General Chemistry Question Thread 2

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dcohen said:
Is it possible the answer's wrong for the fist one? Calculating it out, 11 = -log(x), x = 10^-11. Dilute by 100, x = 10^-13 so unless I'm missing something pH should be 13 as you said. Furthermore, if you dilute an acid how does it become more acidic (9)? It should become more basic (13). Perhaps someone else can chime in.
Click to expand...
Well...no. I hate thinking of basic solutions rather than acidic solutions...so let's put this in terms on an acidic solution. Suppose we had a solution of pH 1 and we dilute it 100 times, would it make sense for the pH to decrease to -1? Similarly, if you have a basic solution, and you dilute it, it cannot become more basic. Diluting a solution will generally get you closer to pH 7 (neutral solution).
 
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thanks for the explanation, but is there a reason why this cannot be derived mathematically the way dcohen is trying to do? It makes sense but I just dont understand why it cant be proven using the H+ and pH relationship.
 
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Franksta1118 said:
thanks for the explanation, but is there a reason why this cannot be derived mathematically the way dcohen is trying to do? It makes sense but I just dont understand why it cant be proven using the H+ and pH relationship.
Click to expand...
I was trying to avoid doing the math because it sucks typing it out...
So, you have an initial conc of [H+]=10^-11. We are going to dilute it 100 times (add lots of H2O). But the problem is that the conc of [H+] in the H2O we are adding is much much more than that in solution. [H+]=10^-7 for the solution. This really ruins our math and makes it much harder to calculate. So, let's go another way with this. If pH=11, what does pOH equal? Well, pH+pOH=14, so pOH=3. Then, the [OH-]=10^-3. Well, If we dilute it with a liquid that has a [OH-]=10^-7, it really wont make a difference on our conc of OH-. Therefore, we can use the [OH-] to determine our new pH. well, diluting the [OH-] by 100 gives us a new [OH-]=10^-5. The [H+] would then equal 10^-9. and the pH=9.
Hope that helps...and im sorry if it confused you further...:luck:
 
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Wow, great explanation. That makes perfect sense. Yea, I see my problem now, I was neglecting the H+ from the dissociation of water which is in fact much more significant in comparison to the H+ in our initial solution. So, is it safe to say that diluting anything with water will always bring the pH closer to that of water? So we can just see that 100 fold dilution changes the pH by 2 units and it MUST get closer to that of water (pH=7). I'm just thinking about this for future questions, to have a quick way of checking myself.
 
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Franksta1118 said:
Wow, great explanation. That makes perfect sense. Yea, I see my problem now, I was neglecting the H+ from the dissociation of water which is in fact much more significant in comparison to the H+ in our initial solution. So, is it safe to say that diluting anything with water will always bring the pH closer to that of water? So we can just see that 100 fold dilution changes the pH by 2 units and it MUST get closer to that of water (pH=7). I'm just thinking about this for future questions, to have a quick way of checking myself.
Click to expand...
EXACTLY!!! that's the way i do it. i never go through the calculations, because it takes way too long. usually, its enough to look through the answer choices and make a reasonable guess. im glad that helped...:luck:
 
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ZnCl3.OH
I will try and take a stab here. I am pretty sure the logic fileserver used is correct. According to my MCAT instructors, anytime an electron is gained or lost you adjust the OS accordingly. Here H has its usual OS of +1 and Cl each is -1, Oxygen is usually a -2 but here it lost an electron so to adjust we must have oxygen being -1. Therefore the OH combined has an OS of zero. So now, Cl3 is 3- and Zn must be 3+ to add up to the total charge of 0. I think the lesson to be learned is that just as gaining electrons to attain an anionic form will decrease the OS by 1, losing an electron will increase the OS by 1. So oxygen goes from being 2- to being 1- (just as it is in any peroxide).
Hope that helps.
 
