General Chemistry Question Thread 2

[gridiron]

This is a continuation of the original General Chemistry Thread.

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[Nma]

Flip the first equation to get CO2 -> C + 02 delat H=394. This gives you the conversion of diamond to graphite. Since heats of reactions are additive in exactly the same way that the reactions they belong to are additive -396 + 394 = -2 per mole x 2 mole = -4

Hope that helps!

I guess u added the enthalpies bcos diamond is changing into graphite? as opposed to if there were 2 reactions w/ given enthalpy values, and we were asked for the enthalpy change for a third one. right?
My confusion lies on the addition/subtraction.... Hreaction= Hproduct - Hreactant, meanwhile u added.

 

[blastula]

You always add them. Since I reversed the first equation delta H changed from -394 to +394.

 

[Nma]

Once again, i dont mean to be so annoying, but what about the above equation for calculating the change in enthalpy? (change of)H= (change of)Hproduct - (change of)Hreactants?
according to this equation, the reactant is subtracted from the product, and so shouldnt graphite be subtracted from diamond?

 

[RSAgator]

Once again, i dont mean to be so annoying, but what about the above equation for calculating the change in enthalpy? (change of)H= (change of)Hproduct - (change of)Hreactants?
according to this equation, the reactant is subtracted from the product, and so shouldnt graphite be subtracted from diamond?

this is a very simple application of hess's law. You're given two reactions with the enthalpy value for the reactions. The question wants you to give the enthalpy change from diamond to graphite, so you want diamond to be on the left side and graphite to be on the right side. You do this by multiplying equation 1 (graphite) by -1 to reverse its direction (as well as its enthalpy value) and then adding the two equations. The steps look something like this:

You have:

(1) Cgraphite(s) + O2(g) ---> CO2(g) (change of)H = -394kJ
(2) Cdiamond(s) + O2(g) ---> CO2(g) (change of)H = -396kJ

You want Cdiamond(s) ---> Cgraphite(s)

so: (1) * -1 ==> CO2(g) ---> O2(g) + Cgraphite(s) (change of)H = 394

note that when the reaction is reversed, so is the enthalpy of the reaction hence 394 not -394. You can do this with any reaction because the energy you get FROM a reaction should be opposite in sign to the reverse reaction. If 394 joules are released in forming a bond (in this case forming the bonds between O2 and Cgraphite to form CO2) then 394 joules must be added to break that same bond (to break CO2 down into O2 and Cgraphite) hence the opposite sign on H with the opposite directionality.

this + (2) gives you (394) + (-396) = -2. 2 mols means -4



if you add the equations you'll see

Cdiamond(s) + O2(g) + CO2(g) ---> Cgraphite(s) + O2(g) + CO2(g)

or

Cdiamond(s) ---> Cgraphite(s)

2

 

[Nma]

Thnx Dcohen (and blastula)...my confusion stemmed from whether we add or subtracted...but i get it now. Thnx again

 

[Nma]

How is it possible that "formation of liquid water" is more exothermic than "formation of water vapor"?

 

[physics junkie]

How is it possible that "formation of liquid water" is more exothermic than "formation of water vapor"?

exothermic - a process that releases energy
endothermic - a process that absorbs energy

Your question doesn't really make sense for the reasons I'll state in the next paragraph. I'll clarify the matter for you as best I can though.

Formation of liquid water from vapor is exothermic. Formation of liquid water from ice is endothermic. Formation of water vapor from liquid water can only be endothermic. Please absorb this and restate your question precisely so we can clear this up for you.

If your question is the one I'm guessing it is you'll be able to tell by this answer: When you vaporize water you are breaking hydrogen bonds. That takes a lot of energy. When you melt ice you are breaking ionic bonds but you are making hydrogen bonds...since breaking bonds is endothermic and making bonds is exothermic the two balance out better so you have a lower amount of latent heat necessary to convert ice into water than you do to convert water into vapor.

