Chemistry Question + Answer

[eng_dude786]

Hey everyone,

 

I just started this thread because I have trouble with chemistry. Hopefully someone can help me out.

 

 

My first Question:

 

If you have sulfuric acid (H2S04) and say its concentration is 10 M, and you are asked to find its pH.

 

Since sulfuric acid is a polyprotic acid, do you ignore the second proton it can donate?

 

I have seen either people ignoring the second proton or multiplying the hydrogen concentration by 2. Which one is right for the MCAT? (hopefully I was not too confusing )

[Jixe]

If they asked you to compute the pH of that solution, they would give you the Ka of the dissociation of HSO4-. If it's very low, then yes, you can ignore its contribution to pH. You would definitely not multiply the pH by two. Increasing pH means greater basicity. Maybe you meant multiplying the [H+] by two? Definitely not again, as HSO4- does not dissociate completely (it is not a strong acid).

[dan0105]

Umm to expand on that,

 

the way H2SO4 works is that, the first proton always comes off because it is a strong acid. Thus 1mole of H2SO4 makes exactly one mole of H+, the second proton however is weak, so you need to do the calculation with ka = [H][sO4]/[HSO4] and then recognize the change etc etc and solve for x. Then you need to add the two hydrogen amounts you have together.

 

USUALLY however, the second amount is very very neglegible and most of the answers are different enough that you dont need to do the second step.

[eng_dude786]

okay...so i think I understand what you guys are saying. When doing pH calculations just treat it like only one proton is dissociated. So if H2S04 is 5 M, then [H+] in water is also 5 moles/liter and not 10 moles/liter.

 

So like when do you take the diprotic-ness of sulfuric acid into account? If you are doing titrations question, would you then say it results in 2 hydrogens (normality or something like that)? (equation--> N1V1=N2V2)

 

thanks again

[goose]

I think its safe to say that for H2S04, you won't have to consider the second proton as - like mentioned above - its dissociation is negligible. The first proton dissociates nearly 100% in aqueous solution, so the pH = -log[H2SO4]

 

You have to consider the "diprotic-ness" of acids when considering buffers, specifically in the case where the first dissociation is not enough to buffer the addition of a base or acid.

[The Law]

Good question eng_dude!

[avenir001]

Good question eng_dude!

 

lol. if u don't have anything meaningful to contribute, then plz shut up.

[The Law]

lol. if u don't have anything meaningful to contribute, then plz shut up.

 

Psh, anything I post automatically counts as meaningful.

[dan0105]

okay...so i think I understand what you guys are saying. When doing pH calculations just treat it like only one proton is dissociated. So if H2S04 is 5 M, then [H+] in water is also 5 moles/liter and not 10 moles/liter.

 

So like when do you take the diprotic-ness of sulfuric acid into account? If you are doing titrations question, would you then say it results in 2 hydrogens (normality or something like that)? (equation--> N1V1=N2V2)

 

thanks again

 

OK, ya, I was talking about just the molarity there. The first H2SO4 breaks down into H and HSO4, so you get 5moles/L H, then the HSO4 breaks down (partially) into some H and SO4, so overall its like 5.1moles/L of H or something, you really need the Ka (or pKa) to figure it out exactly.

 

As for the titration, there are still two hydrogens in every molecule so ALL of them react during a titration with a strong base. Hence, in the formula

NaVa = NbVb or whatever (its just moles acid = moles base), they use N to signify that you DO need to take into account the "di"proticness. So like 2mL of a 1M H2SO4 solution would require 4mL of a 1M NaOH solution to reach equivalence.

[AndyDude]

50g of water at 80C is added to another 150g of water at 25C. No heat is exchanged with the surroundings. What is the final T?

 

I approached this by setting the heats lost and gained equal to each other.

 

Next, m1*c1*dT= m2*c2*dT

 

Since, c1 = c2, the eq'n becomes: 50*(T2-T1) = 150*(T2-T1)

-> T2-80 = 3*(T2-25)

 

T2 for both are equal so, T2 - 3T2 = -75+80C

T2 = +5/-2 = -2.5C

 

But it makes no sense. The correct answer is 39C and rightly so since the mass of 25C is 3x the 80C mass and final T should be closer to 25 than 80 and in-between. Where I get -2.5C is beyond me. Unless I am making a reaaally stupid mistake. I have gone over it more than a few times. Should I be adding the 2 masses together?

 

Any help is greatly appreciated.

