General Chemistry Question Thread 2

PRIVATE MCAT LESSONS. Click to learn more!
This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.
Sort by date Sort by votes
Is a buffer solution ALWAYS a weak acid and its conjugate base

or weak base/conjugate acid....

or can it be something else too?
 
Upvote 0 Downvote
axp107 said:
Is a buffer solution ALWAYS a weak acid and its conjugate base

or weak base/conjugate acid....

or can it be something else too?
Click to expand...

It's the same thing... if we say we have a buffer of "HA and A-" then we can say HA is the weak acid, and A- is the conjugate base of that weak acid. We can also say A- is a weak base, and HA is the conjugate acid of that weak base. It's two different ways of saying the same thing.

Now, a CONJUGATE base just means that it's the same molecule as the acid, with one less H+. Like A- is the conjugate base of HA, since A- is just HA with one less H+. It can not be anything else. A buffer must be a weak acid and its conjugate base (or the same way of saying it, it can be a weak base and its conjugate acid).

So NH4+ and CH3COO- would not be a buffer system, since CH3COO- is not the conjugate base of NH4+. NH4+ and NH3 would be a buffer, since NH3 is the conjugate base of the weak acid NH4+.
 
Upvote 0 Downvote
RPedigo said:
It's the same thing... if we say we have a buffer of "HA and A-" then we can say HA is the weak acid, and A- is the conjugate base of that weak acid. We can also say A- is a weak base, and HA is the conjugate acid of that weak base. It's two different ways of saying the same thing.

Now, a CONJUGATE base just means that it's the same molecule as the acid, with one less H+. Like A- is the conjugate base of HA, since A- is just HA with one less H+. It can not be anything else. A buffer must be a weak acid and its conjugate base (or the same way of saying it, it can be a weak base and its conjugate acid).

So NH4+ and CH3COO- would not be a buffer system, since CH3COO- is not the conjugate base of NH4+. NH4+ and NH3 would be a buffer, since NH3 is the conjugate base of the weak acid NH4+.
Click to expand...


Aah, I see.. thanks. The main reason I asked is b/c: http://www.berkeley-review.com/TBR/bookdetails/passages/genchem1frame.html
Look at the first question. Choices 3 AND 4 are not conjugates... yet we have to pick 1 answer that results in a buffer with pH 9.5.

just another quick Q...

Is a buffer something you ADD to an ongoing titration between acid/base... is it just the half equivalence point during a titration?
 
Upvote 0 Downvote
Members don't see this ad :)
axp107 said:
Aah, I see.. thanks. The main reason I asked is b/c: http://www.berkeley-review.com/TBR/bookdetails/passages/genchem1frame.html
Look at the first question. Choices 3 AND 4 are not conjugates... yet we have to pick 1 answer that results in a buffer with pH 9.5.

just another quick Q...

Is a buffer something you ADD to an ongoing titration between acid/base... is it just the half equivalence point during a titration?
Click to expand...

For the link you posted, recall that HCl is a strong acid and dissociates completely-- so some of the NH3 would be NH4+.. also recall that NaOH dissociates completely, so some of that HCN would become CN-. This creates a buffer solution.
 
Upvote 0 Downvote
So is a buffer solution basically just a weak acid being TITRATED with it's conjugate base... or is it something you actually add to a titration?
 
Upvote 0 Downvote
axp107 said:
So is a buffer solution basically just a weak acid being TITRATED with it's conjugate base... or is it something you actually add to a titration?
Click to expand...

Let's say you have some weak base (A-) in a beaker, and then we take a buret and slowly titrate HCl into it. Since HCl dissociates completely, we're basically just adding H+. After the first drop, we have some HA because some of the A- picked up the H+ from the HCl. Now that we have both HA and A-, we have a buffer solution. In this way, we have made a buffer by titrating a strong acid into a weak base.

Now, the same thing occurs if we have some weak acid (HA) in a beaker, and then we take a buret and slowly titrate NaOH into it. NaOH is a strong base and dissociates completely, so we're adding basically OH-. After the first drop of NaOH added, some of the HA has sacrificed its H+, so we have some unreacted HA but now we also have some conjugate base (A-), so we have a buffer solution.

Thirdly, we could start with a buffer solution already in place (by mixing, say, equal parts acetic acid (HAc) and sodium acetate (Ac-).

If you titrated a weak acid with its conjugate base, you also would have buffer, but that's not typically done I don't think. Hopefully this helps.
 