 

[blastula]

Formation of liquid water, you are going from H2 (g) + O2(g) -> H2O(l) enthalpy = -68 kcal/mole

Formation of water vapor, you are going from H2O(g) -> H2O(l) enthalpy = -10

This is because the bonds in H2O don't release as much energy as H2(g) and O2(g)

Notice that H2 (g) + O2 (g) -> H2O (g) has enthalpy = -58

Hope that clears it up!

 

[armymt]

thanks y'all for your answers... it's just that I got the audio osmosis CD's and like I said , there are stuffy Jon and Jordan is saying which really contradicts my beliefs. lol. I know they are not crack-pots, but just, haha, they're funny.

 

[Nma]

exothermic - a process that releases energy
endothermic - a process that absorbs energy

Your question doesn't really make sense for the reasons I'll state in the next paragraph. I'll clarify the matter for you as best I can though.

Formation of liquid water from vapor is exothermic. Formation of liquid water from ice is endothermic. Formation of water vapor from liquid water can only be endothermic. Please absorb this and restate your question precisely so we can clear this up for you.

If your question is the one I'm guessing it is you'll be able to tell by this answer: When you vaporize water you are breaking hydrogen bonds. That takes a lot of energy. When you melt ice you are breaking ionic bonds but you are making hydrogen bonds...since breaking bonds is endothermic and making bonds is exothermic the two balance out better so you have a lower amount of latent heat necessary to convert ice into water than you do to convert water into vapor.

I stated the question as it was asked in my "EK chemistry" book. I was confused bcos i know liquid could be formed by melting or condensation...they did not specify this in their question...but rather explained "formation of liquid" in their answer as condensation.
Thnx n e ways.

And thnx Blastula

 

[Nma]

Formation of liquid water, you are going from H2 (g) + O2(g) -> H2O(l) enthalpy = -68 kcal/mole

Formation of water vapor, you are going from H2O(g) -> H2O(l) enthalpy = -10

This is because the bonds in H2O don't release as much energy as H2(g) and O2(g)

Notice that H2 (g) + O2 (g) -> H2O (g) has enthalpy = -58

Hope that clears it up!

for formation of water vapor...dont u mean: H2O(l)-> H2O(g)

but i get it now...thnx

 

[werd]

i'll reiterate that NaCl is a molecule. although ice posesses a pattern of predictable secondary organization and interaction, it is still made of H2O molecules. although table salt posesses a crystal lattice structure, it is still made up of NaCl molecules. one may debate whether it is rigorously correct to consider them so, but NaCl an H2O are both considered to be molecules.

werd
bs biochem

 

[blastula]

for formation of water vapor...dont u mean: H2O(l)-> H2O(g)

but i get it now...thnx

Sorry about the confusion but formation of water vapor (heat of vaporization) H2O (l) -> H2O (g) has enthalpy = +10 k cal/mole

Water molecules absorb 10 kcal of heat energy per mole, so the enthalpy increases. Liquids in general increase enthalpy because gas molecules move faster and have more energy (and enthalpy) than molecules in a liquid. The same amount of heat energy is given off again when water vapor condenses:

H2O(g) -> H2O(l) enthalpy = -10

Just know that going from gas to liquid will give off more heat than going from gas to gas. Going from liquid to gas will absorb heat.

 

[NontradICUdoc]

I cannot agree that NaCl is a molecule but rather an ionic compound. Unlike Water, where the oxygen atom and the hydrogen atom share an electron, making them a molecule, Sodium gives up an electron to make its valence configuration 1S2 2S2 2P6 giving it a +1 charge. At the same time, Chlorine takes up that electron to makes its valence configuration 1S2 2S2 2P6 3S2 3P6. Giving it a -1 charge.