 

On a side note I have the equation set right, since dT (of water at 80C initially) = 3x dT of water at 25C. --> 3*(39-25) = (80-39)

[photato]

try converting the temperature to K and see if you get the right answer

[eng_dude786]

u have to check the units of 'c'...how is 'c' defined. If its in kg, or Kelvin then you have to do the appropriate conversions.

 

I think since you crossed out 'c' from both sides you might have overlooked the conversion process (g to kg or C to K)

 

Also, a side check is to to know that the final answer will be hotter than the coldest liquid available.

 

hope it works out!

[unknown user 7]

Does the "eng" stand for engineering or english?

[dan0105]

m1*c1*dT (specific heat? it looks weird without the proper triangle delta...)

anyways, Sure, the specific heat lost equals the amount gained, expect that the formula requires the ACTUAL temperature change, ie 80 degress-x (an unknown amount), not 80-25,because that isn't what the change is, that is just the difference between the two original temps!

 

The actual new temp will be somewhere in between, ie 80-x and x-25 for the other.

 

Therefore, lets do it again with this:

 

->m1c1dT = m2c2dT

->m1c1(80 - x) = m2c2(x - 25) [The hot one gets cool, thus its less than 80, the other gets hotter, so more than 25]

->(50)(80 - x) = (150)(x - 25)

->(80-x) = 3(x - 25)

-> 80 + 75 = 3x

-> 115 = 3x

-> x = 115/3 = 38.333

 

 

You were close, but don't get tricked on what dT actually is. Its 80-x NOT 80-25.

 

cheers

[dan0105]

Uhhh, in reference to the other comment, yes you can cancel those c's just fine.

[AndyDude]

Thanks, dan. I learned something new. Bless the forums!

[Jwells]

 

->m1c1dT = m2c2dT

->m1c1(80 - x) = m2c2(x - 25) [The hot one gets cool, thus its less than 80, the other gets hotter, so more than 25]

->(50)(80 - x) = (150)(x - 25)

->(80-x) = 3(x - 25)

-> 80 + 75 = 3x

-> 115 = 3x

-> x = 115/3 = 38.333

 

cheers

 

I don't mean to be picky, but I will be.

 

80 + 75 = 155, not 115

and you should have 4x when the parenthesises are removed, not 3. This gives X = 155/4 = 38.75

 

It's funny how we manage to make equations work to the right answer when we know the final answer. (Dan's equations are right, just a small arithmetic error).

 

 

To AndyDude, I think I see where you made your mistake. You set up your equation as follows:

 

50*(T2-T1) = 150*(T2-T1) <-- I think when you wrote this you realized T1 on the left is not the same as T1 on the right. Be careful with your notation.

 

and you were thinking that the temperature in both cases should be final minus initial. The problem with this is you are claiming that the Heat lost equals the heat gained. In actuality, the MAGNITUDE of the heat lost equals the MAGNITUDE of the heat gained. We need to be consistent with our notation which means we need to define either heat lost or heat gained as positive, and the other as negative. This makes the equation look like the following: (T2 is still the final temp, TA and TB are the two starting temps)

 

50*(T2-TA) + 150*(T2-TB) = 0 <-- This says Heat Lost plus Heat Gained = 0

OR

50*(T2-TA) = - 150*(T2-TB) <-- notice the negative sign

 

If you push that negative sign inside the parenthesises you get

50*(T2-TA) = 150*(TB-T2)

which is the same as Dan was showing you.

[AndyDude]

Thanks Jwells for the clarification.

[dan0105]

Hahaeh just like on the real mcat, my elementary math skills are pathetic... I have a major problem of figuring out how to solve the question in 5 seconds but then spending 5 minutes trying to actually do the math. Calculators (and spell-check) ruined my life.

 

cheers guys, bring on the next question andy!

[Sergem]

Hi,

 

The way I figured this one out is to do ratios of the mass and their temperatures since we are dealing with the same substance and no heat is exchanged with the surroundings.

 

To find the final temperature of the 200g water, a quarter of it was at 80C and 3/4 of it was at 25C:

 

 

(0.25)(80) + (0.75)(25) = 38.75C

[oohpsjin]

I'm having trouble with orbitals!

E.g. in SF6 how does sulfur hve empty d orbitals available to form an expanded octet, while in OCl6 oxygen does not!?

E.g. in heterocyclic rings such as those that contain nitrogen, how is it that the lone pair electons is p electrons, not s?

you see I'm having trouble with the basics...can anyone suggest some good readings for my problem?

thanks for any help.