Upvote 0 Downvote
Cu + HNO3 -----> Cu(No3)2 + H2O + NO2

I am having trouble balancing this redox reaction
Maybe I got the reaction wrong or whatever I don't know I am lost . Chemistry is not my thing


PLEASE HELP!!!!
 
Upvote 0 Downvote
So is a buffer solution basically just a weak acid being TITRATED with it's conjugate base... or is it something you actually add to a titration?
Click to expand...

I thought buffers were always weak acids because weak acids can accept an H+ as well as give off an H+
 
Upvote 0 Downvote
pakbabydoll said:
Cu + HNO3 -----> Cu(No3)2 + H2O + NO2

I am having trouble balancing this redox reaction
Maybe I got the reaction wrong or whatever I don't know I am lost . Chemistry is not my thing


PLEASE HELP!!!!
Click to expand...

i just googled the answer...no point in wasting time trying to do math.

http://answers.yahoo.com/question/index?qid=20060912203101AAo9UEM

Copper is oxidized by concentrated nitric acid. The equation is
3Cu(s) + 8HNO3(aq) -> 3Cu(NO3)2(aq) + 4H2O(l) + 2NO(g)

3 copper atoms go from the 0 to the +2 state, and 2 nitrogen atoms go from the +5 to the +2 state.
 
Upvote 0 Downvote
Can someone help with this question?

I don't understand how the quantity of NaOH added influences the buffer between HCO3-, and CO3 2-

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


a) One-half equivalent of NaOH with one equivalent of H2CO3.
b) One and one-half equivalents of NaOH with one equivalent of H2CO3.
c) One and one-half equivalents of NaOH with one equivalent of H3PO4.
d) Two and one-half equivalents of NaOH with one equivalent of H3PO4.

Given:

Acid |Ka1 value| | Ka2 value | | Ka3 value |

H3PO4 |6.9 x 10-3| |6.2 x 10-8| |4.8 x 10-13|
H2CO3 |4.3 x 10-7| |1.5 x 10-11|
 
Upvote 0 Downvote
axp107 said:
I don't understand how the quantity of NaOH added influences the buffer between HCO3-, and CO3 2-
Click to expand...

Because as you add a strong base (NaOH), the base is going to remove a proton (H+) from the HCO3-. That means if you have a bunch of HCO3- and add NaOH, eventually more and more of the HCO3- will become CO3 2-. So by adding NaOH, you can serve to make H2CO3 into HCO3- and also HCO3- into CO3 2-. This way, adding equivalents of NaOH can make a specific buffer-system ratio.

Does this make sense?
 
Upvote 0 Downvote
In the Princeton Review's Physical Science Review book it says that the molecules of real gas have volume, which reduces the effective volume of the container, making V(real) < V(ideal).

However, in Examkracker's General Chemistry book it says that the volume of molecules of real gas need to be added to the ideal volume so that V(real) > V(ideal).

Which is correct?
 
Upvote 0 Downvote
Members don't see this ad :)
Hi guys...

Got a question about determining atomic radius based on Zeff because it involves cations/anions:

Al3+
Al
S
S2-

Please put them in order from the smallest to the biggest radius?
Can someone write an explanation as well?

Thank you
 
Upvote 0 Downvote
spoudaios said:
Al3+
Al
S
S2-

Please put them in order from the smallest to the biggest radius?
Can someone write an explanation as well?
Click to expand...

S2- > Al > S > Al3+

1) Atomic radii decrease going from left to right in the period table. I'm assuming you already know this. Keep in mind that although the radii decrease from left to right, they do not decrease significantly. Only a little bit.

2) If you add electrons to an element, the ionic radius will be much bigger. This is because you have more electrons than protons. The nucleus has a positive charge, which keeps electrons close. If you add more electrons, the nucleus' positive charge has to be distributed over more electrons. The nucleus will have a "looser grip" over the electrons because of this. This means electrons will stray further away from the nucleus. Hence, a much bigger radius.

3) If you take away electrons from an element, the ionic radius will be much smaller. Having more protons than electrons will cause the nucleus to pull the remaining electrons more "tightly." This will significantly decrease the size of the radius.

Now, going back to your original question:

1) Al > S
2) S2- >> S
3) Al >> Al3+

S2- >> Al > S >> Al3+
 
Upvote 0 Downvote
More question:

What about this: S < Al3+ < Al < S2-

S has 16 protons
Al has 13 protons

How do you compare between S and Al3+? One has more protons and the other one loses electron.
What about S2- and Al ?

