General Chemistry Question Thread

[QofQuimica]

All users may post questions about MCAT, DAT, OAT, or PCAT general chemistry here. We will answer the questions as soon as we reasonably can. If you would like to know what general chemistry topics appear on the MCAT, you should check the MCAT Student Manual (http://www.aamc.org/students/mcat/studentmanual/start.htm)

Acceptable topics:
-general, MCAT-level gen chem.
-particular MCAT-level gen chem problems, whether your own or from study material
-what you need to know about gen chem for the MCAT
-how best to approach to MCAT gen chem passages
-how best to study MCAT gen chem
-how best to tackle the MCAT physical sciences section

Unacceptable topics:
-actual MCAT questions or passages, or close paraphrasings thereof
-anything you know to be beyond the scope of the MCAT

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If you really know your gen chem, I can use your help. If you are willing to help answer questions on this thread, please let me know. Here are the current members of the General Chemistry Team:

-QofQuimica (thread moderator): I have my M.S. in organic chemistry and I'm currently finishing my Ph.D., also in organic chemistry. I have several years of university general chemistry TA teaching experience. In addition, I teach general chemistry classes through Kaplan for their MCAT, DAT, OAT, and PCAT courses. On the MCAT, I scored 14 on PS, 43 overall.

-Learfan: Learfan has his Ph.D. in organic chemistry and several years worth of industrial chemistry experience. He scored 13 on the PS section of the MCAT, and 36 overall.

-Sparky Man: Sparky Man has his Ph.D. in physical chemistry. He scored 14 on the PS section of the MCAT, and 36 overall.

-GCT: GCT scored in the 99th percentile on the PCAT. He has also taught introductory physics and general chemistry.

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[commuter9]

thanks Q and heymanooh. it now makes sense why O2- and N3- are such strong bases.

 

[commuter9]

Which of the following cations is the strongest acid in aqueous solution?

A. Fe3+
B. Ca2+
C. Cu2+
D. Zn2+

Answer: A The greater the charge of the cation and the smaller the size, the stronger the acidic solution formed.

My question is why does size matter for an acid? The only case where I thought it mattered was when we were looking at periodic trends and we say that the bigger the radius the longer the H-X bond will be so the easier it would be to break the bond (as in the case with HI and that is why it is a stronger acid than HF).

 

[heymanooh1]

Which of the following cations is the strongest acid in aqueous solution?

A. Fe3+
B. Ca2+
C. Cu2+
D. Zn2+
Answer: A The greater the charge of the cation and the smaller the size, the stronger the acidic solution formed.

My question is why does size matter for an acid? The only case where I thought it mattered was when we were looking at periodic trends and we say that the bigger the radius the longer the H-X bond will be so the easier it would be to break the bond (as in the case with HI and that is why it is a stronger acid than HF).

For this problem, remember the Lewis definition for an acid and base. You'll notice one of the cations has a larger charge density compared to the others i.e. it has larger overall charge and this charge is "spread out" over a smaller size/volume. What the greater charge density does in aqueous solution is that it polarizes the O-H bond to a greater extent in a water molecule since the oxygen atom will be more "attracted" to the cation with greater charge density. As a consequence of this polarization, the H atom becomes more acidic relative to a free water molecule.

You are correct in your summary for the H-X acidity trends. But with H-X, it's more a question of knowing how the size of an atom, X in this case, will affect bond strength and how that will be reflected in the acidity of an aqueous solution. In cations, or more likely ionic salts, it's the charge/charge density of the cation that affects acidity in aqueous solution.

 

[rcd]

How does the number of OH-'s that can bind to a cation affect the cation's acidity?

For instance, Fe3+. With each OH- bond, the Fe's oxidation state remains the same, so FeOH{2+} is attracted to an OH- just as much as an Fe3+ cation is attracted to an OH-, correct?

But is Fe3+ is more acidic then FeOH{2+} since Fe3+ needs to bind one more OH- then FeOH{2+} to get to the stable product (Fe(OH)3)?

Fe3+ + H2O <> Fe(OH){2+} + H+ k1
Fe(OH){2+} + H2O <> Fe(OH)2{+} + H+ k2=k1
Fe(OH){+} + H2O <> Fe(OH)3 + H+ k3=k1
---k1=k2=k3
Fe{3+} + 3H2O <> Fe(OH)3 + 3H+
k=k1*k1*k1
vs.
FeOH{2+} + 2H2O <> Fe(OH)3 + 2H+
k=k1*k1

 

[commuter9]

For this problem, remember the Lewis definition for an acid and base. You'll notice one of the cations has a larger charge density compared to the others i.e. it has larger overall charge and this charge is "spread out" over a smaller size/volume. What the greater charge density does in aqueous solution is that it polarizes the O-H bond to a greater extent in a water molecule since the oxygen atom will be more "attracted" to the cation with greater charge density. As a consequence of this polarization, the H atom becomes more acidic relative to a free water molecule.