These 2 ions now attract each because they are of opposite charges. In addition, once table salt is added to water, the Na and the Cl dissociate and are now floating around the water as Na cations and Cl anions. This is why it is not a molecule. If you take a molecule like Glucose (C6H12O6) and add it to water, it does not dissociate. It may make hydrogen bonds if the intramolecular bonds are polar, but glucose and other molecules do not dissociate when dissolved in a solvent. Salts do because they are not molecules. They are atoms held together by attractive forces and not covalent bonds.

 

[werd]

it's unusual for a disagreement like this to arise... so i did some looking. from what i can tell, neither of us is actually wrong. many texts emphasize that ionic compounds should not be referred to as molecules. in contrast, some routinely refer to salts as "..nacl molecules..". one can even find discussion of "nacl molecules" in reputable chemistry journals (ie http://www.springerlink.com/content/x22141kgp3235053/).

it appears that neither/both is the right answer. it may come down to semantics of "molecule" vs. "molecular compound." interesting finding... but i don't think this stuff is gonna show up on an mcat

 

[armymt]

lol..it's no big deal, I was just wondering about something Jon and Jordan said..it's all good thanks

 

[Thewrongstuff]

I'm like totally confused on this too

Campbell says salt is not a molecule because it doesn't have a definite size / number of atoms and that in order for something to be a molecule it needs to be covalently bonded (hence no anions or cations)

.... Go figure...I always thought it to be a molecule

 

[majik1213]

exothermic - a process that releases energy
endothermic - a process that absorbs energy

no:

exothermic - a process that releases heat (enthalpy, deltaH > 0)
endothermic - a process that absorbs heat (deltaH < 0)
exergonic - a process that releases energy (Gibbs-free energy, deltaG > 0)
endergonic - a process that absorbs energy (deltaG < 0)

 

[BerkReviewTeach]

I'm like totally confused on this too

Campbell says salt is not a molecule because it doesn't have a definite size / number of atoms and that in order for something to be a molecule it needs to be covalently bonded (hence no anions or cations)

.... Go figure...I always thought it to be a molecule

Then I wonder how Campbell would define a molecular solid such as sodium metal. It is not an ionic compound, as all Na atoms are neutral. However, it has no definite size, because the number of metal atoms can vary. But, the metallic bonding holding the sodium atoms together is definitely covalent in nature.

While this is interesting to ponder, it really is semantics and not pertinent to the MCAT. Cool little digression, but I somehow doubt it works towards getting better at the MCAT.

 

[Vihsadas]

Well here's one question which I think could definitely come up on the MCAT (I fabricated this question):

Is the chemical formula for solid table salt (NaCl) a:

a) molecular formula
b) polymer unit in a polymeric formula
c) empirical formula
d) structural formula

I think that the only correct answer would be c (right?), and to me, this kind of question seems like completely fair game for the MCAT.

 

[armymt]

who's campbell? the soup guy?

 

[stirfryboy]

I need help understanding this equation as it relates to force attraction of an ionic bond:

=R(n+e)(n-e)/d2

can't find it anywhere
any help would be appreciated

 

[RSAgator]

I need help understanding this equation as it relates to force attraction of an ionic bond:

=R(n+e)(n-e)/d2

can't find it anywhere
any help would be appreciated

honestly not sure what you've written (can you define variables? is that 2*d or d^2?). However, it looks somewhat analogous to an equation for electrostatic force. Is that maybe (n*(positive charge on an electron))*(n*(negative charge on an electron))?

F=k*e1*e2/d^2 is the equation i'm thinking.

 

[BerkReviewTeach]

bump... and a little house cleaning to keep space on screen for current questions.

 

[phospho]

did a few searches, but couldn't find an answer for this.

Let's say you have the # of moles. You multiply that # with Avogadro's #, and your answer is the # of atoms.

But your answer could also be the number of molecules.

So does that mean that the # of atoms equals the # of molecules?

Sorry if it's s stupid question. I'm just not getting this.

Thanks in advance.