I agree that:
Al > Al3+
S2- > S
Al > S

It's pretty straightforward if we compare ions between the same atoms, but when it compares with different atoms, i got confused.




MundaneMD said:
S2- > Al > S > Al3+

1) Atomic radii decrease going from left to right in the period table. I'm assuming you already know this. Keep in mind that although the radii decrease from left to right, they do not decrease significantly. Only a little bit.

2) If you add electrons to an element, the ionic radius will be much bigger. This is because you have more electrons than protons. The nucleus has a positive charge, which keeps electrons close. If you add more electrons, the nucleus' positive charge has to be distributed over more electrons. The nucleus will have a "looser grip" over the electrons because of this. This means electrons will stray further away from the nucleus. Hence, a much bigger radius.

3) If you take away electrons from an element, the ionic radius will be much smaller. Having more protons than electrons will cause the nucleus to pull the remaining electrons more "tightly." This will significantly decrease the size of the radius.

Now, going back to your original question:

1) Al > S
2) S2- >> S
3) Al >> Al3+

S2- >> Al > S >> Al3+
Click to expand...
 
Upvote 0 Downvote
spoudaios said:
More question:

What about this: S < Al3+ < Al < S2-

S has 16 protons
Al has 13 protons

How do you compare between S and Al3+? One has more protons and the other one loses electron.
What about S2- and Al ?

I agree that:
Al > Al3+
S2- > S
Al > S

It's pretty straightforward if we compare ions between the same atoms, but when it compares with different atoms, i got confused.
Click to expand...

Well, it's a bit tricky comparing different ions. Notice, in my explanation, I wrote that:

1) Al > S
2) S2- >> S
3) Al >> Al3+

The negative ion will be much bigger than the neutral atom. The positive ion will be much smaller than the neutral atom. Comparing neutral atoms across the period, the size differences are not that big. This is why the Al > S, but S2- >> Al > S >> Al3+.

Classify it like this:
Negative ions will be very big, neutral atoms will be close to each other, and positive ions will be very small.

Negative ions (very big) >> Neutral atoms >> Positive ions (very small)

Between the neutral atoms, you can use the general periodic trend of radii decreasing across periods, and increasing down the groups. I don't think the MCAT requires you to differentiate between two positive ions or two negative ions. You only need to know that negative ions are very big, and positive ions are very small.


This powerpoint presentation from a chemistry course might be helpful because they have listed all the ionic radii values:
http://osf1.gmu.edu/~jschreif/211/overheads/Ebbing9.ppt
 
Upvote 0 Downvote
I always freak out when I see a compound like this:

CuCO3 . Cu(OH)2

what exactly is that? How do I treat problems/passages like that?
 
Upvote 0 Downvote
CuCO3.Cu(OH)2

Don't get scared, just imagine that it is like the separate compounds (CuCO3, and Cu(OH)2), both being held together temporarily. There is a weak bond holding them together.

What happens to weak bonds if you apply stress (e.g. heat) to them? They break.

CuCO3.Cu(OH)2 + heat --> CO2 + H2O + 2CuO

Since the bonds are weak, it is also vulnerable to attack from other compounds.

The questions on the MCAT only require you to know that a compound like this would dissociate into stable compounds.

For example, it might say a solid and gas evolved during the heating of this sample. What is the gas? CO2. What is the solid? CuO.

Something like that.
 
Upvote 0 Downvote
I feel like an idiot. I haven't done any aqueous chem since high school and I just started studying. Anyway, here's the question:

If you add 500 ml of 0.2M sodium hydroxide to 500 ml of 0.4M of an unknown weak acid, the final pH is 6.2. What is the pK of the weak acid?

So, you have .1 moles of NaOH and .2 moles of weak acid. The NaOH would subtract .1 mol from the acid and form .1 mol of conjugate base, right? So, you end up with .1 mol of weak acid and .1 mol of conjugate base, so the pK = pH = 6.2?

is that right?

thanks :oops:
 
Upvote 0 Downvote
Yeah, acid-base chemistry took me the longest time to get down pat. The trick is to know exactly when to use each equation, and more importantly, when you don't need to use an equation and can just think through the problem to answer (i.e. half-equivalence point and all the things that happen at that point)
 
Upvote 0 Downvote
"The question states:..."

"The book says..."