You are correct in your summary for the H-X acidity trends. But with H-X, it's more a question of knowing how the size of an atom, X in this case, will affect bond strength and how that will be reflected in the acidity of an aqueous solution. In cations, or more likely ionic salts, it's the charge/charge density of the cation that affects acidity in aqueous solution.

thanks what you said makes sense...but i'm not sure about the next post...it seems more confusing

 

[commuter9]

in the PR book it says that regardless of whether strong or weak, mixing molar equivalents of any acid with any base will lead to a complete neutralization reaction...is that true? do equal moles of HF and NH3 neutralize each other? (Is this because M1V1=M2V2 and it doesn't matter if the acids are strong or weak?)

 

[mtrl1]

For this problem, remember the Lewis definition for an acid and base. You'll notice one of the cations has a larger charge density compared to the others i.e. it has larger overall charge and this charge is "spread out" over a smaller size/volume. What the greater charge density does in aqueous solution is that it polarizes the O-H bond to a greater extent in a water molecule since the oxygen atom will be more "attracted" to the cation with greater charge density. As a consequence of this polarization, the H atom becomes more acidic relative to a free water molecule.

You are correct in your summary for the H-X acidity trends. But with H-X, it's more a question of knowing how the size of an atom, X in this case, will affect bond strength and how that will be reflected in the acidity of an aqueous solution. In cations, or more likely ionic salts, it's the charge/charge density of the cation that affects acidity in aqueous solution.

You're almost right. But the reason for the acidity trends has to do with the stability of the conjugate base as well as polarizibility. Organic and General chem teachers argue about this topic for days. O-chem teachers prefer the base stability explanation because it helps explain nucelophile trends for the halides. A larger base (i.e. I-) has more "room" for the negative charge; (i.e. more entropy) which plays a role in driving the thermodynamics of the reaction.

 

[heymanooh1]

You're almost right. But the reason for the acidity trends has to do with the stability of the conjugate base as well as polarizibility. Organic and General chem teachers argue about this topic for days. O-chem teachers prefer the base stability explanation because it helps explain nucelophile trends for the halides. A larger base (i.e. I-) has more "room" for the negative charge; (i.e. more entropy) which plays a role in driving the thermodynamics of the reaction.

Yeah, OK. When I see a question on the MCAT that has a list of cations, asking for which one leads to a more acidic solution, I'll make sure to look for the thermodynamically more stable conjugate base or the base that's a better nucleophile as the answer

 

[QofQuimica]

in the PR book it says that regardless of whether strong or weak, mixing molar equivalents of any acid with any base will lead to a complete neutralization reaction...is that true? do equal moles of HF and NH3 neutralize each other? (Is this because M1V1=M2V2 and it doesn't matter if the acids are strong or weak?)

I think what is confusing you is what happens to the salt that forms after you mix the acid and base together. Any time you have a weak acid or a weak base, you have the potential for there to be an equilibrium at the end (i.e., buffering, where some of the salt hydrolyzes back to its original ion). It is difficult to predict what will happen when you mix a weak acid with a weak base because it varies depending on the relative strengths of each species. In other words, you can have one possible scenario where the weak acid is stronger than the weak base, and in that case you will wind up with an acidic solution afterward. Conversely, if the weak acid is weaker than the weak base, you will wind up with a basic solution afterward. You will only get a neutral solution if the acid and base are of the same strength.

 

[commuter9]

I think what is confusing you is what happens to the salt that forms after you mix the acid and base together. Any time you have a weak acid or a weak base, you have the potential for there to be an equilibrium at the end (i.e., buffering, where some of the salt hydrolyzes back to its original ion). It is difficult to predict what will happen when you mix a weak acid with a weak base because it varies depending on the relative strengths of each species. In other words, you can have one possible scenario where the weak acid is stronger than the weak base, and in that case you will wind up with an acidic solution afterward. Conversely, if the weak acid is weaker than the weak base, you will wind up with a basic solution afterward. You will only get a neutral solution if the acid and base are of the same strength.

So Q are you saying then that the book is wrong, strength of the acids and bases do matter in neutralization?

 

[killinsound]

Quick question regarding sort of the same topic. When you are titrating a weak acid with say... a stronger base, what's the reason that the pH is greater than 7. I mean it makes sense, but what is really "happening". Is it because the HA -> A-, so the weak acid leaves a conjugate base that reacts with water to leave some OH-?

In EK, it says that the pH of the equivalence point can only be calculated through the Kb. Why not through the Ka? Is it assuming that you are titrating with a base? I'm really confused about this one.

Also, I'm not understanding how a substance that is at a point above melting point has a liquid vapor pressure that is greater than the solid vapor pressure.

 

[rcd]

Quick question regarding sort of the same topic. When you are titrating a weak acid with say... a stronger base, what's the reason that the pH is greater than 7. I mean it makes sense, but what is really "happening".