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the 2nd law of thermodynamics essentially states that all spontaneous processes go towards an increase in entropy.

i'm confused. if a process is spontaneous, then delta G MUST be negative. Delta G could be negative EVEN if entropy is negative, if we have a very negative delta H.

can someone please let me know what they think?

thank you very much:):luck:
 
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phospho said:
the 2nd law of thermodynamics essentially states that all spontaneous processes go towards an increase in entropy.

i'm confused. if a process is spontaneous, then delta G MUST be negative. Delta G could be negative EVEN if entropy is negative, if we have a very negative delta H.

can someone please let me know what they think?

thank you very much:):luck:
Click to expand...
Be careful when using the 2nd law. It says that everything tends to disorder, yes. But your system itself doesnt have to. For a system, the change in entropy could be negative.
Even then, if entropy is positive for the system, delta G could still be positive--> if you have a really endothermic rxn (delta H is very positive). The only times you can determine with certainty whether a rxn is spontaneous or not spontaneous is if delta S and delta H had opposite signs.
Hope this helps?...:luck:
 
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jgberken said:
Be careful when using the 2nd law. It says that everything tends to disorder, yes. But your system itself doesnt have to. For a system, the change in entropy could be negative.
Even then, if entropy is positive for the system, delta G could still be positive--> if you have a really endothermic rxn (delta H is very positive). The only times you can determine with certainty whether a rxn is spontaneous or not spontaneous is if delta S and delta H had opposite signs.
Hope this helps?...:luck:
Click to expand...

makes perfect sense, thank you very much! :)
 
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i'm sorry for being so annoying with my questions - i'm just trying to tie up my loose ends before i put away the review material for good :oops:

We know that adding solute to something pure will weaken its intermolecular forces. We also know that weaker intermolecular forces means we will have a higher vapor pressure. However, why is it that whenever we add solute to something, the vapor pressure actually decreases?

After googling the concepts, it seems that this is because the molecules of the solute somehow replace the molecules of the solvent on the surface. I'm confused - it seems too simplistic of an approach. What if we add the solute and then stir the solute-solvent mixture? Wouldn't that eliminate the "problem" of having the solute linger at the surface?

I guess my real question is this: Before today, I thought that the fact that we weakened the intermolecular forces by putting the solute in the solvent should dominate over the effect of "solute molecules lingering at the surface". Can someone give me a logical explanation of why solute lowers the vapor pressure of a solvent?

thanks in advance for your time, and again, sorry to be so annoying...:luck:
 
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phospho said:
i'm sorry for being so annoying with my questions - i'm just trying to tie up my loose ends before i put away the review material for good :oops:

We know that adding solute to something pure will weaken its intermolecular forces. We also know that weaker intermolecular forces means we will have a higher vapor pressure. However, why is it that whenever we add solute to something, the vapor pressure actually decreases?

After googling the concepts, it seems that this is because the molecules of the solute somehow replace the molecules of the solvent on the surface. I'm confused - it seems too simplistic of an approach. What if we add the solute and then stir the solute-solvent mixture? Wouldn't that eliminate the "problem" of having the solute linger at the surface?

I guess my real question is this: Before today, I thought that the fact that we weakened the intermolecular forces by putting the solute in the solvent should dominate over the effect of "solute molecules lingering at the surface". Can someone give me a logical explanation of why solute lowers the vapor pressure of a solvent?

thanks in advance for your time, and again, sorry to be so annoying...:luck:
Click to expand...
Whoa...we're not decreasing the intermolecular forces by adding a solute, we're increasing them. We know that a solute has a charge. We also know that ion-dipole forces are stronger than dipole-dipole forces. Therefore, we are strengthening the intermolecular forces by adding a solute. And I completely disagree with the solute accumulating at the surface. That's soo wrong. You are right, it's stupid.
When a solute is dissolved in a solvent. You need to add a greater amount of heat to get the solvent in the gas phase, and, therefore, the vapor pressure decreases.
:luck:
 
