bme post

[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)

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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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[medworm]

Thanks for the prompt reply. So in other words, pH can be directly calculated when the acid is strong, dissociates 100% as in HCl. Whereas when the substance in soln is no longer strong, doesn't dissociate 100% (as indicated by a pKa >4 or 5), we would need to know the relative amount of acid and conjugate base to determine pH. Does that sum it up??

What about the instance where you have 0.00000001 HCl? Is the pH still 2? Thanks.

 

[QofQuimica]

Thanks for the prompt reply. So in other words, pH can be directly calculated when the acid is strong, dissociates 100% as in HCl. Whereas when the substance in soln is no longer strong, doesn't dissociate 100% (as indicated by a pKa >4 or 5), we would need to know the relative amount of acid and conjugate base to determine pH. Does that sum it up??

Pretty much, except that the cutoff between strong and weak acids is generally a pKa of zero, not four or five. HCl has a pKa around -7.

What about the instance where you have 0.00000001 HCl? Is the pH still 2? Thanks.

If you mean 0.00000001 M, then no. 0.00000001 = 1 x 10^-8, so you couldn't get a pH of 2. But this brings up a great trick question, because you wouldn't get a pH of 8, either. (Doesn't make sense for an acidic solution to have a pH of 8, right?) Can you figure out what the pH of a 0.00000001 M HCl solution would be? Try to solve it, and I'll post the answer tomorrow if you need help.

 

[medworm]

Good point about the pKa. Just not used to seeing small pKa values. So would 0.00000001 M HCl be pH 7 b/c there's still not a single [OH-] in soln? Another way of looking at it is ... there's one lonely proton embedded amongst 10^8 other molecules. Kinda like your chances of winning the lotto.

 

[Nutmeg]

Good point about the pKa. Just not used to seeing small pKa values. So would 0.00000001 M HCl be pH 7 b/c there's still not a single [OH-] in soln? Another way of looking at it is ... there's one lonely proton embedded amongst 10^8 other molecules. Kinda like your chances of winning the lotto.

You need to think about the definition of pH. If pH = -log[H+], then what does that say about the concentration of H+ when the pH is neutral (ie 7)?

 

[QofQuimica]

Good point about the pKa. Just not used to seeing small pKa values. So would 0.00000001 M HCl be pH 7 b/c there's still not a single [OH-] in soln? Another way of looking at it is ... there's one lonely proton embedded amongst 10^8 other molecules. Kinda like your chances of winning the lotto.

Yes, the pH would be approximately seven. It's not one molecule in 10^8 molecules, though, because you have a second source of H+ ions. Remember that you don't only have your tiny HCl concentration of 1 x 10^-8 M. You also have H+ present due to the autoionization of water:

2 H2O <-> (H3O+) + (OH-)

At 25C, you'll have a Keq (Kw) value of 1 x 10^-14, making the concentration of H+ (and OH-) 1 x 10^-7 M. So when you add the two sources of H+ together, you end up with 1.1 x 10^-7 M, which gives you a pH slightly less than 7. (The actual value is 6.96.)

Moral of the story: always think before you calculate. This question tricks students if they plug and chug without considering what answer value is reasonable. As we already discussed, it is not reasonable to have a pH of 8 for an acidic solution, so you should realize that there must be some additional source of protons present.

 

[QofQuimica]

You need to think about the definition of pH. If pH = -log[H+], then what does that say about the concentration of H+ when the pH is neutral (ie 7)?

Oops, sorry, Nutmeg. I didn't see your new clue until after I posted.

 

[medworm]

Ohmigod -- then that also explains why adding water changes the pH of a non-buffered soln and doesn't affect the pH of a buffered soln. Did I just see the light here?

WTH, why didn't they explain it to me that way in G Chem class?

 

[Mister Pie]

I'm a little confused about how to apply these solubility rules. A practice test I took involved looking at how BaF2 affected a particular colligative property. Based on these solubulity rules, F is a halogen, and shouldn't be soluble with barium. However, the answer has barium fluoride dissociating into 3 ions, so there's a factor of three involved. Did I apply these rules incorrectly or are is the answer key wrong?

Here is SilvrGrey330's extremely clever mnemonic to remember salt solubility rules.