Also, I'm not understanding how a substance that is at a point above melting point has a liquid vapor pressure that is greater than the solid vapor pressure.

It's just because you get a basic salt.

The stronger thing, base in this case, dissociates and never reforms.

ex) NaOH -> Na+ + OH-
That never happens in the reverse.

But with a weak acid/base (acid in your example)
CH3COOH <> CH3COO- + H+
That does happen in the reverse.

So, 1M NaOH neutralizes 1M CH3COOH completely, but you have salt leftover:

NaOH + CH3COOH <> HOH + CH3COONa

That salt dissociates to some degree, and you have CH3COO- and Na+ in solution. Na+ never forms NaOH because NaOH is a strong base. But CH3COOH will form from CH3COO-, lowering [H+] in solution and therefore raising pH.

That gives you a basic equivalence point.

 

[killinsound]

Woops my fault, I think you answered after I edited my post.

When EK says that you can only use Kb to calculate the equivalence point of a strong base weak acid titration, are they assuming that the titrant is base and that the solution is acidic?

In other words, could you use Ka to calculate the equivalence point if the titrant was acid?

 

[QofQuimica]

Woops my fault, I think you answered after I edited my post.

When EK says that you can only use Kb to calculate the equivalence point of a strong base weak acid titration, are they assuming that the titrant is base and that the solution is acidic?

In other words, could you use Ka to calculate the equivalence point if the titrant was acid?

Usually the strong base or acid is the titrant. Most of the time you'll be titrating a weak acid with a strong base like NaOH. You could do it backwards (i.e., titrate a weak base with a strong acid like HCl), but then you'd probably be given the weak base's Kb. If the acid is strong, it doesn't really have a measurable Ka. In other words, dissociation is so nearly complete that Ka is some really, really huge number.

 

[QofQuimica]

It's just because you get a basic salt.

The stronger thing, base in this case, dissociates and never reforms.

ex) NaOH -> Na+ + OH-
That never happens in the reverse.

But with a weak acid/base (acid in your example)
CH3COOH <> CH3COO- + H+
That does happen in the reverse.

So, 1M NaOH neutralizes 1M CH3COOH completely, but you have salt leftover:

NaOH + CH3COOH <> HOH + CH3COONa

That salt dissociates to some degree, and you have CH3COO- and Na+ in solution. Na+ never forms NaOH because NaOH is a strong base. But CH3COOH will form from CH3COO-, lowering [H+] in solution and therefore raising pH.

That gives you a basic equivalence point.

Right. I'm talking about what happens AFTER the neutralization. Don't get all hung up about this little detail, commuter. What you're liable to be asked about is what kind of solution you wind up with after the neutralization is complete, and THAT is what depends on the strength of the acid and base.

 

[killinsound]

Usually the strong base or acid is the titrant. Most of the time you'll be titrating a weak acid with a strong base like NaOH. You could do it backwards (i.e., titrate a weak base with a strong acid like HCl), but then you'd probably be given the weak base's Kb. If the acid is strong, it doesn't really have a measurable Ka. In other words, dissociation is so nearly complete that Ka is some really, really huge number.

Oops, I think I phrased my question incorrectly.

If you were to titrate weak base with a strong acid, would you use the Ka of the weak base to figure out the equivalence point.

For instance NH3 + HCL -> NH4+ + CL-

Since the strong acid dissociates completely and the NH3 is all reacted up, you would use the Ka of NH4+ to calculate the pH at equivalence point. (In other words, you can use Ka to find the pH and not just Kb as how EK is sort of implying).

 

[QofQuimica]

Oops, I think I phrased my question incorrectly.

If you were to titrate weak base with a strong acid, would you use the Ka of the weak base to figure out the equivalence point.

For instance NH3 + HCL -> NH4+ + CL-

Since the strong acid dissociates completely and the NH3 is all reacted up, you would use the Ka of NH4+ to calculate the pH at equivalence point. (In other words, you can use Ka to find the pH and not just Kb as how EK is sort of implying).

I see what you're saying. Yeah, you could do it because the pKa and pKb of the conjugate acid/base pair are related, but it will take more math. (The pKa and pKb of an aqueous acid-base conjugate pair will add up to pKw, which is 14 at 25 C.) So, I'd go with the Kb if you're titrating a weak base with a strong acid.

 

[killinsound]

I see what you're saying. Yeah, you could do it because the pKa and pKb of the conjugate acid/base pair are related, but it will take more math. (The pKa and pKb of an aqueous acid-base conjugate pair will add up to pKw, which is 14 at 25 C.) So, I'd go with the Kb if you're titrating a weak base with a strong acid.

Wouldn't the Ka (in this case) be LESS math? The weak base goes to its conjugate acid. The conjugate acid then would be used to determine the the concentration of H+ through the use of Ka... Is there a shorter method I'm unaware of?