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jgberken said:
Whoa...we're not decreasing the intermolecular forces by adding a solute, we're increasing them. We know that a solute has a charge. We also know that ion-dipole forces are stronger than dipole-dipole forces. Therefore, we are strengthening the intermolecular forces by adding a solute. And I completely disagree with the solute accumulating at the surface. That's soo wrong. You are right, it's stupid.
When a solute is dissolved in a solvent. You need to add a greater amount of heat to get the solvent in the gas phase, and, therefore, the vapor pressure decreases.
:luck:
Click to expand...


thanks for explaining this to me :)

You are saying that adding solute would increase the intermolecular forces. That definitely makes sense when we bring up the fact that boiling point elevation takes place in the presence of a solute in a pure solvent.

However, when we add solute to a pure solvent, we decrease its freezing point, which means we are decreasing its melting point. When something takes less heat to break up its solid molecules into a liquid, doesn't that mean that it has weaker intermolecular forces?

thank you! :love:
 
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phospho said:
thanks for explaining this to me :)

You are saying that adding solute would increase the intermolecular forces. That definitely makes sense when we bring up the fact that boiling point elevation takes place in the presence of a solute in a pure solvent.

However, when we add solute to a pure solvent, we decrease its freezing point, which means we are decreasing its melting point. When something takes less heat to break up its solid molecules into a liquid, doesn't that mean that it has weaker intermolecular forces?

thank you! :love:
Click to expand...
Exactly! The difference between freezing/melting and boiling is that freezing/melting requires crystallization. Crystallization needs the molecules to be stacked up perfectly with one another. Boiling just requires getting it out of the liquid phase. So, if we have a solute (ie salt) in a solvent (ie H20), it would be harder for the water molcules to crystallize and stack up against one another in the presence on the Na+ and Cl-. For this reason, melting point and freezing point decrease (lower temperatures) in the presence of a solute.
Does that make sense or was that horrible confusing...?
 
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jgberken said:
Exactly! The difference between freezing/melting and boiling is that freezing/melting requires crystallization. Crystallization needs the molecules to be stacked up perfectly with one another. Boiling just requires getting it out of the liquid phase. So, if we have a solute (ie salt) in a solvent (ie H20), it would be harder for the water molcules to crystallize and stack up against one another in the presence on the Na+ and Cl-. For this reason, melting point and freezing point decrease (lower temperatures) in the presence of a solute.
Does that make sense or was that horrible confusing...?
Click to expand...

that was perfect!! makes much more sense now, especially if we consider the kinetic theory stuff and how lowering temp lowers kinetic energy which lowers the speed which makes everything clump together a bit more which lowers the volume.

i feel like my questions are really stupid and that I may need more time with general chemistry than i thought i would. Someone like me who is "preparing to put away the notes" should know this stuff cold. PS is scaring the crap out of me. :oops:

thank you so so much for being patient with me
 
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Wait... just wanted to make a small correction. It is boiling point elevation and freezing point depression. So, freezing goes down and boiling goes up.
 
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I just realized that all my confusion started when I read in the Kaplan book this:

"The melting/freezing point of an alloy will be lower than that of either of the component metals, because the new bonds are weaker".
 
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tncekm said:
Wait... just wanted to make a small correction. It is boiling point elevation and freezing point depression. So, freezing goes down and boiling goes up.
Click to expand...

yep, i figured thats what it was, thank you for the heads up though... BP goes up since we need more heat to put stuff in the vapor form, and we need lower temp to make stuff stack together better to crystallize...
 
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phospho said:
yep, i figured thats what it was, thank you for the heads up though... BP goes up since we need more heat to put stuff in the vapor form, and we need lower temp to make stuff stack together better to crystallize...
Click to expand...
Bingo. You've got it.
 
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phospho said:
I just realized that all my confusion started when I read in the Kaplan book this:

"The melting/freezing point of an alloy will be lower than that of either of the component metals, because the new bonds are weaker".
Click to expand...