C A S H n Gia

Read it as "Cashin' Gia"...how to remember that? well the story is...im a pimp...and gia is my hoe, and i need to get my cash from her. hence...Cashing from gia.

C is clorates, A is acetates, S is sulfates, H is halogens, n is Nitrates, and Gia is Group I A metals. ---> THESE ARE ALL SOLUBLE, XCEPT

for S: Ca, Ba, Sr.....just remember the tv network CBS
for H: Ca, Ba, Sr + Happy...whats happy? Hb Ag Pb ...mercury, silvr and lead...add a py to the end and all the first letters spell HAPPY

and if its not part of CASHnGIA...its insoluble.

 

[hippocampus]

1)what is equivalent? how does that relate to titration? For example, if they said "add one quivalent of NaOH," what does that mean?

2)Also, what is the significance of NV = NV?

3) Why do they like to say for mol/L that something is kept constant? Like if you were comparing 0.01 M of A with 0.01 M of B, you would keep something the same because it doesn't really change?

 

[QofQuimica]

I'm a little confused about how to apply these solubility rules. A practice test I took involved looking at how BaF2 affected a particular colligative property. Based on these solubulity rules, F is a halogen, and shouldn't be soluble with barium. However, the answer has barium fluoride dissociating into 3 ions, so there's a factor of three involved. Did I apply these rules incorrectly or are is the answer key wrong?

Salt solubility is a complex equilibrium, and it isn't an either-or phenomenon like a simplified set of rules makes it out to be. Books may differ on how detailed they get about the rules for predicting it, and also where they draw the line between calling a salt "soluble" versus calling it "insoluble." BaF2 is one of those salts that falls in the gray area. Its Ksp at 25 C is 2.4 x 10^-5. So it's actually probably most accurate to call it partially soluble. (It's one of the more soluble of the insoluble salts; in comparison, BaSO4 has a Ksp of 1.5 x 10^-9 and Ba3(PO3)2 has a Ksp of 6 x 10^-39; now THAT is an insoluble salt.) However, your book apparently wants you to assume that barium halide salts are soluble. In any case, try to keep in mind that the solubility rules are guides to help you predict the relative solubilities of different salts, and take them all with a grain of (ahem!) salt.

 

[QofQuimica]

1)what is equivalent? how does that relate to titration? For example, if they said "add one quivalent of NaOH," what does that mean?

It tells you the mole ratio being used. Most acids are monoprotic, so you need one mole of base to neutralize one mole of acid. This means that the most common acid/base mole ratio in titrations is 1:1. For example, if I have 2.5 moles of a monoprotic acid, then I want to add one equivalent of NaOH to neutralize it, meaning I must add 2.5 moles of NaOH. On the other hand, if I have a diprotic acid, then I must use TWO equivalents of NaOH to completely neutralize it. So 2.5 moles of diprotic acid would be neutralized by five moles of NaOH.

2)Also, what is the significance of NV = NV?

You are probably already familiar with this law written in the form M1V1 = M2V2. Basically, the law states that the moles of compound 1 must equal to the moles of compound 2. How do you get the moles? Multiplying the concentration (M) by the volume (V) gives you the moles on both sides of the equation.

The way that you have written the equation uses normality (N) instead of molarity, but the basic idea is the same. Normality is a measure of how many equivalents of protons (for an acid) or hydroxides (for a base) you have per molecule. For monoprotic acids and monobasic bases, molarity and normality will be the same. So a 1 M HCl or NaOH solution is also 1 N. For diprotic acids or dibasic bases, however, there are two equivalents of protons or hydroxides per molecule. So a 1 M H2SO4 or Ca(OH)2 solution will be 2 N. Here, you must remember to divide the normality by two to get the molarity if you measure the normality for diprotic or dibasic species. For monoprotic acids and monobasic bases, you can just convert the normality directly to molarity.

3) Why do they like to say for mol/L that something is kept constant? Like if you were comparing 0.01 M of A with 0.01 M of B, you would keep something the same because it doesn't really change?

I'm not exactly sure what you're asking here, but I'm guessing that what you mean is that the number of moles stays the same on both sides. That is, if you have 0.5 moles of your acid, you must have 0.5 moles of base to neutralize it (assuming they are monoprotic and monobasic as discussed previously). Of course, if you change the volume of the solution, you will change the molarity as well. However, the number of moles of acid or base remains constant, regardless of the solution volume you have it dissolved in.