You can't post other people's copyright protected materials.
 
Upvote 0 Downvote
LaCasta said:
The question states:...
Click to expand...

I can see your confusion, just remember that on an empty stomach your pH will be <2, so if you ingest Magnesium Hydroxide (equivalent to a tums or another antacid), it will neutralize more of the Hydrogen Ions most likely leaving the pH of the stomach >2, so LESS aspirin will be protonated when compared to an empty stomach.

Just remember, the lower the pH in the stomach the more protonated something will become. So, if you increase the pH as in this scenario the substance (aspirin in this situation) will become less protonated.

Hope this helps.
 
Upvote 0 Downvote
Thanks

I guess if you consider the pH as a product sort of, as in the number of H+ in the solution, then yeah it makes sense, a higher pH would shift the equation of aspirin to the left.

Thanks again
 
Upvote 0 Downvote
your original thinking was actually correct as well.... "aspirin -" is deprotonated aspirin. there will be more of it as the equation is pulled to the right, and less protonated aspirin (the "aspirin" on the left of the equation)... less protonated aspirin is choice D.

anything that increases the pH will affect weak acids or bases in the same way... higher pH leads to less protonated species and more deprotonated species, whether that particular "species" is an acid or a base. lower pH causes more protonated species and less deprotonated species at equilibrium.

i think of it like this...

increase the pH --> there's more OH- flying around --> OH- takes (steals?!) protons from things --> less protonated species and more deprotonated species

decrease the pH --> there's more H3O+ (or just H+ if you prefer) flying around --> things are less inclined to give up H+ cuz there's a lot of it already in solution --> more protonated species and less deprotonated species

werd
 
Upvote 0 Downvote
werd said:
your original thinking was actually correct as well.... "aspirin -" is deprotonated aspirin. there will be more of it as the equation is pulled to the right, and less protonated aspirin (the "aspirin" on the left of the equation)... less protonated aspirin is choice D.

anything that increases the pH will affect weak acids or bases in the same way... higher pH leads to less protonated species and more deprotonated species, whether that particular "species" is an acid or a base. lower pH causes more protonated species and less deprotonated species at equilibrium.

i think of it like this...

increase the pH --> there's more OH- flying around --> OH- takes (steals?!) protons from things --> less protonated species and more deprotonated species

decrease the pH --> there's more H3O+ (or just H+ if you prefer) flying around --> things are less inclined to give up H+ cuz there's a lot of it already in solution --> more protonated species and less deprotonated species

werd
Click to expand...

Thanks

You're right, I think I just misinterpreted the answers, makes sense
 
Upvote 0 Downvote
axp107 said:
Can someone help with this question?

I don't understand how the quantity of NaOH added influences the buffer between HCO3-, and CO3 2-

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


a) One-half equivalent of NaOH with one equivalent of H2CO3.
b) One and one-half equivalents of NaOH with one equivalent of H2CO3.
c) One and one-half equivalents of NaOH with one equivalent of H3PO4.
d) Two and one-half equivalents of NaOH with one equivalent of H3PO4.

Given:

Acid |Ka1 value| | Ka2 value | | Ka3 value |

H3PO4 |6.9 x 10-3| |6.2 x 10-8| |4.8 x 10-13|
H2CO3 |4.3 x 10-7| |1.5 x 10-11|
Click to expand...
also, could someone explain to me what you're looking for here? what's the answer? how do the Kas come into play? What's the value for NaOH (Kb?)
 
Upvote 0 Downvote
MundaneMD said:
A reaction's order is determined by counting the number of reactants involved in the reaction. If only one reactant is required in a reaction, it is first order. If two reactants are required, it is a second order reaction.

Here are a few examples:

A + B --> C
In this case, you have two reactants forming a single product. The formation of C depends on both the presence of A and B. This makes it a second order reaction.


C --> D + E
Now C is decomposing into D, and E. The formation of D and E depend on how fast C can break down. Since it only depends on one reactant, the reaction is first order.


2D --> F
You need 2 D's to make one F, so this means that the reaction is second order.


Now pretend that the individual reactions have different speeds:
A + B --> C is fast
C --> D + E is slow
2D --> F is very fast

The overall reaction,
A + B --> E + 1/2F
would be limited by the slowest step, which is the decomposition of C.
But since the decomposition of C is a first order reaction, the overall reaction is also first order.

Also, if you have the rate equation in this format:
rate = k[X][Y]
the highest exponent for X and Y also gives you the reaction rate.