 

[QofQuimica]

Wouldn't the Ka (in this case) be LESS math? The weak base goes to its conjugate acid. The conjugate acid then would be used to determine the the concentration of H+ through the use of Ka... Is there a shorter method I'm unaware of?

Yes, if you are given the Ka, that will be the shortest way. But if you have to solve for the Ka from the Kb, I think that's going to be harder in general than just using the Kb. Either way, you don't want to spend a lot of time calculating on MCAT problems, 1) because you don't have a calculator, and 2) because you just don't have the time.

 

[killinsound]

Yes, if you are given the Ka, that will be the shortest way. But if you have to solve for the Ka from the Kb, I think that's going to be harder in general than just using the Kb. Either way, you don't want to spend a lot of time calculating on MCAT problems, 1) because you don't have a calculator, and 2) because you just don't have the time.

Thanks Q!


Last "sort of question" of the day (I swear). I'm having trouble making sense of the relation of solid vapor pressure to liquid vapor pressure to gas vapor pressure.

For instance, I kind of understand why solid and liquid VP's are equal to each other at melting point, but I REALLY don't understand why liquid VP is higher than solid VP above melting piont. I guess I'm not making sense of what is meant by solid, and liquid VP, but I'm pretty sure I understand what gas VP is.

 

[swifteagle43]

Thanks Q!


Last "sort of question" of the day (I swear). I'm having trouble making sense of the relation of solid vapor pressure to liquid vapor pressure to gas vapor pressure.

For instance, I kind of understand why solid and liquid VP's are equal to each other at melting point, but I REALLY don't understand why liquid VP is higher than solid VP above melting piont. I guess I'm not making sense of what is meant by solid, and liquid VP, but I'm pretty sure I understand what gas VP is.

I am a humble MCAT student. I think what you might be looking for is that there are more molecules at a higher velocity in the liquid than the solid. The solid is not able to exert much vapor pressure since it is so tightly packed and at lower energy. It isn't able to push out much. Liquids on the other hands can push out against the atmosphere from the inside. Hence they are able to have a higher vapor pressure.

 

[commuter9]

Solution A contains a buffer formed of 2M acetic acid and .8M potassium acetate. Solution B contains a buffer formed of .2M acetic acid and .08M potassium acetate. How do the solutions compare?

A. The pH and buffer capacity of both solutions are the same
B. The pH of solution A is lower than that of solution B, but the buffer capacity is the same
C. The pH of solution A is the same as that of solution B, but solution A has a great buffer capacity
D. The pH of solution A is lower than that of solution B, and solution A has a greater buffer capacity than solution B

The answer is C. I understand that the pH would be the same since the pKa is the same and the ratio of base to acid is the same, but I guess I'm not exactly understanding how buffering capacity comes into henderson hasselbach (or does it not?). Is the buffering capacity great just because there is a higher concentration of acid and conjugate base? Does pKa determine buffering capacity at all? I guess I'm trying to tie it together since I seem to be missing pieces...

 

[commuter9]

1 more question:

Which of the following are characteristics of an ideal solution?
I. Solute molecules do not interact with each other.
II. Solvent molecules do not interact with each other.
III. Solvent-solute interactions are similar to solute-solute and solvent-solvent interactions.

A. I only
B. III only
C I and II only
D. I, II, and III

The answer is B. The explanation EK gives is that ideal solutions obey raoult's law. The solvent and solutes make similar bonds and are similar size and shape. I thought though with an ideal solution the solutes and the solvent molecules shouldn't interact with each other right (stick together etc)?
Am I getting this mixed up w/ the ideal gas law? A gas is a solution though. I think I have confused myself further.

 

[rcd]

Solution A contains a buffer formed of 2M acetic acid and .8M potassium acetate. Solution B contains a buffer formed of .2M acetic acid and .08M potassium acetate. How do the solutions compare?

A. The pH and buffer capacity of both solutions are the same
B. The pH of solution A is lower than that of solution B, but the buffer capacity is the same
C. The pH of solution A is the same as that of solution B, but solution A has a great buffer capacity
D. The pH of solution A is lower than that of solution B, and solution A has a greater buffer capacity than solution B

The answer is C. I understand that the pH would be the same since the pKa is the same and the ratio of base to acid is the same, but I guess I'm not exactly understanding how buffering capacity comes into henderson hasselbach (or does it not?). Is the buffering capacity great just because there is a higher concentration of acid and conjugate base? Does pKa determine buffering capacity at all? I guess I'm trying to tie it together since I seem to be missing pieces...

It's all in the formula, if you have a large [acid] or [base] then changing either one of them by adding an acid or base would change the ratio [base]/[acid] relatively less. So the log of that ratio would change by less, meaning the pH changes by less, meaning yeah, more acid+base means better buffering capacity.

pKa does affect buffering capacity, the closer the solution pH is to the pKa, the greater the buffering capacity.