 

[hippocampus]

^^ for what you said above, were you just comparing strong acid and strong bases, or everything?

lets say you had 1M of acetic acid, which is monoprotic, does that mean you need 1M, 1 equivalent, of NaOH to neutralize it?

Also, I read somewhere something about if you start with the same moles, the number of equivalents will be the same. And then they relate this to the equivalence point. How can that be? It shouldn't be the same for 1M of HCl and 1M HC2H3O2 right?

Thanks

 

[QofQuimica]

^^ for what you said above, were you just comparing strong acid and strong bases, or everything?

lets say you had 1M of acetic acid, which is monoprotic, does that mean you need 1M, 1 equivalent, of NaOH to neutralize it?

Also, I read somewhere something about if you start with the same moles, the number of equivalents will be the same. And then they relate this to the equivalence point. How can that be? It shouldn't be the same for 1M of HCl and 1M HC2H3O2 right?

Thanks

No, this is true for all acids; it doesn't matter whether the acid is strong or weak as long as the base is strong enough to deprotonate it completely. But the acid and base don't need to be the same molarities. You can use 0.5 M NaOH to neutralize 1 M acetic acid if you want; you'll just need twice the volume of base to end up with the same number of moles of acid and base, that's all. The moles of acid must be equal to the moles of base at the equivalence point; you can adjust the volumes of either (usually your base) if they are different molarities to make the moles equal.

 

[Mister Pie]

Salt solubility is a complex equilibrium, and it isn't an either-or phenomenon like a simplified set of rules makes it out to be. Books may differ on how detailed they get about the rules for predicting it, and also where they draw the line between calling a salt "soluble" versus calling it "insoluble." BaF2 is one of those salts that falls in the gray area. Its Ksp at 25 C is 2.4 x 10^-5. So it's actually probably most accurate to call it partially soluble. (It's one of the more soluble of the insoluble salts; in comparison, BaSO4 has a Ksp of 1.5 x 10^-9 and Ba3(PO3)2 has a Ksp of 6 x 10^-39; now THAT is an insoluble salt.) However, your book apparently wants you to assume that barium halide salts are soluble. In any case, try to keep in mind that the solubility rules are guides to help you predict the relative solubilities of different salts, and take them all with a grain of (ahem!) salt.

Thanks! That explained it for me. Stuff like this bothers me though: am I to assume that BaF2 should be soluble on the test or not? (I realize you're not psychic, but what would you do?)

Another unrelated question: I always thought that the ionization energy for an atom that was half filled would be greater than the atoms that were directly around it. However, I was taking another practice test and they seemed to go purely by ionization trends. I'm still unclear about which imparts more stability, I guess.

 

[QofQuimica]

Thanks! That explained it for me. Stuff like this bothers me though: am I to assume that BaF2 should be soluble on the test or not? (I realize you're not psychic, but what would you do?)

I think I would go with it being soluble in a colligative properties context, and not assume it was insoluble unless the question asked me to calculate the Ksp. The thing is, the MCAT only tests you at freshman general chemistry level. You don't need upper level coursework or a Ph.D. in chemistry; you just need to be very comfortable with freshman level gen chem and sophomore organic, and that's it.

Another unrelated question: I always thought that the ionization energy for an atom that was half filled would be greater than the atoms that were directly around it. However, I was taking another practice test and they seemed to go purely by ionization trends. I'm still unclear about which imparts more stability, I guess.

I think that this is probably the same kind of issue as before. I agree with you that there should be exceptions for half-filled and filled subshells. We do a problem in class about oxygen having a lower IE than nitrogen. But maybe that book doesn't go to this level of detail.

 

[mostwanted]

Hi Q, can you please explain the circumstances for U or E= Q-W and U = Q+W, like wht's the difference and does the Q value just represent the enthalpy, and does this equation say anything about gibbs free energy?

 

[hippocampus]

when they say the boiling point increased, they mean from 95C to 150C.

when they say the melting point increases, does it mean:
a) -14C to -256
b) -14C to 76C


Which one? they keep changing it...

 

[QofQuimica]

Hi Q, can you please explain the circumstances for U or E= Q-W and U = Q+W, like wht's the difference and does the Q value just represent the enthalpy, and does this equation say anything about gibbs free energy?