For our example,
rate = k[C]
so the exponent in this case is 1, confirming that its a first order reaction.

(note: I wrote this right after waking up, so I'm sorry if there are any mistakes. I'll fix them later.)
Click to expand...
hmm, i'm not exactly sure where ur getting things from...it may be that this is a general approach to rate law questions but I have issues with some parts so hear me out.

The idea that the number of reactants will determine ur rate is not necessarily true. For example, in your A + B --> C reaction, it can very easily be a first order reaction if one of the reactants is zero order and the other is first order. By that, I mean that the rate of the reaction is only affected by the concentration of one of the reactants. If you were to find that doubling both of the concentrations quadruples the rate of the reaction...then you know it's an overall 2nd order reaction...but you won't know if both reactants are first order, or if one reactant is second order and the other is zero order. If I remember correctly, the only way to determine the order of a reaction is to experimentally observe the kinetics of a reaction while adjusting the concentration of the reactants.

The whole thing with the rate limiting step seems on par with what i've studied.

When you're given rate law equations, I remember the overall rate of the reaction being the sum of the exponents, not the highest (you might be talking about something different though).

ex:
rate = k[x]^1[y]^2...the overall reaction rate is 3.

Lemme know what u think.
 
Upvote 0 Downvote
axp107 said:
Can someone help with this question?

I don't understand how the quantity of NaOH added influences the buffer between HCO3-, and CO3 2-

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


a) One-half equivalent of NaOH with one equivalent of H2CO3.
b) One and one-half equivalents of NaOH with one equivalent of H2CO3.
c) One and one-half equivalents of NaOH with one equivalent of H3PO4.
d) Two and one-half equivalents of NaOH with one equivalent of H3PO4.

Given:

Acid |Ka1 value| | Ka2 value | | Ka3 value |

H3PO4 |6.9 x 10-3| |6.2 x 10-8| |4.8 x 10-13|
H2CO3 |4.3 x 10-7| |1.5 x 10-11|
Click to expand...

The ideal buffer solution should have equal amounts of some conjugate base and its corresponding conjugate acid. However, the answer choices are not that simple.

Now if the concentrations of the conjugate base and the conjugate acid are equal, we will recognize from the David Hasslehoff equation:

pH = pKa + log[(conjugate base)/(conjugate acid)]

That the pH is actually equal to the pKa, if [conjugate base] = [conjugate acid] since the log of 1 = 0.

The Ka values are provided so that we can determine which is likely to provide the appropriate pH level, and we are looking for one that will provide a pH of approximately 10.83

Since pH = pKa given ideal buffer condition, we just have to convert the Ka expressions to pKa, and whichever gives a pH of 10.83 should be used as a buffer.

The second Ka expression for carbonic acid, if its anti-log were calculated, actually gives 10.83 for the pKa. So we know that carbonic acid will work (Kaplan gives a good method for approximating antilogs). Given the narrow list of answer choices, it looks like we can narrow everything down to B. However, this would be an incomplete explanation of the problem.

We have to determine that answer B provides equal concentrations of the conjugate base and the conjugate acid. In other words, will 1.5 equivalents of NaOH and 1 equivalent of H2CO3 provide equal concentrations of the buffer solution, which for the second dissociation constant, should consist of 50% HCO3- and 50% of the carbonate anion (-2 charge)?

It might help to break down the problem into little bits.

Let's first react 1 equivalent of NaOH and 1 equivalent of H2CO3. The result is a complete conversion of H2CO3 to its conjugate base, that is, HCO3-.

Since we have hypothetically reacted all of the H2CO3, we have none of that left.

So, now we still have 0.5 equivalents of NaOH remaining. In addition, we have 1 equivalent of HCO3-.

Let's hypothetically react the remaining 0.5 equivalents of NaOH with 0.5 equivalents of HCO3-. The OH- will deprotonate the acid, leaving us with 0.5 equivalents of the carbonate anion.

After this is done, we have 0 equivalents of NaOH, but 0.5 equivalents of the carbonate ion.

Now recall that we didn't use all of the HCO3-. We only used 0.5 equivalents of it. Thus, we have 0.5 equivalents of HCO3 remaining.

So in the end, we have 0.5 equivalents of HCO3, and 0.5 equivalents of the carbonate anion (CO3). These meet the criterion for a good buffer solution.