 

[estairella]

Am I getting this mixed up w/ the ideal gas law? A gas is a solution though. I think I have confused myself further.

You are confusing properties of an ideal gas with properties of an ideal solution. You are correct that an ideal gas is a solution, but (obviously) not all solutions are gases.

Ideal gas: there are no interactions between any of the molecules in a gas (purely kinetic collisions, equal size and shape, etc.)

Ideal solution: "Solvent-solute interactions are similar to solute-solute and solvent-solvent interactions."

I've already told you that in an ideal gas there are no interactions. So when we apply the ideal solution statement, we are effectively saying 0 = 0 = 0. No contradiction!

 

[commuter9]

thanks rcd and trozman for the explanations.

two more questions:
Ca(s) + F2(g) CaF2(s)
The reaction above is an example of all of the following reaction types EXCEPT
A. precipitation
B. combination
C. oxidation-reduction
D. formation

Answer is A: Although a solid is formed in the reaction, it is not formed from dissolved species, so the reaction is not an example of a precipitation reaction. Therefore, (A) is correct.

What does this explanation mean?



All of the following are state functions EXCEPT:
A. Volume
B. Entropy
C. Work
D. Enthalpy

Answer is C Work. When the magnitude of a quantity depends only on the initial and final states of the system, it is known as a state function. Work depends on the path taken, so (C) is correct.

I have been noticing that in chemistry work depends on path, but in physics it doesn't right? In physics work is path independent since it is F*d and so we are only concerned with displacement. In thermo though work is path dependent?

 

[Whwsf]

Hi. I was going over the AAMC test 7R test when i stumbled upon question 28 in PS. I dont know if i'm allowed to post this but if anyone can explain to me the answer that would be great.... The explanation i have just doesnt make any sense.

 

[lisichka]

Hello,
I have a question about the following Kaplan question lol sorry for being repetetive:

What is the effect of increasing the conc. of reactants in a voltaic cell?
answer: rate increases,voltage increases, spontaneity increases.

I understand all of their explanations about increase in electron flow, blah, blah, but......
I thought emf or voltage was an INTRINSIC quality,meaning it is not dependant on the amount, so how come in the above question, emf does depend on the amount (reactant concentration). I am thinking about the case when the number of moles in the whole reaction doubles, number of electron doubles, but emf is still the same since it is intrinsic, but delta G, on the other hand, doubles, since it is extrinsic.

any ideas about my concept discrepancy? thank you very much

 

[rcd]

Hello,
I have a question about the following Kaplan question lol sorry for being repetetive:

What is the effect of increasing the conc. of reactants in a voltaic cell?
answer: rate increases,voltage increases, spontaneity increases.

I understand all of their explanations about increase in electron flow, blah, blah, but......
I thought emf or voltage was an INTRINSIC quality,meaning it is not dependant on the amount, so how come in the above question, emf does depend on the amount (reactant concentration). I am thinking about the case when the number of moles in the whole reaction doubles, number of electron doubles, but emf is still the same since it is intrinsic, but delta G, on the other hand, doubles, since it is extrinsic.

any ideas about my concept discrepancy? thank you very much

emf[o] is for standard conditions at 1 M concentrations. You can add whatever coefficients you want, ex) 2A -> 2B or 1A -> 1B, but emf[o] is only valid when you've got 1M of A (ie you can double a reaction but emf[o] stays constant because you're not changing concentrations in that case). So that's where the illusion of emf being an intrinsic property comes in. Actual emf changes with concentration, and all the other stuff in an equation the MCAT would likely provide.

 

[lisichka]

emf[o] is for standard conditions at 1 M concentrations. You can add whatever coefficients you want, ex) 2A -> 2B or 1A -> 1B, but emf[o] is only valid when you've got 1M of A (ie you can double a reaction but emf[o] stays constant because you're not changing concentrations in that case). So that's where the illusion of emf being an intrinsic property comes in. Actual emf changes with concentration, and all the other stuff in an equation the MCAT would likely provide.

why doubling a reaction--> means NO concentration change , therefore
no emf standard change? going from 1M to 2M is a concentration change though. ?????????????????

 

[rcd]

why doubling a reaction--> means NO concentration change , therefore
no emf standard change? going from 1M to 2M is a concentration change though. ?????????????????

You're absolutely right, going from 1M to 2M is a concentration change.

But concentration is 100% independent of coefficients. Savvy?

You can write:
A -> B
2A -> 2B
3A -> 3B

All you want, but you're not changing concentrations.

 

[lisichka]

You're absolutely right, going from 1M to 2M is a concentration change.

But concentration is 100% independent of coefficients. Savvy?