Yeah, I can see that getting confusing. I'm confused just looking at it :hardy:
 
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i have what will probably be a ridiculously easy question for some of you, but gen chem is not my forte...

i'm calculating ΔT for the reaction Q= MCΔT but I have 2 different initial temperatures of solutions...an aqueous KOH solution at 23C and an aqueous HCl solution at 21C. What do I use as the initial temp? I'm adding the KOH solution to the HCL solution.

thanks for any help!
 
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If you're calculating ΔT and you already have mass (m) and specific heat (c) --or heat capacity m*c = C-- and you've got the value of heat evolved (Q), then just rearrange to:

ΔT=Q/(mc) and solve for ΔT.

I'm not quite sure why you would have two different solutions, and two different temperatures in the question...Maybe if you posted the question in its entirety it would be easier for us to help you out with.
 
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When using Q=mc(delta T), its best to try and use a positive value for your change in temp. So in this case if you have one solution at 23 degrees and one at 21 the final temperature will obviously fall somewhere between the two once the heat is transferred. Therefore, for the solution that is initially at 21 degrees delta T= Tfinal-21 so that its a positive value. For the solution initially at 23 degrees delta T= 23 - Tfinal so that it too is a positive value. This will ensure that you obtain the correct solution.
 
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This is a question from a practice exam so please keep that in mind when answering this:

What was the most likely identity of the precipitate that formed when NaCl(aq) was added to Cd(NO3)2(aq)?

A) Cd2Cl
B) CdCl
C) CdCl2
D) CdCl3

How do you solve this quickly on the MCAT and predict the product? I never understood how you can predict products accurately.
 
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This is the same type of question I think, from the same passage:

When a strip of Cu is placed into H2O(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 likely identity of the gas?

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

How do you predict these things?!
 
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w1ll said:
This is a question from a practice exam so please keep that in mind when answering this:

What was the most likely identity of the precipitate that formed when NaCl(aq) was added to Cd(NO3)2(aq)?

A) Cd2Cl
B) CdCl
C) CdCl2
D) CdCl3

How do you solve this quickly on the MCAT and predict the product? I never understood how you can predict products accurately.
Click to expand...

my guess is that the the correct answer is choice [C]
Given that you know: NaCl --> Na+ and Cl-
then...
-having Cd2Cl gives Cd an oxidation number of +1/2 .... unlikely. Strike out [A]

-Cd(NO3)2 was the original compound, so Cd must have at least an oxidation # of +2, which would mean Cl- is actually ox# -2...its not. strike out choice

-CdCl3 implies that Cd must have ox#=3+. If it did, it would make NO3 have a charge of -1.5 (its easier just to remember NO3 has a 1- charge). Strike out choice [D]
 
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w1ll said:
This is the same type of question I think, from the same passage:

When a strip of Cu is placed into H2O(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 likely identity of the gas?

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

How do you predict these things?!
Click to expand...

1) HNO3(aq) will dissociate into H+ and NO3-
2) Cu will give off 2e- to become Cu2+
3) the 2e- will be attracted to 2H+ to form H2 (a diatomic molecule)
Check: H2 is gas. problem works out.

thats my best go at it. hope it helps.
-matt
 
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jgberken said:
Whoa...we're not decreasing the intermolecular forces by adding a solute, we're increasing them. We know that a solute has a charge. We also know that ion-dipole forces are stronger than dipole-dipole forces. Therefore, we are strengthening the intermolecular forces by adding a solute. And I completely disagree with the solute accumulating at the surface. That's soo wrong. You are right, it's stupid.
When a solute is dissolved in a solvent. You need to add a greater amount of heat to get the solvent in the gas phase, and, therefore, the vapor pressure decreases.
:luck:
Click to expand...

the solute doesnt necessarily have to have a charge.
i think of it like this: I am boiling water for my spaghetti (the uncharged flavor)
once it starts to boil, i add the strands of spaghetti--then the water seems to stop boiling...
why?
recalling the definition of boiling point, we know that its the temperature at which the vapor pressure = atmospheric pressure.
and since more H2O molecules are interacting with the spaghetti, it means that less are allowed to enter the vapor phase.
this means we have to provide more energy before we can make the vapor pressure = atmospheric pressure, therefore, BP elevation.
 