To reiterate, we know that the pH will be 10.83 because we are using the second dissociation constant for carbonic acid and not its first. Everything plugs into the Hasslehoff equation finely.

Let me know if anything was unclear.
 
Upvote 0 Downvote
axp107 said:
Can someone help with this question?

I don't understand how the quantity of NaOH added influences the buffer between HCO3-, and CO3 2-

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


a) One-half equivalent of NaOH with one equivalent of H2CO3.
b) One and one-half equivalents of NaOH with one equivalent of H2CO3.
c) One and one-half equivalents of NaOH with one equivalent of H3PO4.
d) Two and one-half equivalents of NaOH with one equivalent of H3PO4.

Given:

Acid |Ka1 value| | Ka2 value | | Ka3 value |

H3PO4 |6.9 x 10-3| |6.2 x 10-8| |4.8 x 10-13|
H2CO3 |4.3 x 10-7| |1.5 x 10-11|
Click to expand...

Buffer automatically means pka = pH... which one of the pka values correspond to ~ 10.8? H2CO3's ka2

Automatically eliminate C) and D)

Half an equivalent of NaOH would yield and equal amount of H2CO3 and HCO3- ... but that corresponds to ka1 NOT ka2. Eliminate A. Your answer is B.

Honestly, half of the questions on the MCAT, you can get by elimination... but if you're wondering why it's B)... 1 and a half equivalents of NaOH would cause an equal amount of HCO3- and CO3 2- to exist... corresponding to ka2.
 
Upvote 0 Downvote
thevillageidiot said:
In the Princeton Review's Physical Science Review book it says that the molecules of real gas have volume, which reduces the effective volume of the container, making V(real) < V(ideal).

However, in Examkracker's General Chemistry book it says that the volume of molecules of real gas need to be added to the ideal volume so that V(real) > V(ideal).

Which is correct?
Click to expand...

The PR statement is correct. If you read the EK statement carefully, it is correct too. Although it appears at first glance to contradict, it is not contradictory to the PR statement. They are both referring to the container volume relative to the free space within the container, but defining them differently. It's a matter of what the "ideal volume" is. Let's ignore both statements and look at the Real gas equation for a moment.

What Van der Waals has essentially done is taken the ideal gas equation PV = nRT and subsituted a modified real pressure and modified real volume into it to make the real gas equation.

So, it is basically set up that Pideal = Pcontainer - nb, where nb is essentially the volume of the molecules in the container. The true empty space is what we are now considering to be the ideal volume in the van der Waals equation. The empty space is less than the container volume. But that is not the ideal volume per say. The ideal volume is the container volume, because that is what an ideal gas would occupy.

I'm not sure I've helped at all with this explanation other than to say it's about perspective.
 
Upvote 0 Downvote
Richter915 said:
hmm, i'm not exactly sure where ur getting things from...it may be that this is a general approach to rate law questions but I have issues with some parts so hear me out.

Lemme know what u think.
Click to expand...

Yea, you're correct. Thanks for catching my mistake. It seems I'm a bit rusty on rate laws. I deleted my post and posted a link for the correct way to do the problems.

http://www.wwnorton.com/college/chemistry/gilbert/concepts/chapter14/ch14_2.htm

Once again, sorry about that.
 
Upvote 0 Downvote
smeagol said:
The ideal buffer solution should have equal amounts of some conjugate base and its corresponding conjugate acid. However, the answer choices are not that simple.

Now if the concentrations of the conjugate base and the conjugate acid are equal, we will recognize from the David Hasslehoff equation:

pH = pKa + log[(conjugate base)/(conjugate acid)]

That the pH is actually equal to the pKa, if [conjugate base] = [conjugate acid] since the log of 1 = 0.

The Ka values are provided so that we can determine which is likely to provide the appropriate pH level, and we are looking for one that will provide a pH of approximately 10.83

Since pH = pKa given ideal buffer condition, we just have to convert the Ka expressions to pKa, and whichever gives a pH of 10.83 should be used as a buffer.

The second Ka expression for carbonic acid, if its anti-log were calculated, actually gives 10.83 for the pKa. So we know that carbonic acid will work (Kaplan gives a good method for approximating antilogs). Given the narrow list of answer choices, it looks like we can narrow everything down to B. However, this would be an incomplete explanation of the problem.

We have to determine that answer B provides equal concentrations of the conjugate base and the conjugate acid. In other words, will 1.5 equivalents of NaOH and 1 equivalent of H2CO3 provide equal concentrations of the buffer solution, which for the second dissociation constant, should consist of 50% HCO3- and 50% of the carbonate anion (-2 charge)?