 

[armymt]

did a few searches, but couldn't find an answer for this.

Let's say you have the # of moles. You multiply that # with Avogadro's #, and your answer is the # of atoms.

But your answer could also be the number of molecules.

So does that mean that the # of atoms equals the # of molecules?

Sorry if it's s stupid question. I'm just not getting this.

Thanks in advance.

no, good question though. I too had trouble understanding this before, but basically, avogardo # is :

1) # of atoms in a single molecule in your sample

OR

2) # of mols in a single molecule in your sample

avogardo # is usually used as a conversion aid in order to eliminate all unwanted units and get the desired unit.

Sorry, I hope I helped a bit. So yes, just use it for the # of atom or the # of mols.

 

[iA-MD2013]

no, good question though. I too had trouble understanding this before, but basically, avogardo # is :

1) # of atoms in a single molecule in your sample

OR

2) # of mols in a single molecule in your sample

avogardo # is usually used as a conversion aid in order to eliminate all unwanted units and get the desired unit.

Sorry, I hope I helped a bit. So yes, just use it for the # of atom or the # of mols.

that is a good question, everyone wonders that at some point
yeah, exactly! try to think of avogadro's number just as a number. for example, we can talk about a dozen molecules, a dozen moles, or a dozen atoms. similarly we talk about 6.022x10^23.
so, you could use this number to determine the number of molecules in a mol or the number of atoms in a mol.
i hope that helped too...

 

[armymt]

that is a good question, everyone wonders that at some point
yeah, exactly! try to think of avogadro's number just as a number. for example, we can talk about a dozen molecules, a dozen moles, or a dozen atoms. similarly we talk about 6.022x10^23.
so, you could use this number to determine the number of molecules in a mol or the number of atoms in a mol.
i hope that helped too...

oh yeah, I am sorry, I was wrong. the person above me is right.

here is a revision of my answer:

avogardo # is:

1) # of atoms in a single mole in your sample

OR

2) # of molecules in a single mole in your sample

I'm sorry , thanks

 

[iA-MD2013]

oh yeah, I am sorry, I was wrong. the person above me is right.

here is a revision of my answer:

avogardo # is:

1) # of atoms in a single mole in your sample

OR

2) # of molecules in a single mole in your sample

I'm sorry , thanks

lol...i knew that was what you meant to say. you just mixed it up
but yes, what armymt said.

 

[phospho]

lol...i knew that was what you meant to say. you just mixed it up
but yes, what armymt said.

no, good question though. I too had trouble understanding this before, but basically, avogardo # is :

1) # of atoms in a single molecule in your sample

OR

2) # of mols in a single molecule in your sample

avogardo # is usually used as a conversion aid in order to eliminate all unwanted units and get the desired unit.

Sorry, I hope I helped a bit. So yes, just use it for the # of atom or the # of mols.

that is a good question, everyone wonders that at some point
yeah, exactly! try to think of avogadro's number just as a number. for example, we can talk about a dozen molecules, a dozen moles, or a dozen atoms. similarly we talk about 6.022x10^23.
so, you could use this number to determine the number of molecules in a mol or the number of atoms in a mol.
i hope that helped too...

Thank you both for your replies. They make a lot of sense. I've been googling this and I couldn't seem to get a hold of the right info. Thank you so much!

 

[armymt]

Thank you both for your replies. They make a lot of sense. I've been googling this and I couldn't seem to get a hold of the right info. Thank you so much!

you're welcome.

 

[armymt]

lol...i knew that was what you meant to say. you just mixed it up
but yes, what armymt said.

haha

 

[fileserver]

New question, copied from Boneshaw35

"if you have ZnCl3.OH what is the oxidation number of Zn. Didn't know how to handle the .OH part"

How do you approach the radical oxidation number?

 

[Kaustikos]

New question, copied from Boneshaw35

"if you have ZnCl3.OH what is the oxidation number of Zn. Didn't know how to handle the .OH part"

How do you approach the radical oxidation number?