This is the equation for the First Law of Thermodynamics, where U (or E) is the internal energy, Q is the heat, and W is the work. There are two conventions, one with a minus sign (physics convention) and one with a plus sign (chemistry convention). The two equations work the same way, but you must be mindful of the different sign conventions for work for each equation. In both cases, heat being released (an exothermic reaction) gives you a negative Q, and heat being absorbed (an endothermic reaction) gives you a positive Q. But the sign conventions for work are opposite for the two equations. If you use the chemistry form (U = Q + W, which I prefer), then work done ON the system ("endowork") is positive, and work done BY the system ("exowork") is negative. If you would like to use the physics form (U = Q - W), then work done ON the system is negative, and work done BY the system is positive.

This equation does not tell you about spontaneity by itself, because it does not take into account entropy.

 

[QofQuimica]

when they say the boiling point increased, they mean from 95C to 150C.

when they say the melting point increases, does it mean:
a) -14C to -256
b) -14C to 76C


Which one? they keep changing it...

It can't be A, because going from -14 to -256 is a DECREASE, not an increase. Remember that -1 is the GREATEST negative integer, not the smallest like +1 is for the positive integers!

 

[mostwanted]

how do you get lewis structures for molecules in which carbon is not the central atom? for example N2H2 i know the structure now, but i thought u subtract 1 from 14 (total number of valence elecrons) when there is an N bond? i guess u only do that when carbon is the central atom

 

[QofQuimica]

how do you get lewis structures for molecules in which carbon is not the central atom? for example N2H2 i know the structure now, but i thought u subtract 1 from 14 (total number of valence elecrons) when there is an N bond? i guess u only do that when carbon is the central atom

I would add together all of the valence electrons: 10 for the two Ns, and 2 for the two Hs to give 12 total. In order to get an octet for each N, you'd put a double bond between them. I'm sorry, but I don't know what you're talking about as far as subtracting 1 from 14 for a N.

 

[mostwanted]

i'm sorry Q i meant N2H4, see what i mean now?

and one more thing, i'm a bit confused about the volume aspect of the real gas deviations. For example, at very high pressure, the gas liquifies, i'm guessing the volume of the real gas would be higher than predicted by the ideal gas rules, because real gas molecules have some volume, but wouldn't it also be less due to physical interactions between the molecules. for ideal gas deviations i'm having trouble with the aspect of volume of the container and the volume of the gas itself. can you please explain a little?

I would add together all of the valence electrons: 10 for the two Ns, and 2 for the two Hs to give 12 total. In order to get an octet for each N, you'd put a double bond between them. I'm sorry, but I don't know what you're talking about as far as subtracting 1 from 14 for a N.

 

[QofQuimica]

i'm sorry Q i meant N2H4, see what i mean now?

and one more thing, i'm a bit confused about the volume aspect of the real gas deviations. For example, at very high pressure, the gas liquifies, i'm guessing the volume of the real gas would be higher than predicted by the ideal gas rules, because real gas molecules have some volume, but wouldn't it also be less due to physical interactions between the molecules. for ideal gas deviations i'm having trouble with the aspect of volume of the container and the volume of the gas itself. can you please explain a little?

Ok, then, for N2H4 (hydrazine), you will have 14 valence electrons: 5 from each N, and 1 from each H. You don't need to subtract or add anything unless your molecule has a charge on it. (Subtract one from the total number of valence electrons for a positive charge, and add one for a negative charge.) Maybe you're confusing summing the valence electrons for Lewis structures with the rules for calculating units of unsaturation from organic chem?

Check this explanation of deviations from the ideal gas law from the Explanations thread.

 

[MDhopeful023]

Hi guys....

I was wondering why BH3 is not tetrahedral... I figured that it has lone pairs that stick out of the plane therefore it is sp3 geometry... however the book says that it's sp2...

can anyone explain this?

thanks alot.

 

[Andrew99]

Everytime I think I understand this, I get it mixed up.

Does Ecell = Eox + Ered?

Is the reduction half-reaction the one with the highest numerical voltage?

If so, you make the other cell an oxidation and switch the sign on the voltage (if necessary) and then add the two up correct?

Is this procedure different in voltaic and electrolytic cells?