It might help to break down the problem into little bits.

Let's first react 1 equivalent of NaOH and 1 equivalent of H2CO3. The result is a complete conversion of H2CO3 to its conjugate base, that is, HCO3-.

Since we have hypothetically reacted all of the H2CO3, we have none of that left.

So, now we still have 0.5 equivalents of NaOH remaining. In addition, we have 1 equivalent of HCO3-.

Let's hypothetically react the remaining 0.5 equivalents of NaOH with 0.5 equivalents of HCO3-. The OH- will deprotonate the acid, leaving us with 0.5 equivalents of the carbonate anion.

After this is done, we have 0 equivalents of NaOH, but 0.5 equivalents of the carbonate ion.

Now recall that we didn't use all of the HCO3-. We only used 0.5 equivalents of it. Thus, we have 0.5 equivalents of HCO3 remaining.

So in the end, we have 0.5 equivalents of HCO3, and 0.5 equivalents of the carbonate anion (CO3). These meet the criterion for a good buffer solution.

To reiterate, we know that the pH will be 10.83 because we are using the second dissociation constant for carbonic acid and not its first. Everything plugs into the Hasslehoff equation finely.

Let me know if anything was unclear.
Click to expand...
great explanation, thank you.
 
Upvote 0 Downvote
i had a question on one of the problems from passage that i just completed and wanted to see what you guys think. the main jist of the passage is that we're heating carbon dioxide and solid carbon to give carbon monoxide (CO2(g) + C(s) = 2CO(g) ) and we're doing this in a sealed furnace in order to keep constant pressure.
The question asks if we were to stabilize the system at 1200K and inject a sample of helium into the furnace, what would happen to the amount of carbon dioxide in the system?
A) it should increase
B) it should decrease
C) it should be completely converted to carbon monoxide
D) it will remain the same

The answer key says the correct answer is D, so my question has to do with Le Chatelier's principle and why it doesn't apply here. In the explanation it basically says that although helium raises the pressure in the furnace, the partial pressures of the reacting gases are unchanged (Dalton's Law) and therefore keep on behaving as if the helium weren't present so the amount of carbon dioxide will remain the same. What I don't understand is why the increase in pressure doesn't drive the equilibrium toward the side with the fewest moles of gas (CO2's side) and therefore increase the amount of carbon dioxide. If anyone has any insight into this, I'd really appreciate it. Thanks!
 
Upvote 0 Downvote
nepatslvr said:
i had a question on one of the problems from passage that i just completed and wanted to see what you guys think. the main jist of the passage is that we're heating carbon dioxide and solid carbon to give carbon monoxide (CO2(g) + C(s) = 2CO(g) ) and we're doing this in a sealed furnace in order to keep constant pressure.
The question asks if we were to stabilize the system at 1200K and inject a sample of helium into the furnace, what would happen to the amount of carbon dioxide in the system?
A) it should increase
B) it should decrease
C) it should be completely converted to carbon monoxide
D) it will remain the same

The answer key says the correct answer is D, so my question has to do with Le Chatelier's principle and why it doesn't apply here. In the explanation it basically says that although helium raises the pressure in the furnace, the partial pressures of the reacting gases are unchanged (Dalton's Law) and therefore keep on behaving as if the helium weren't present so the amount of carbon dioxide will remain the same. What I don't understand is why the increase in pressure doesn't drive the equilibrium toward the side with the fewest moles of gas (CO2's side) and therefore increase the amount of carbon dioxide. If anyone has any insight into this, I'd really appreciate it. Thanks!
Click to expand...

Helium is an inert gas.. it doesn't affect anything. It's just one of those "exceptions" to the LeChatlier's Principle that you've gotta memorize.
 
Upvote 0 Downvote
bozz said:
Helium is an inert gas.. it doesn't affect anything. It's just one of those "exceptions" to the LeChatlier's Principle that you've gotta memorize.
Click to expand...

Ah, ok, thanks. Blah, another exception to memorize, so exciting.
 
Upvote 0 Downvote
What is the end point of the titration in terms of acids and their conjugate base?
 
Upvote 0 Downvote
everythings either an acid or a base

half way point: equal quantities of both
 
Upvote 0 Downvote
Can you guys help me solve these dilution problems. These problems are pretty different from the ones I've done in G-Chem. I guess the percentage is throwing me off as I am confused on how to approach it. Any help is appreciated. Thanks.