Isn't it +3? I thought the ".OH" had little to do with oxidation numbers, but I could be wrong. I was never an expert on oxidation numbers.

 

[fileserver]

This problem is similar to the one that appeared on the MCAT. Unfortunately, I don't have the answer. If we ignore the radical, then Zn would be +4, =) , because O has -2, H +1, Cl -1 X3. So I still don't know if we should ignore the radical?

 

[Kaustikos]

This problem is similar to the one that appeared on the MCAT. Unfortunately, I don't have the answer. If we ignore the radical, then Zn would be +4, =) , because O has -2, H +1, Cl -1 X3. So I still don't know if we should ignore the radical?

edit, nvm

http://forums.studentdoctor.net/showthread.php?t=451992&page=2 shows what it should be. (+3)

 

[fileserver]

I assume that molecule is essentially Zn with three Cl in polar covalent bonds around it, with a charge of +, and OH-. Zn has a very low electronegativity so its bonds with both Cl and O are polar covalent or ionic. The period is kinda weird - I've never seen that before.

For oxidation numbers, just do Zn last. O is 1- and each Cl is 1-, so that's 4-. H is 1+, so for the compound to be neutral (though technically it splits into its ions readily), Zn must be 3+.

Thank you for reply and the reference, Kaustikos .

So, O normal is -2; but because of the radical, O is now -1 (because it lost an electron), which makes Zn +3. Anyone want to support this? I hate to see one like this on my next MCAT and put down the wrong answer.

 

[phospho]

Thank you for reply and the reference, Kaustikos .

So, O normal is -2; but because of the radical, O is now -1 (because it lost an electron), which makes Zn +3. Anyone want to support this? I hate to see one like this on my next MCAT and put down the wrong answer.

bump

i'd also like to know the answer to fileserver's question.

thanks!

 

[phospho]

Was also wondering if someone could help me out here.

Ca(OH)2 has approximately the same Ksp as CaSO4. Which of them has the greater solubility in terms of mol L-1?


(A) They both have the same solubility.[x]
(B) Ca(OH)2[x]
(C) CaSO4[x]
(D) It depends on the temperature at the time. [x]

INCORRECT:
Your Answer: C
Correct Answer: B

Here's how I thought about this:

Let's assume that Ksp = 0.5 for example

that would mean that for CaSO4, s=the square root of 0.5 which means s=0.7.

For Ca(OH)2, s=the cube root of (Ksp/4), so s= 0.5

since "s" is higher for CaSO4, then the solubility of CaSO4 would be greater, which means the answer is C. Why is it telling me that my answer is wrong though? Am I missing something?

Thanks in advance

 

[tncekm]

I think that's worded strangely. But, the only thing I can think of is that Ca(OH)2 is correct since Ca(OH)2 will dissociate into 3 ions (vs 2 for CaSO4).

 

[RSAgator]

CaSO4 ---> Ca + SO4
Ca(OH)2 ---> Ca + 2OH

The first one goes to 2s, the second to 3s. I agree it was worded strangely. You almost got it, just had to take it a step further.

 

[fileserver]

Was also wondering if someone could help me out here.

Ca(OH)2 has approximately the same Ksp as CaSO4. Which of them has the greater solubility in terms of mol L-1?


(A) They both have the same solubility.[x]
(B) Ca(OH)2[x]
(C) CaSO4[x]
(D) It depends on the temperature at the time. [x]

INCORRECT:
Your Answer: C
Correct Answer: B

Here's how I thought about this:

Let's assume that Ksp = 0.5 for example

that would mean that for CaSO4, s=the square root of 0.5 which means s=0.7.

For Ca(OH)2, s=the cube root of (Ksp/4), so s= 0.5

since "s" is higher for CaSO4, then the solubility of CaSO4 would be greater, which means the answer is C. Why is it telling me that my answer is wrong though? Am I missing something?