You can write:
A -> B
2A -> 2B
3A -> 3B

All you want, but you're not changing concentrations.

i c
thanks

 

[chem neub]

What pressure would have to be applied to steam at 350C to condense the the steam to liquid water ( &#916;Hvap=40.7 kJ/mol)?
Thanx

second Q
&#916;Hvap 20 kJ/mol C(s) 3.0J/gC
&#916;Hfus 5.0 kJ/mol C(l) 2.5J/gc
bp 75C C(g) 1.0J/gc
mp-15C
Givin the above for sub X, calculate the Energy that must be removed to convert substance X from a gas @ 100C to a solid @ -50C. Assume X has a polar mass of 75.0g/mol
Much appreciated ( even tips to help start would be great because im clueless)

 

[rcd]

What pressure would have to be applied to steam at 350C to condense the the steam to liquid water ( &#916;Hvap=40.7 kJ/mol)?
Thanx

second Q
&#916;Hvap 20 kJ/mol C(s) 3.0J/gC
&#916;Hfus 5.0 kJ/mol C(l) 2.5J/gc
bp 75C C(g) 1.0J/gc
mp-15C
Givin the above for sub X, calculate the Energy that must be removed to convert substance X from a gas @ 100C to a solid @ -50C. Assume X has a polar mass of 75.0g/mol
Much appreciated ( even tips to help start would be great because im clueless)

I suspect q1 came from your chem book and not MCAT prep material since it requires a long-winded equation.

q1 - Find an equation relating temperature, dH of vaporization, pressure, and R. I halfway remember what it is but I'm not sure. It's in your book.

q2 - You never gave the mass of X (which you need), so I'm just going to call it 1g. That means you have 1g / (75g/mol) 1/75 mol of X.

These are stepwise problems. You have some X @ 100C. So first you get it to its bp 75C. dH = (100-75)C * 1.0J/gC * 1g. So you lost 25J energy in the first step. Now you you make the gas liquid. 1/75 mol * 20kJ / mol * 10^3J / kJ. You add that to the previous dH. I'm not going to do the rest of the problem, but that should give you the general idea. The easiest ways to mess these problems up are to 1) use specific heats that are for different states (ex gas) for temperature changes then the state you're in (ex liquid), and 2) not convert everything to the same unit.

 

[commuter9]

what is the difference between heat of hydration and heat of solution? why is heat of hydration negative while heat of solution is positive? i though that heat of solution is the heat given off or absorbed when solutes dissolve in a solution. is this right?

 

[ladoo007]

Hello,

My question is:

When a strip of Cu is placed into H20 (l), no change is observed. However, when a strip of Cu is placed into a solution of HNO3 (aq), a gas evolves. What is the most likley identity of the gas?

A) NO (g)
B) CO2 (g)
C) H2 (g)
D) O3 (g)

Answer: I put that it is C, but the answer is A. I understand that the HNO3 must react with the copper and since the copper is oxidized the HNO3 should be reduced. But I wrote out a reaction with HNO3 and Cu that results in Cu(NO3)2 and H2 (g) and it balances and the H is reduced while the Cu is oxidized and it seems like a viable answer. How come H2 could not have been the identity of the gas that evolved?

 

[ladoo007]

One more question...

I was doing a test and the answer to one of the questions said that THF (which has a ether as part of a 5-member ring with the other 4 members being CH2s) can hydrogen bond... i thought that oxygens in ethers cannot hydrogen bond because there is no hydrogen attached? I realized that this oxygen still has two long pairs of electrons and so perhaps can still be on the receiving end of a hydrogen bond... but does this classify it as a compound that can "hydrogen bond"?

Thanks for the help!

 

[spicedmanna]

One more question...

I was doing a test and the answer to one of the questions said that THF (which has a ether as part of a 5-member ring with the other 4 members being CH2s) can hydrogen bond... i thought that oxygens in ethers cannot hydrogen bond because there is no hydrogen attached? I realized that this oxygen still has two long pairs of electrons and so perhaps can still be on the receiving end of a hydrogen bond... but does this classify it as a compound that can "hydrogen bond"?

Thanks for the help!

Yes, I think you are on the right track with your reasoning. There are hydrogen bond donors and acceptors. Keep in mind that "hydrogen bonding" is a strong intermolecular force resulting from the electrostatic attraction of X-H, where X is a highly electronegative element such as F, N, and O, with another moeity that contains available lone pairs of electrons. What is thought to occur is that X pulls electron density away from the H, imparting on it a relatively strong delta-positive charge. This attracts, electrostatically, the available, electron-rich lone-pairs of an acceptor moiety that is in proximity to the electron-poor H. So, yes, THF can participate in hydrogen bonding, not in and of itself, but in the presence of a suitable donor.

I think your test question might have been worded better. THF can participate in a hydrogen bond, but does not hydrogen bond with another THF molecule.

 

[endofevangelion]

Alright this has been confusing me for so long: consider you have a semi-permeable membrane separating two sides of a container containing solution. The left side has a greater amount of solute than the right, so naturally we know by osmosis the water is going to flow from the right side to the left side. However, which direction is the osmotic pressure greater?