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w1ll said:
This is the same type of question I think, from the same passage:

When a strip of Cu is placed into H2O(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 likely identity of the gas?

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

How do you predict these things?!
Click to expand...

Believe it or not, I think it's actually A. I remember having this problem, and the solution said NO gas would evolve because NO3 oxidizes Cu, and the product is NO.

MY original answer was C as well, but all the solution said was that it was possible...but much less plausible.
 
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MrMattOglesby said:
1) HNO3(aq) will dissociate into H+ and NO3-
2) Cu will give off 2e- to become Cu2+
3) the 2e- will be attracted to 2H+ to form H2 (a diatomic molecule)
Check: H2 is gas. problem works out.

thats my best go at it. hope it helps.
-matt
Click to expand...

If this was the case why would there be no gas evolved from Cu in H2O? H2O is a combination of H+ and OH- so theoretically if the gas was H2 you would have it evolve from both. Can't be CO2 because no carbon so that leaves you with O3 and NO. My initial hunch was NO but I can't really reason why not O3. Possibly because it's a redox reaction and for some reason nitrogen is a better electron acceptor than the oxygen.
 
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dcohen said:
If this was the case why would there be no gas evolved from Cu in H2O? H2O is a combination of H+ and OH- so theoretically if the gas was H2 you would have it evolve from both. Can't be CO2 because no carbon so that leaves you with O3 and NO. My initial hunch was NO but I can't really reason why not O3. Possibly because it's a redox reaction and for some reason nitrogen is a better electron acceptor than the oxygen.
Click to expand...
nice...good call! I was able to cross out CO2 for the same reason and O3 becuase O3 would mean that we are oxidizing the HNO3 and therefore reducing the Cu. But, the Cu could not and would not be reduced. It is oxidized...so we must be releasing something reduced: either H2 or NO. and as you said, it's NO.
good question. I remember seeing it somewhere, no idea where.
 
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MrMattOglesby said:
my guess is that the the correct answer is choice [C]
Given that you know: NaCl --> Na+ and Cl-
then...
-having Cd2Cl gives Cd an oxidation number of +1/2 .... unlikely. Strike out [A]

-Cd(NO3)2 was the original compound, so Cd must have at least an oxidation # of +2, which would mean Cl- is actually ox# -2...its not. strike out choice

-CdCl3 implies that Cd must have ox#=3+. If it did, it would make NO3 have a charge of -1.5 (its easier just to remember NO3 has a 1- charge). Strike out choice [D]
Click to expand...


In this problem, the reason why Cd keeps its oxidation number is because it is aq to aq where as the other problem was a redox reaction?
 
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any replies to that last question? Also got another:

Given:
Ag+ + e- → Ag Eo = +0.80 V
Cu+ + e- → Cu Eo = +0.52 V
Pb2+ + 2 e- → Pb Eo = -0.13 V
Zn2+ + 2 e- → Zn Eo = -0.76 V

With which of the above metals can copper form a galvanic cell in which copper is reduced?

A) With silver only
B) With lead only
C) With lead and zinc
D) With silver and zinc
 
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w1ll said:
any replies to that last question? Also got another:

Given:
Ag+ + e- → Ag Eo = +0.80 V
Cu+ + e- → Cu Eo = +0.52 V
Pb2+ + 2 e- → Pb Eo = -0.13 V
Zn2+ + 2 e- → Zn Eo = -0.76 V

With which of the above metals can copper form a galvanic cell in which copper is reduced?

A) With silver only
B) With lead only
C) With lead and zinc
D) With silver and zinc
Click to expand...