Thanks in advance

You forgot to multiply the s by 2.
Ksp of ca(oh)2 = x(2x)^2=4x^2; after you solve for x, you get x. But [OH]=[2x] as in the equation, so you have to multiply by 2.

GJ for knowing ksp is less than 1. Notice the answer would different if Ksp is bigger than 1.

x (2x)^2 = 4x^3 = 4000 x=10 2*10 = 20
x x = 4000 x =63

4x^3=4/1000 x=.1 =>2*.1=.2
x^2=4/1000 x=.063

PS: I still want my question answer! See above

 

[BerkReviewTeach]

Was also wondering if someone could help me out here.

Ca(OH)2 has approximately the same Ksp as CaSO4. Which of them has the greater solubility in terms of mol L-1?


(A) They both have the same solubility.[x]
(B) Ca(OH)2[x]
(C) CaSO4[x]
(D) It depends on the temperature at the time. [x]

INCORRECT:
Your Answer: C
Correct Answer: B

Here's how I thought about this:

Let's assume that Ksp = 0.5 for example

that would mean that for CaSO4, s=the square root of 0.5 which means s=0.7.

For Ca(OH)2, s=the cube root of (Ksp/4), so s= 0.5

since "s" is higher for CaSO4, then the solubility of CaSO4 would be greater, which means the answer is C. Why is it telling me that my answer is wrong though? Am I missing something?

Thanks in advance

Your numbers are absolutely correct. But, different numbers, wuld yield a different result. Consider the following cases:

Case 1:
Ksp = 4 x 10exp-12
SCaSO4 = 2 x 10exp-6
SCa(OH)2 = 1 x 10exp-4
Result is that Ca(OH)2 has a greater molar solubility

Case 2:
Ksp = 4 x 10exp-6
SCaSO4 = 2 x 10exp-3
SCa(OH)2 = 1 x 10exp-2
Result is that Ca(OH)2 has a greater molar solubility by a smaller amount than case 1

Case 3 (your example:
Ksp = 0.50
SCaSO4 = 0.71
SCa(OH)2 = 0.50
Result is that CaSO4 has a greater molar solubility

As Fileserver pointed out, if Ksp is greater than 1, then the trend continues with CaSO4 having the greater molar solubility.

Your logic is correct for your example. So, the conclusion is that based on the information listed in the question, the answer cannot be determined.

If you happen to be familiar with Ksp values for calcium salts, they are typically in the 10exp-5 to 10exp-12 range at ambient temperature. Just for your sanity, the Ksp values are roughly:

CaSO4 Ksp = 6 x 10exp-5
Ca(OH)2 Ksp = 3 x 10exp-6

They differ by a factor of roughly 20, so they are not exactly approximately the same. But the question makes a good point that you need to consider molar solubility separately from Ksp.

You forgot to multiply the s by 2.
Ksp of ca(oh)2 = x(2x)^2=4x^2; after you solve for x, you get x. But [OH]=[2x] as in the equation, so you have to multiply by 2.

For molar solubility, you do not need to multiply by 2. You are solving for the concentration of the salt on the whole, not of the cation or anion. You are correct if the question were asking for the concentration of hydroxide.

 

[fileserver]

Oh, I see. Molar solubility is the amount that the salt is willing to dissociate, so it's just x. Thank you, BerkReviewTeach, for point that out.

BerkReviewTeach, do you happen to know how to deal with radical when calculating oxidation number, =) ?

 

[phospho]

BerkReviewTeach and Fileserver, this makes a lot of sense now. I'm very glad I asked the question. Thank you both so much

Fileserver, I will keep bumping this thread till someone answers us!

 

[Franksta1118]

Hey these two questions have really been bugging me:

1. Say we have a solution of pH 11, then if I make a 100 fold dilution with pure water what will be the pH? Using the equations for conc H+ I find that the conc of H+ decreases by 100 times so then the pH should be 13. But, the answer is that the pH is 9. I don't get how this works out mathematically???