[dan0105]

E.g. in heterocyclic rings such as those that contain nitrogen, how is it that the lone pair electons is p electrons, not s?

 

I already put a link down earlier, but: http://mymcat.com/w/index.php?title=Aromaticity_Introduction

 

 

a single bond occurs from two s orbitals overlapping (and its called a sigma bond). A double bond then occurs when a second pair or orbitals overlap a bit, but it can't be two more s orbitals because they would be on top of the old ones, so p orbitals are used (which stick out to the sides). p orbitals sticking out on each of the atoms will overlap to produce the second bond (a pi bond). And this is the double bond. In conjugated systems, there are a bunch of these pi bonds in a alternating fashion, so it ends up the p orbitals and swap their current partner-pair with the one on the other side and the whole thing gets into a big pi bonding mess (see benzene).

 

When nitrogen and sulfur and oxygen and co get into the mess, they only way the can contribute to the pi mess is if they do it through a p orbital (because again s orbits wouldn't be sticking out to the side like the others and so wouldn't come into contact.) The real trick is deciding when the nitrogen is going to donate into a p or leave its electrons in a lone pair. Read the article...

[oohpsjin]

E.g. in heterocyclic rings such as those that contain nitrogen, how is it that the lone pair electons is p electrons, not s?

 

I already put a link down earlier, but: http://mymcat.com/w/index.php?title=Aromaticity_Introduction

 

 

a single bond occurs from two s orbitals overlapping (and its called a sigma bond). A double bond then occurs when a second pair or orbitals overlap a bit, but it can't be two more s orbitals because they would be on top of the old ones, so p orbitals are used (which stick out to the sides). p orbitals sticking out on each of the atoms will overlap to produce the second bond (a pi bond). And this is the double bond. In conjugated systems, there are a bunch of these pi bonds in a alternating fashion, so it ends up the p orbitals and swap their current partner-pair with the one on the other side and the whole thing gets into a big pi bonding mess (see benzene).

 

When nitrogen and sulfur and oxygen and co get into the mess, they only way the can contribute to the pi mess is if they do it through a p orbital (because again s orbits wouldn't be sticking out to the side like the others and so wouldn't come into contact.) The real trick is deciding when the nitrogen is going to donate into a p or leave its electrons in a lone pair. Read the article...

 

Thanks for the help with the heterocylics. But I'm still confused about orbitals in general that may be solved with this question: OF5 cannot exist because there are no empty d orbitals for O. However, compounds such as IBr, UF6, NaLiCO3 can exist. Is it because they have empty d orbitals? but WHERE? For instance aren't the d orbitals filled in IBr for both I and Br?

[The Law]

I'm having trouble with orbitals!

E.g. in SF6 how does sulfur hve empty d orbitals available to form an expanded octet, while in OCl6 oxygen does not!?

E.g. in heterocyclic rings such as those that contain nitrogen, how is it that the lone pair electons is p electrons, not s?

you see I'm having trouble with the basics...can anyone suggest some good readings for my problem?

thanks for any help.

 

Okay, think of it this way... elements in the third row of the periodic table have d-orbitals available for extra bonds. Elements in 2nd row and lower do not.

 

You may be wondering why this is the case. It has to do with quantum numbers. When does the d-orbital appear? The orbitals in the order they fill up are:

 

1s2 2s2 2p6 3s2 3p6 4s2 3d10

 

The primary quantum number is designated by the number in front of the letter. The d-orbital begins to appear at 3 (so this is the minimum energy level required for them).

 

So now, if we look at oxygen. Oxygen is at the second row. It only has s and p available. Sulfur is in the third row, so it has s, p, and d available.

 

When does this come in handy? It comes in handy when looking at Lewis structures. The "octet" rule applies to those in the 2nd row and earlier. Atoms higher than that, have extra orbitals available (the d-ones). So if an atom is 2nd row and earlier, it can have max 4 bonds (or less if hydrogen/helium)... if it is 3rd row+ it can have more than 4 bonds.

 

Hope this helps...

[goose]

Having a little trouble with electrochemistry..

 

Those passages with rechargeable batteries kill me. Here's what I understand:

Oxidation always at anode, reduction always at cathode therefore electrons always flow from anode to cathode. Where this gets me is when a discharging battery (galvanic) switches to a charging battery (electrolytic).

 

Since the anode of a discharging cell (sp. the metal strip) is oxidized and loses mass, recharging must require the reverse of this reaction (the same metal strip is reduced). Does this mean that the identities of that cathode and anode switch when the cell is recharging?