 

[QofQuimica]

I was wondering why BH3 is not tetrahedral... I figured that it has lone pairs that stick out of the plane therefore it is sp3 geometry... however the book says that it's sp2...

can anyone explain this?

thanks alot.

No, there are no lone pairs on BH3. Take a look at a periodic table, and you will find B in group III, right above Al. BH3 is a Lewis acid, meaning that it is electron deficient and doesn't have a complete octet. Since there are only three groups around it, then its electronic geometry as well as its molecular geometry are trigonal planar, and it has sp2-hybridization. The same is true for carbocations, incidentally.

 

[QofQuimica]

Everytime I think I understand this, I get it mixed up.

Does Ecell = Eox + Ered?

Is the reduction half-reaction the one with the highest numerical voltage?

If so, you make the other cell an oxidation and switch the sign on the voltage (if necessary) and then add the two up correct?

Is this procedure different in voltaic and electrolytic cells?

My advice is to dispense with using equations altogether, as this is likely to just confuse you and cause you to mix up the signs. I wrote a previous post explaining how to calculate Ecell without memorizing a formula.

 

[MDhopeful023]

No, there are no lone pairs on BH3. Take a look at a periodic table, and you will find B in group III, right above Al. BH3 is a Lewis acid, meaning that it is electron deficient and doesn't have a complete octet. Since there are only three groups around it, then its electronic geometry as well as its molecular geometry are trigonal planar, and it has sp2-hybridization. The same is true for carbocations, incidentally.

Hi QofQuimica... you mean B is in group 13 right? can it also be explained via electronegativity then (looking at the whole trends in the periodic table)?

 

[Nutmeg]

Hi QofQuimica... you mean B is in group 13 right? can it also be explained via electronegativity then (looking at the whole trends in the periodic table)?

There are apparently two different numbering schemes. One numbers them 1-18, the other numbers them I-VIIIA and I-VIIIB. THis chart shows both shcemes--I learned the latter in school.

I don't know what you mean about the electronegativity--I can't quite see how that would have an effect. The geometry comes out of the total electron configuration. Electronegativity effects things like the size of the orbitals, but not the shapes.

 

[QofQuimica]

There are apparently two different numbering schemes. One numbers them 1-18, the other numbers them I-VIIIA and I-VIIIB. THis chart shows both shcemes--I learned the latter in school.

Yes, this was what I meant, group IIIA, which is group III of the main group elements (as opposed to the Bs, which are all transition elements.)

Nutmeg, I have to get you to teach me how to insert images into the posts like that. Where do you store the images that you insert? I guess you have to have your own website or something? Because I've tried to do it straight from the web, but it doesn't work.

 

[Nutmeg]

Yes, this was what I meant, group IIIA, which is group III of the main group elements (as opposed to the Bs, which are all transition elements.)

Nutmeg, I have to get you to teach me how to insert images into the posts like that. Where do you store the images that you insert? I guess you have to have your own website or something? Because I've tried to do it straight from the web, but it doesn't work.

Put the letters "IMG" in the place of the word "corn", and write it like this:

[corn]http://umbra.nascom.nasa.gov/eit/images/eclipse/williams/Williams_College_wl.jpg[/corn]

To get the specific URL of the image you're trying to link, right-click the image and select "properties." It will give the address (URL) of the image itself. Or, if you find the image using google or yahoo image search, then you can click "see full-size image" next to the thumbnail, and the address in the address box is the URL of the image alone.

 

[POnuts]

I'm having trouble figuring out this partial pressure problem..

Suppose that CH4(g) reacts completely with O2(g) to form CO2(g) and H2O(g) with a total pressure of 1.2 torr. What's the partial pressure of H2O(g)?

The answer says its .8 torr....How do I find the mole fraction of H20 to get this answer??

Thanks for any help

 

[CanuckAnatomist]

Well, assuming a complete reaction to just CO2 and H2O,

1. Balance the equation. That gives you 2 H2O's on the right hand side.
2. X(H2O) = n (H2O) / n (total)
3. n(H2O)=2
4. to your n (total) of 3 (product side, 2 H2O + 1 CO2)
5. X(H2O)=2/3
6. P(H2O) = 2/3 x Ptot = 0.8 torr

kosher?