A few websites I found said that since pressure is a force, the osmotic pressure must be greater on the right side because there is a net drive of water toward the left.

However, my gen chem book says that osmotic pressure is the force OPPOSING the incoming water, and therefore the left side must have a greater osmotic pressure than the right

Can someone clear this up?

 

[commuter9]

Yeah that's basically it but watch out on how you define "the atmosphere".Vapor pressure is not necessarily part of atmospheric pressure. One can create are an atmosphere consisting of different components such as methane gas, sulfur gas, carbon monoxide etc. and place it over liquid water at 25 degrees Celsius. Before equilibrium is reached, let's say the total pressure of the "atmosphere"/different gases above liquid water = 0.6 atm. As equilibrium is established, the "atmospheric" pressure will eventually rise above 0.6 atm because liquid water develops a vapor pressure, adding to the initial value of 0.6 atm.

To clarify a bit more. Fill a flask with water at a given temperature, evacuate and seal it from atmospheric/external pressure. Initially, the vapor pressure above the liquid will be zero. Letting the flask equilibrate at the given temperature, there will be a distribution of liquid molecules that can go into the vapor phase because there's sufficient energy. At the same time there will be gas molecules losing energy and returning to the liquid phase. Eventually these two rates equal each other, thus establishing equilibrium. The resulting vapor pressure above liquid water will be a different value from atmospheric because it has been sealed off.

Don't confuse the vapor pressure of a liquid being equal to atmospheric/ external pressure. This is true when a liquid has reached it boiling point temperature. The boiling point temperature of a liquid depends on atmospheric/external pressure and the substance(s) that makes up the liquid. For example, at a sea level pressure of 1 atm, the boiling point of water is 100 degrees Celsius, therefore the vapor pressure of water = 1 atm. At a higher altitude in Denver, CO, the boiling point of liquid water will be 94 degrees Celsius because the external/atmospheric pressure will be lower (~0.8 atm). The vapor pressure of boiling water at this elevation will also be ~0.8 atm. Now if you measure the vapor pressures of liquid water at both elevations at 25 degrees Celsius, they will be the same since the distribution of water molecules in the vapor phase will be identical and yet the atmospheric pressure is different.

The linear relationship(s) between (vapor) pressure and temperature are valid over a given region as observed on their respective phase diagrams and dependent on the substance.

I had a vapor pressure question and so I looked at heymanooh's explanation (which makes sense to me), but I was still confused. Here's the sitch:

the question is under what conditions does a solid sublimate?
the answer is when the vapor pressure of the solid is greater than the partial pressure of its vapor above it.
isn't the vapor pressure the partial pressure of the vapor above it? i mean i thought that vapor pressure is the pressure the gas molecules that have escaped the solid or liq on the solid or liq. what is this other partial pressure they are talking about?

then the next question asks under what conditions does a liquid boil (which i think is when vapor pressure= atm pressure), but the answer is when the vapor pressure of teh liquid is greater than the total pressure of the gas above it.
why is this answer different from the one above it? in this case they seem to be talking about atm pressure by calling it "total pressure", but in the case before they are talking about something else.

very confused on a topic i thought i knew...

 

[lisichka]

It's all in the formula, if you have a large [acid] or [base] then changing either one of them by adding an acid or base would change the ratio [base]/[acid] relatively less. So the log of that ratio would change by less, meaning the pH changes by less, meaning yeah, more acid+base means better buffering capacity.

pKa does affect buffering capacity, the closer the solution pH is to the pKa, the greater the buffering capacity.

don;t we have to increase both acid and base in order to increase buffering capacity? i think if only acid or base is increased, then only pH changes without any changes in buffering capacity????????????????

please, anyone, correct me if i am wrong or just clarify this point.
thanks a bunch

 

[BrokenGlass]

Message Deleted.

 

[DiverDoc]

Which formula ?! KE= 1/2 mv^2=3/2 k T (k being the Boltzmann constant) OR K.E. = 3/2 RT. If not the same, when do I use them?

 

[spicedmanna]

Which formula ?! KE= 1/2 mv^2=3/2 k T (k being the Boltzmann constant) OR K.E. = 3/2 RT. If not the same, when do I use them?

k = R/Na, where Na = avogadro's number

So, your first formula can be related to the second by multiplying it by Na. Thus, your first equation is the average KE per molecule, the second, per mole.

 

[DiverDoc]

k = R/Na, where Na = avogadro's number

So, your first formula can be related to the second by multiplying it by Na. Thus, your first equation is the average KE per molecule, the second, per mole.

Makes sense now. Thanks alot

 

[m0nkeymanxx]

Hey guys,
This is nothing from the mcats but i thought you guys do help with some homework.