1) Make 500 milliliters of a 20% solution of glucose (weight/volume). The molecular weight of glucose is 180.2 g/mole.

2) Make 60 milliliters of 0.2% NaCl (weight/volume) from 1% NaCl.
 
Upvote 0 Downvote
pnoybballin said:
Can you guys help me solve these dilution problems. These problems are pretty different from the ones I've done in G-Chem. I guess the percentage is throwing me off as I am confused on how to approach it. Any help is appreciated. Thanks.

1) Make 500 milliliters of a 20% solution of glucose (weight/volume). The molecular weight of glucose is 180.2 g/mole.

2) Make 60 milliliters of 0.2% NaCl (weight/volume) from 1% NaCl.
Click to expand...

What are the four answer choices for each? Solving questions like these outright is not as useful as learning how to estimate an answer and zero in on the best choice in the process.

The first question involves finding the mass of water needed to make the 500 mL of solution and then determining the appropriate mass of glucose to make the 20% solution. The problem with the question is that it would help to know the density of the final solution. Otherwise, you have to assume that the volume of the sucrose is negligible once it has dissolved into water (which is likely not the case). The answer choices could help determine whether they care about the density of the solution or not.
 
Upvote 0 Downvote
"The Answer to Life, the Universe, and Everything is........ 42."

"I checked it very thoroughly and that quite definitely is the answer. I think the problem, to be quite honest with you, is that you've never actually known what the question is."
 
Upvote 0 Downvote
Percentage weight per volume [ % (w/v)]: grams of solute in a total volume of 100 mL.

Example: a solution 25% (w/v) is made by using 25 grams of solute and adding solvent until the total volume is 100 mL.

When the solvent is not specified, it is assumed to be water.

Now just do your algebra :)

I know, I don't like dilution problems either. Try this one on for size (these aren't MCAT questions by the way):

Answer the questions below concerning the following recipe for 10x TBE (Tris-Borate-EDTA) electrophoresis buffer:

750 mL dd H2O
108.0 g Tris base
55.0 g Boric Acid
40 mL 0.5 M EDTA, pH 8.0

Adjust pH to 8.3 with NaOH (sodium hydroxide). Adjust volume to 1 L.


Notice that Tris, Tris base, Trizma, and Trizma base are all the same reagent, namely Tris (Hydroxymethyl) Aminomethane (C4H11NO3).

Questions:

1) The MW of Tris is 121.2. What is the millimolar concentration of Tris base in a 10x TBE buffer?

2) The MW of Boric Acid is 61.83. What is the millimolar concentration of boric acid in a 1x TBE buffer?

3) What is the molar concentration of EDTA in 10x TBE buffer?

4) How much 10 mM EDTA do you need to make 10x TBE? Discuss your results.

5) How much 1x TBE can you make if you have only 23g of Tris base?

6) What recipe would you use to make 1 L of 1X TBE starting with a solution of 10X TBE?
 
Upvote 0 Downvote
NickMB said:
I know, I don't like dilution problems either. Try this one on for size (these aren't MCAT questions by the way):

Answer the questions below concerning the following recipe for 10x TBE (Tris-Borate-EDTA) electrophoresis buffer:

750 mL dd H2O
108.0 g Tris base
55.0 g Boric Acid
40 mL 0.5 M EDTA, pH 8.0

Adjust pH to 8.3 with NaOH (sodium hydroxide). Adjust volume to 1 L.
Click to expand...
I'm sorry, but I'm completely perplexed by this question. And it is part of the titrations I'm most worried about. How do you calculate the amount of base (or acid) to add to a solution to get to X pH?
 
Upvote 0 Downvote
You must log in or register to reply here.
Next unread thread

Similar threads

T
  • Question
Simple Chemistry Question ***Frustrated
Replies
1
Views
1K
sf26613
S
treeclimbingmonkey
  • Question
General ATP/GTP question
Replies
4
Views
1K
treeclimbingmonkey
treeclimbingmonkey
K
**SPOILER** AAMC Chemistry QPack Passage 3
Replies
3
Views
10K
BerkReviewTeach
B
highlander11
  • Question
General Study Questions
Replies
2
Views
782
AdamJ2020
AdamJ2020
elloL
  • Question
TBR gen chem part II Titration curves review #25
Replies
0
Views
5K
elloL
elloL
Top