide OH-, causing H+ of H2CO3 to form H2O. Now Half of H2CO3 will convert into its conjugate base and remaining half will remain H2CO3, thus creating a buffer. Tell me if I am right

 

[iA-MD2013]

Now OP... for practice, answer this question

If you get this question, you understand buffers completely.

Use this passage: http://www.berkeley-review.com/TBR/bookdetails/passages/genchem1frame.html

To make a buffer at pH = 10.83, which of the following should be mixed?

1. One-half equivalent of NaOH with one equivalent of H2CO3.
   [*]One and one-half equivalents of NaOH with one equivalent of H2CO3.
   [*]One and one-half equivalents of NaOH with one equivalent of H3PO4.
   [*]Two and one-half equivalents of NaOH with one equivalent of H3PO4.

Now before you start going, "there's something wrong with the question! wtf is NaOH doing? It can't be! THIS IS MADNESS"... think about what is happening

OK, so this entire problem involves application of henderson-hasselbach...and it takes a while.
1. adding 0.5 eq of OH- would deprotonate half of the carbonic acid. So, we have 0.5 eq HCO3- and 0.5 eq H2C03. at this pt, pH=pKa (1:1 acid to conj. base). according to the chart, this would be pH~ 6.5. NO
2. Now, 1 eq OH- deprotonates all the H2CO3 to HCO3-. the other 0.5 eq OH- deprotonates the 0.5 eq HCO3- to CO3(2-). so, in this solution, we have 0.5 eq HCO3- and 0.5 eq CO3(2-). Once again, henderson hasselbach tells us that pH=pKa. This would be about 10.8 (-log(1.5x10^-11)). YAY!!!
I am so happy that this ended up being the ans because im typing in the dark and it is incerdibly annoying.
Ans= 2.

 

[Sam212]

OK, so this entire problem involves application of henderson-hasselbach...and it takes a while.
1. adding 0.5 eq of OH- would deprotonate half of the carbonic acid. So, we have 0.5 eq HCO3- and 0.5 eq H2C03. at this pt, pH=pKa (1:1 acid to conj. base). according to the chart, this would be pH~ 6.5. NO
2. Now, 1 eq OH- deprotonates all the H2CO3 to HCO3-. the other 0.5 eq OH- deprotonates the 0.5 eq HCO3- to CO3(2-). so, in this solution, we have 0.5 eq HCO3- and 0.5 eq CO3(2-). Once again, henderson hasselbach tells us that pH=pKa. This would be about 10.8 (-log(1.5x10^-11)). YAY!!!
I am so happy that this ended up being the ans because im typing in the dark and it is incerdibly annoying.
Ans= 2.

Damnit, you are right and I read the Ka2 value as Ka1..... lesson learned: read carefully!

 

[iA-MD2013]

Damnit, you are right and I read the Ka2 value as Ka1..... lesson learned: read carefully!

ive misread things so many times...but never learn my lesson!!! just hope it doesnt happen when it really matters

 

[Kaustikos]

OK, so this entire problem involves application of henderson-hasselbach...and it takes a while.
1. adding 0.5 eq of OH- would deprotonate half of the carbonic acid. So, we have 0.5 eq HCO3- and 0.5 eq H2C03. at this pt, pH=pKa (1:1 acid to conj. base). according to the chart, this would be pH~ 6.5. NO
2. Now, 1 eq OH- deprotonates all the H2CO3 to HCO3-. the other 0.5 eq OH- deprotonates the 0.5 eq HCO3- to CO3(2-). so, in this solution, we have 0.5 eq HCO3- and 0.5 eq CO3(2-). Once again, henderson hasselbach tells us that pH=pKa. This would be about 10.8 (-log(1.5x10^-11)). YAY!!!
I am so happy that this ended up being the ans because im typing in the dark and it is incerdibly annoying.
Ans= 2.

- edit - nvm, I jumped the gun.

 

[Vihsadas]

Now OP... for practice, answer this question

If you get this question, you understand buffers completely.

Use this passage: http://www.berkeley-review.com/TBR/bookdetails/passages/genchem1frame.html

To make a buffer at pH = 10.83, which of the following should be mixed?

1. One-half equivalent of NaOH with one equivalent of H2CO3.
2. One and one-half equivalents of NaOH with one equivalent of H2CO3.
3. One and one-half equivalents of NaOH with one equivalent of H3PO4.
4. Two and one-half equivalents of NaOH with one equivalent of H3PO4.

Now before you start going, "there's something wrong with the question! wtf is NaOH doing? It can't be! THIS IS MADNESS"... think about what is happening

love it. haha.