I'm having trouble figuring out this partial pressure problem..

Suppose that CH4(g) reacts completely with O2(g) to form CO2(g) and H2O(g) with a total pressure of 1.2 torr. What's the partial pressure of H2O(g)?

The answer says its .8 torr....How do I find the mole fraction of H20 to get this answer??

Thanks for any help

 

[QofQuimica]

Put the letters "IMG" in the place of the word "corn", and write it like this:

[corn]http://umbra.nascom.nasa.gov/eit/images/eclipse/williams/Williams_College_wl.jpg[/corn]

To get the specific URL of the image you're trying to link, right-click the image and select "properties." It will give the address (URL) of the image itself. Or, if you find the image using google or yahoo image search, then you can click "see full-size image" next to the thumbnail, and the address in the address box is the URL of the image alone.

 

[POnuts]

Well, assuming a complete reaction to just CO2 and H2O,

1. Balance the equation. That gives you 2 H2O's on the right hand side.
2. X(H2O) = n (H2O) / n (total)
3. n(H2O)=2
4. to your n (total) of 3 (product side, 2 H2O + 1 CO2)
5. X(H2O)=2/3
6. P(H2O) = 2/3 x Ptot = 0.8 torr

kosher?

oh...can't believe I didn't balance the rxn..thanks for the input

 

[MDhopeful023]

There are apparently two different numbering schemes. One numbers them 1-18, the other numbers them I-VIIIA and I-VIIIB. THis chart shows both shcemes--I learned the latter in school.

I don't know what you mean about the electronegativity--I can't quite see how that would have an effect. The geometry comes out of the total electron configuration. Electronegativity effects things like the size of the orbitals, but not the shapes.

Thanks alot

 

[hippocampus]

if you keep adding water to an acidic buffer solution, how come the pH doesn't change?

for example, if the water was the titrant, and the acidic buffer soln is the soln ur titrating

 

[mostwanted]

hey Q, can you please explain how acid base indicators work, and how do you pick a good indicator, and can you please give examples of some common indicators?

 

[Mutt]

I have an Electrochemistry question that is driving me nuts.

Here's the skinny:

It's an electrolytic cell. The half reactions are as follows:

.5 I2 + e- ---> I- Eo = 0.54V

H2O + e- ---> .5 H2 + OH- Eo= -0.83V

What is the potential of the cell under standard conditions?

I know the EMF equation of EMF= Ecathode - Eanode but my trouble is in assigning which is the reduction and which is the oxidation.

Here is my reasoning: The most positive Eo should be the one more likely to be reduced which should be the formation of I- because it's Eo is +0.54V, but I know that in the hydrolysis of water, the production of H2(g) is a reduction process. Yet, it's Eo is too negative...

How do I reconcile this?

 

[QofQuimica]

if you keep adding water to an acidic buffer solution, how come the pH doesn't change?

for example, if the water was the titrant, and the acidic buffer soln is the soln ur titrating

Because the pH of a buffering system depends on the ratio of the concentrations of the conjugate base and the conjugate acid. Adding water does lower the molarity of HA, but it also lowers the molarity of A-, so the ratio of [A-]/[HA] stays constant, and the pH stays constant as well. pH of a buffer only depends on pKa and the ratio of [A-]/[HA], not on the solution volume. You can easily see this by examining the Henderson-Hasselbalch equation.

 

[QofQuimica]

hey Q, can you please explain how acid base indicators work, and how do you pick a good indicator, and can you please give examples of some common indicators?

Here is a brief explanation of indicators. You should try to choose an indicator with an endpoint that is close to your titration's equivalence point. The most common indicator used in general chemistry is phenolphthalein, which turns pink around pH 9.

 

[QofQuimica]

Here is my reasoning: The most positive Eo should be the one more likely to be reduced which should be the formation of I- because it's Eo is +0.54V, but I know that in the hydrolysis of water, the production of H2(g) is a reduction process. Yet, it's Eo is too negative...

How do I reconcile this?

Ecell is supposed to be negative for an electrolytic cell. Don't rely on this formula; you're much better off if you understand what is going on. Check out this post for more help on calculating EMF.