1. An aqeuous solution of ethylene glycol used as automobile engine collant is 40.0% HOCH2CH2OH by mass and has a density of 1.05g/ml. What are the molarity, molality and mole fraction?
I figured out molarity and molality. 6.77M 10.7m
any help with the mole fraction. is it possible to find without look up the molecular weight of ethylene glycol

2. Without detailed calculations determine which of the following aqueous solution has the greates mole percent CH3CH2OH: a. 0.5 M CH3CH2OH, d=.944g/ml; b. 5% CH3CH2OH; c. 0.5m CH3CH2OH; 5.0% CH3CH2OH by volume (d=0.991g/ml). The density of the pure liquid CH3CH2OH is .789g/ml

 

[gridiron]

thanks rcd and trozman for the explanations.

two more questions:
Ca(s) + F2(g) CaF2(s)
The reaction above is an example of all of the following reaction types EXCEPT
A. precipitation
B. combination
C. oxidation-reduction
D. formation

Answer is A: Although a solid is formed in the reaction, it is not formed from dissolved species, so the reaction is not an example of a precipitation reaction. Therefore, (A) is correct.

What does this explanation mean?



All of the following are state functions EXCEPT:
A. Volume
B. Entropy
C. Work
D. Enthalpy

Answer is C Work. When the magnitude of a quantity depends only on the initial and final states of the system, it is known as a state function. Work depends on the path taken, so (C) is correct.

I have been noticing that in chemistry work depends on path, but in physics it doesn't right? In physics work is path independent since it is F*d and so we are only concerned with displacement. In thermo though work is path dependent?

For the first question, you should attack it using process of elimination. D is correct because you are forming a substance from two elements. C is correct because the oxidation/reduction states of the elements change from 0to different values in the product. B is correct because you are combining two elements to form the product. That leaves A, the answer. A is incorrect because of solubility rules. Remember, all compounds which contain alkali cations are soluble. Thus, calcium fluoride, although a solid, is not a precipitate.

In thermodynamics, a state function by definition is a property of the system that depends on the current state of the system and does not concern how the system got to the current state. Mechanical work, PV work, is a transition between states. That is why, the first law is actually written using the lower case delta implying small incremental changes. For the MCAT, do not worry too much as to what this means. Just remember, in thermodynamics, PV work is path dependent. In physics, work is path dependent because displacement is a vector--it depends on how something arrives at its final state. Again, don't worry too much about the details.

 

[PoorSanDiegan]

Freezing point depression question:

A 1.0 molal solution of choloracetic acid (ClCH2COOH) freezes at -1.93 celsius. Use these data to find the Ka of chloroacetic acid. Assume molarities equal the molalities.

The answer is .0227.

I think you use Kf = 1.86 somehow? This is a question out of a chemistry book. Do I need more information?

 

[killinsound]

Just say there is a polyprotic acid:

H2SO4

First Ka is the dissociation of H2SO4 -> H+ and HSO4-

Second HSO4 -> H+ + SO4-

And then it asked which quantities are equal...

They answered that H+ of the first rxn and H+ of the second rxn are equal! How can this be?

Are they assuming that since the 2nd reaction is so infavorable that basically the 1st rxn H+ will equal the 2nd rxn H+ because basically the 2nd rxn is creating negligible amounts of protons?

 

[Nevadanteater]

Freezing point depression question:

A 1.0 molal solution of choloracetic acid (ClCH2COOH) freezes at -1.93 celsius. Use these data to find the Ka of chloroacetic acid. Assume molarities equal the molalities.

The answer is .0227.

I think you use Kf = 1.86 somehow? This is a question out of a chemistry book. Do I need more information?

i did this problem three times and I couldn't get the answer .0227...
so, i looked up the actual Ka of CAA and it is .0014, which is the answer i keep coming up with.

-----------------------------------------

the key to this problem is that freezing point depression is dependent on the amount of solute disolved not the identity.

We expect a 1m solution of CAA to freeze at -1.86 because of the Kf value. But we see that it actually freezes at a slightly lower temperature. this tells us that CAA is splitting up resulting in more solute particles in solution and thus a slightly larger effect on freezing point.

remember:

freezing point depression = m * Kf * i

where i is the number of particles per molal in solution.

for example, NaCl completely solvates into Na+ and Cl- ions, thus i = 2 and doubling the effect of dissolving a 1m solution.

so in this case the CAA is splitting up, but (as with all weak acids) incompletely.

HA ----> H+ + A-

the value for i is the addition of the molarity/molality of all three of these

i = [HA] + [H+] + [A-]

also, using the freezing point depression we solve for i

i = m * Td / deltaT

i = 1.0376

now we know we start with [HA] = 1

and for every molecule of HA that splits up we lose one HA but gain one H+ and one A-

so if we set x = each molecule of HA that splits up we get

[HA] = 1-x
[H+] = x
[A-] = x

from here we can solve for [H+] and then using the definition of an acid dissassociation constant solve for Ka.

does that make sense?

N/A