So what you have here is a list of reduction potentials. As the value increases, a species is more and more likely to be reduced. If you want copper to be reduced, it must have a higher reduction potential that the other species in the redox reaction. The only species with lower reduction potentials than copper are zinc and lead. Either lead or zinc could serve as the anode. Your answer is C.
 
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Also to simplify a bit, a galvanic cell has a positive reduction potential. You know copper gains electrons because it's reduced so it's fine as it is written there (positive reduction potential).

To find which ones would make a galvanic cell with copper you need to reverse the equations so the electrons are on the other side. If you reverse the equation you also need to reverse the sign on the reduction potential. Lead and zinc are the only 2 that turn out to have a positive reduction potential, silver has a negative one. Thus C is the answer.
 
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Why would something like say Ba(OH)2 be more soluble in 1.0M HCl than 1.0m NaOH versus maybe Hf? or MgCl?

I know we aren't supposed to post actual MCAT questions so I changed the substances a bit.
 
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This looks like a Le Chatlier's principle question.

Ba(OH)₂ will dissociate into Ba² + 2OH⁻ in an aqueous solution.

Ba(OH)₂ --> Ba² + 2OH⁻

If you've got a 1.0M NaOH solution you've got additional OH⁻ ions all over the place that are limiting the amount of Ba(OH)₂ that can dissolve, driving the reaction toward the left.

If you're adding it to a 1.0M HCl solution not only with the hydroxide ions dissociate, they'll leave the equilibrium expression entirely as they form water, driving the reaction toward the right.

MgCl will include only spectator ions in the Ba(OH)₂ dissociation, so they will not effect it. However, that being said, not being effected means more is dissolved that in the presence of NaOH.

So, for its solubility you'd see this trend:

Ba(OH)₂ is more soluble in HCl than MgCl and more soluble in MgCl than NaOH and this is all because of what Le Chatlier has taught us about how the ions from HCl, MgCl, and NaOH will effect (or not) Ba(OH)₂'s dissociation.
 
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tncekm said:
MgCl will include only spectator ions in the Ba(OH)₂ dissociation, so they will not effect it. However, that being said, not being effected means more is dissolved that in the presence of NaOH.

So, for its solubility you'd see this trend:

Ba(OH)₂ is more soluble in HCl than MgCl and more soluble in MgCl than NaOH and this is all because of what Le Chatlier has taught us about how the ions from HCl, MgCl, and NaOH will effect (or not) Ba(OH)₂'s dissociation.
Click to expand...

Magnesium hydroxide is very insoluble in water with a ksp of 1.5 x 10^-11. This will shift the reaction to the right as the hydroxide is sequestered by the magnesium making barium hydroxide more soluble. To get a trend in solubility you would have to know how much magnesium chloride your starting with, but if it was 1 M solution for each it would probably go:

MgCl2 > HCl > HF > NaOH

with barium hydroxide being the most soluble in MgCl2 and least soluble in NaOH.
 
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SketchLazy said:
Magnesium hydroxide is very insoluble in water with a ksp of 1.5 x 10^-11. This will shift the reaction to the right as the hydroxide is sequestered by the magnesium making barium hydroxide more soluble. To get a trend in solubility you would have to know how much magnesium chloride your starting with, but if it was 1 M solution for each it would probably go:

MgCl2 > HCl > HF > NaOH

with barium hydroxide being the most soluble in MgCl2 and least soluble in NaOH.
Click to expand...
Okay, gotcha. So, the methodology in my answer was correct (it was a Le Chatlier's principle question), but this would have required outside knowledge of the solubility of Mg(OH)2 to get the entire trend developed. I'd imagine that if a question like that were to come up there would be a passage that would give information on the solubility of Mg(OH)2, right?
 