 

[medworm]

Hi Q, can you clarify for me how HYDRIDES work? First the structure.. it is a Hydrogen atom with 2 electrons, therefore overall -1 charge? So does that make it a radical/ highly reactive/ Nucleophile? Or should I just think of it like a Cl- counterANION?? For instance in CaH2.. it is a solid, b/c it is attached to a metal? Are they not soluble then?

Also, how do I elucidate the 3D structure (bonding) of a HYDRATE? I know the H20 inserts into the little crevices of the core atom, but the # of H20 groups needed to match its oxidation state?

Sorry I'm rambling.

 

[QofQuimica]

Hi Q, can you clarify for me how HYDRIDES work? First the structure.. it is a Hydrogen atom with 2 electrons, therefore overall -1 charge?

Yes.

So does that make it a radical/ highly reactive/ Nucleophile?

A hydride can serve as a nucleophile or as a base. The most common hydride nucleophiles are NaBH4 and LAH, and the most common hydride base is NaH. Hydrides are not radicals; radicals have an unpaired electron. The reactivity of hydrides varies.

Or should I just think of it like a Cl- counterANION??

I don't understand this question. You mean that it's just a spectator ion? Not very likely; hydrides react with water to form H2 gas.

For instance in CaH2.. it is a solid, b/c it is attached to a metal? Are they not soluble then?

Hydrides will dissolve in some solvents. Keep in mind that the calcium in your example isn't neutral calcium metal; it's a Ca2+ cation.

Also, how do I elucidate the 3D structure (bonding) of a HYDRATE? I know the H20 inserts into the little crevices of the core atom, but the # of H20 groups needed to match its oxidation state?

I assume you're asking about complex ions here? This is definitely beyond the scope of the MCAT, but the most common ligand configurations are octahedral (six ligands) or tetrahedral (four ligands).

 

[N1DERL&]

HI Q!

Would you please explain the relationship between specific heat and the change in temperature: inversely proportional or directly proportional. I'm getting confused. Thank you!!!

 

[QofQuimica]

HI Q!

Would you please explain the relationship between specific heat and the change in temperature: inversely proportional or directly proportional. I'm getting confused. Thank you!!!

Hmm, I don't know if I should answer questions for you any more; you said I was crazy.

If you look at the formula relating c and (delta T), you'll see that they are multiplied together, making them inversely proportional:

Q = m*c*(delta T)

If I rearrange the equation to solve for (delta T), then it makes the relationship even more obvious:

(delta T) = Q/(m*c)

(delta T)

1/c

 

[N1DERL&]

Just for that I think I should vote for you again!

So does that mean if something has a high specific heat, it can absorb more heat so the change in temperature will be small?? And vice versa?

Thank you!

 

[QofQuimica]

Just for that I think I should vote for you again!

So does that mean if something has a high specific heat, it can absorb more heat so the change in temperature will be small?? And vice versa?

Thank you!

Exactly right. My work here is obviously done. I am going to retire from this forum, and I hereby bequeath the Gen Chem thread to N1DERL&.

 

[GCT]

Hi Q, can you clarify for me how HYDRIDES work? First the structure.. it is a Hydrogen atom with 2 electrons, therefore overall -1 charge? So does that make it a radical/ highly reactive/ Nucleophile? Or should I just think of it like a Cl- counterANION?? For instance in CaH2.. it is a solid, b/c it is attached to a metal? Are they not soluble then?

You should be more specific with your questions. Hydrides are commonly used as nucleophilic agents, for instance sodium hydride is sometimes used to deprotonate beta carbonyl compounds, such as enolates when using sodium ethanoate for instance may pose problems related to transesterification.

Also, how do I elucidate the 3D structure (bonding) of a HYDRATE? I know the H20 inserts into the little crevices of the core atom, but the # of H20 groups needed to match its oxidation state?
Sorry I'm rambling.

You'll need to refer to crystal field theory of transition metals, you find find it in a standard general chemistry text; attraction between positively charged metal and negative motif of the ligand, repulsion between the lone pairs of the ligand and the electrons in the d orbitals. Each structural/arrangement/# of ligands relates to a different explanation in persepective of the theory.

 

[N1DERL&]

Exactly right. My work here is obviously done. I am going to retire from this forum, and I hereby bequeath the Gen Chem thread to N1DERL&.

oh lordy! that deserves another vote for the crazy one!

Thanks for the help Q-ness~!