2. How can I determine the oxidation states of C and N in cyanide (CN-) with some kind of rule. The rule I use with electronegativities and such leads me to the answer that C has +2 and N is -2 which cant be right due to the -1 overall charge...I looked this up and the answer is C +2 and N -3 but how can we get this with some rule???

 

[phospho]

Hey these two questions have really been bugging me:

1. Say we have a solution of pH 11, then if I make a 100 fold dilution with pure water what will be the pH? Using the equations for conc H+ I find that the conc of H+ decreases by 100 times so then the pH should be 13. But, the answer is that the pH is 9. I don't get how this works out mathematically???

2. How can I determine the oxidation states of C and N in cyanide (CN-) with some kind of rule. The rule I use with electronegativities and such leads me to the answer that C has +2 and N is -2 which cant be right due to the -1 overall charge...I looked this up and the answer is C +2 and N -3 but how can we get this with some rule???

i suck at gen chem, but i'll give them a shot

1. not sure which equations you're talking about, but I don't personally see why you would need anything other than the -log [conc].

think of it this way: you did a 10^2 fold dilution. You reduced the pH by -log(10^-2), which is equal to 2. And 11-2=9

2. I don't know if there is a rule. However, if I were to come across this, I would have left the C till the end. With N having 5 valence shell electrons, it's obviously -3. So, -3+x=-1.

i hope this helps. I'm pretty sure others will know more about these than I do

 

[RSAgator]

Hey these two questions have really been bugging me:

1. Say we have a solution of pH 11, then if I make a 100 fold dilution with pure water what will be the pH? Using the equations for conc H+ I find that the conc of H+ decreases by 100 times so then the pH should be 13. But, the answer is that the pH is 9. I don't get how this works out mathematically???

2. How can I determine the oxidation states of C and N in cyanide (CN-) with some kind of rule. The rule I use with electronegativities and such leads me to the answer that C has +2 and N is -2 which cant be right due to the -1 overall charge...I looked this up and the answer is C +2 and N -3 but how can we get this with some rule???

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.

 

[travelbug73]

Hey these two questions have really been bugging me:

1. Say we have a solution of pH 11, then if I make a 100 fold dilution with pure water what will be the pH? Using the equations for conc H+ I find that the conc of H+ decreases by 100 times so then the pH should be 13. But, the answer is that the pH is 9. I don't get how this works out mathematically???

2. How can I determine the oxidation states of C and N in cyanide (CN-) with some kind of rule. The rule I use with electronegativities and such leads me to the answer that C has +2 and N is -2 which cant be right due to the -1 overall charge...I looked this up and the answer is C +2 and N -3 but how can we get this with some rule???

Ok, I'm not an expert either but this is how I look at them:

1. If you have a solution of pH 11, you have a higher [OH-] than [H+], so addition of water will dilute your [OH-] and bring up your [H+]. Addition of pure water to an acidic or basic solution will tend to make the solution more neutral though this is an inefficient way of doing it. For example, you will need 10^7 mL of water to change the pH of a solution from 14 to 7 at 25C. Hence you use acids to neutralize bases and vice-versa.

2. Oxidation state of N = -3 (most times), so oxidation state of carbon would be +3. But, there is an overall negative charge on cyanide, so, the oxidation state of C = +2.

Please correct me if I'm wrong on either or both.

 

[Kaustikos]

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.

No, you're just getting confused is what it looks like. The water will cause an increase in hydrogen concentration, decreasing the pH. Think of water as teh acid titrating the base. If it were placed in HCL, it would increase the pH and therefor decrease the [H] to satisfay its own pH. In this case, you're adding water to a basic solution, and so it would try to increase [H] to decrease the pH to its level.

Your math is fine you're just going the wrong way.