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SketchLazy said:
Magnesium hydroxide is very insoluble in water with a ksp of 1.5 x 10^-11. This will shift the reaction to the right as the hydroxide is sequestered by the magnesium making barium hydroxide more soluble. To get a trend in solubility you would have to know how much magnesium chloride your starting with, but if it was 1 M solution for each it would probably go:

MgCl2 > HCl > HF > NaOH

with barium hydroxide being the most soluble in MgCl2 and least soluble in NaOH.
Click to expand...
How would the solubility of Magnesium Hydroxide affect the initial solubility of Ba(OH)? I mean, I read what you typed but it just seems that you would have to have Magnesium Hydroxide form in order to even put that into the equation of affecting solublity, right?

I'm just confused as to how the solublity of a product that did not even form works in this case?
 
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Kaustikos said:
How would the solubility of Magnesium Hydroxide affect the initial solubility of Ba(OH)? I mean, I read what you typed but it just seems that you would have to have Magnesium Hydroxide form in order to even put that into the equation of affecting solublity, right?

I'm just confused as to how the solublity of a product that did not even form works in this case?
Click to expand...

If you don't have the solubility constant or the concentration of MgCl2, then you wouldn't be able to figure out a trend. But let's just say you are given the Ksp's of both Barium hydroxide and magnesium hyrdoxide. The Ksp for magnesium hyroxide is lower than the Ksp for Barium hydroxide. So if there is any hydroxide present in solution, it will react with the magnesium ions and precipitate as Mg(OH)2 when the MgCl2 is added. The key to understand here is that magnesium hydroxide will form once you add MgCl2 even if magnesium hydroxide wasn't in the solution to begin with.

It's the same idea as when you add acid to the solution. The reason H+ reacts so readily with OH- is because Kw is 10^14. Since the K is so high, you're more likely to have water in solution than H+ and OH-. For magnesium hydroxide, the same idea applies. The K for it's formation is the inverse of the Ksp, ~6*10^10, meaning there will mostly be magnesium hydroxide rather than Mg+2 and OH- ions in solution. Just like how the H+ reacts with the OH- to form H20 shifting the equation to the right, the Magnesium reacts with the hydroxide to form magnesium hydroxide. Either way the hydroxide is sequestered because of a high equilibrium constant favoring either H2O or Mg(OH)2.
 
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tncekm said:
Okay, gotcha. So, the methodology in my answer was correct (it was a Le Chatlier's principle question), but this would have required outside knowledge of the solubility of Mg(OH)2 to get the entire trend developed. I'd imagine that if a question like that were to come up there would be a passage that would give information on the solubility of Mg(OH)2, right?
Click to expand...

Yeah, your reasoning was right on. I can't say for sure, but I think they expect you to know basic solubility rules. It didn't come up on the official test for me, but solubility rules did come up on my practice tests.
 
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if you increase the volume in a reaction at equilibrium, it will decrease in pressure according to Le Chatelier's Principle.
and conversely, if you increase the pressure, then volume is decreased.
now;
how do i know which direction equilibrium shifts when i do either of these to a system?

here is an example equation to show off your knowledge with:

N2(g) + 3H2(g) --> 2 NH3(g)

1) which direction will the reaction shift when you have an increase in volume? why?
2) which direction will the reaction shift when you have an increase pressure? why?
 
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MrMattOglesby said:
if you increase the volume in a reaction at equilibrium, it will decrease in pressure according to Le Chatelier's Principle.
and conversely, if you increase the pressure, then volume is decreased.
now;
how do i know which direction equilibrium shifts when i do either of these to a system?

here is an example equation to show off your knowledge with:

N2(g) + 3H2(g) --> 2 NH3(g)

1) which direction will the reaction shift when you have an increase in volume? why?
2) which direction will the reaction shift when you have an increase pressure? why?
Click to expand...


Pretty sure that if you increase the pressure your reaction shifts to the side with fewer moles of gas, that is it would move to the product (2 mol NH3 vs. 4 mol H2 + N2). Assuming a raise in volume constitutes a reduction in pressure, I would think the opposite would occur. Your reaction would move to the right side.
 
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