SAT II Chem from Barron's





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Discus: SAT/ACT Tests and Test Preparation: June 2004 Archive: SAT II Chem from Barron's
By Spang (Spang) on Monday, May 31, 2004 - 01:57 pm: Edit

Some questions...

Atoms of different elements can have the same mass number because the atoms of each element have a characteristic number of protons in the nucleus. (T T), Explain + are isotopes of elements considered elements?

13 C 12 and 12 C 12 are isotopes of the element carbon because isotopes of an element have the same number of protons in the nucleus. Is the atomic number supposed to be 6 instead of 12?

Hydrosulfuric acid is often used in qualitative tests because H2S reacts with many metallic ions to give colored precipitates. Isn't H2S called Hydrogen Sulfide and not Hydrosulfuric acid??

The weakes of the bonds between molecules are coordinate covalent bonds because coordinate covalent bonds represent the weak attractive force of the electrons of one molecule for the positively charged nucleus of another.

If you titrate 1 M H2SO4 solution against 50 mm of 1 M NaOH solution, what volume of H2SO4 in mm will be needed for neutralization?

A)10, B)25, C)40, D)50, E)100

Barrons says 25, but isn't that assuming both acidic protons come off of H2SO4, which isn't true since HSO4- is a weak acid? So wouldn't the answer actually be between 50 and 25 mm, hence 40 mm?

Thanks!

By Tri_N (Tri_N) on Monday, May 31, 2004 - 02:50 pm: Edit

No, the answer is 25.

HSO4 is a strong acid, not a weak acid. Here's a rule of thumb: the more oxygen an acid has, the stronger its acidity. If this is SO3, you might have a case.

By Spang (Spang) on Monday, May 31, 2004 - 03:41 pm: Edit

So you are saying that H2SO4 AND HSO4- are both strong acids?

HSO4- <--> H+ + SO4-, Ka= 1.2*10^-2..., isn't this weak?

By Aab123 (Aab123) on Monday, May 31, 2004 - 04:33 pm: Edit

The answer is 25 becuase h2so4 has two H+'s, whereas NaOH has 1. So itll need half.

By Tri_N (Tri_N) on Monday, May 31, 2004 - 04:34 pm: Edit

No, it's not.

HSO4 is a strong acid. If you're wondering about the Ka, weak acids have Ka with 10^-4 or less.

By Spang (Spang) on Monday, May 31, 2004 - 11:03 pm: Edit

Oh... I thought that all acids with a Ka were weak. Ok, thanks for clarifying that then! Does anybody know about the other questions?

By Smazn (Smazn) on Tuesday, June 01, 2004 - 12:00 am: Edit

Strong acids don't have Ka's. HSO4 is not a strong acid. It is the conjugate base of a strong acid. There are only Ka's for weak acid. Ka's for strong acids don't exist.

By Tri_N (Tri_N) on Tuesday, June 01, 2004 - 12:35 am: Edit

Smazn, do you want to make a bet?

HSO4 isn't a strong acid that will dissociate completely but it's certainly isn't a weak acid.

Here's your reasoning:

HSO4 is a conjugate base of a strong acid. Therefore, it's a weak base. On the contrary, it's an acid. Sadly, you're wrong.

By Smazn (Smazn) on Tuesday, June 01, 2004 - 03:05 am: Edit

Isn't HSO4 amphoteric (can act as acid or base)?

By Basketball2k79 (Basketball2k79) on Tuesday, June 01, 2004 - 09:37 am: Edit

i hear barrons is top prep for sat IIS

By Spang (Spang) on Tuesday, June 01, 2004 - 02:59 pm: Edit

It's just weird for H2SO4 to have two full acidic protons rather than say 1.5 or something cuz I am almost 100% sure that HSO4- does not fully dissociate. But anyways, any other replies to the questions in the first post?

By Kewlkiwi102 (Kewlkiwi102) on Tuesday, June 01, 2004 - 03:38 pm: Edit

OK the important thing about H2SO4 in this question is that the NaOH will pull off the H+ that normally would not ionize, so thats why its 25. And HSO4- is amphoteric. Its behaviour would depend on the environment.

I have to run but Ill explain the others later.
~kiwi

By Erics123 (Erics123) on Tuesday, June 01, 2004 - 05:56 pm: Edit

HSO4- is NOT a strong acid!!! A strong acid has Ka >>>> 1. However, an acid-base reaction always
goes to completion, so all of the acid/base will be consumed (assuming stoichiometric quantities). This is because the reaction is highly exothermic (water has very low energy). On the other hand, the ionization reaction is not so exothermic.

In terms of equilibria, HSO4 + NaOH -> NaSO4 + H20, and Kc=[NaSO4]/[HSO4] (don't count the liquids), Kc=1 (you will see why) so if NaSO4 and HSO4 are equal initially, you can remove all of the NaSO4 and HSO4 and Kc is still equal to 1. (Obviously 0/0 is not really equal to one, but . . .)

The same thing holds for a precipitation reaction, by the same logic.

H2S is hydrosulfuric acid when in aqueous solution, hydrogen sulfide otherwise. This is the same thing with hydrofluoric or hydrochloric acid, which are hydrogen fluoride and hydrogen chloride when in gaseous form.

Coordinate covalent bonds are not bonds between molecules, so that part is false. They do result from said attraction (in reality the electron cloud is between both molecules, so it doesn't really matter where they came from).

The atomic number is supposed to be six.

Atoms of different elements can have the same mass number -- Carbon-12 has 6 protons 6 neutrons, whereas Boron-12 has 5 protons 7 neutrons (I don't think Boron-12 is very stable, but that's irrelevant). An isotope of an element is considered the same element (chemically) as a different isotope of that element.

By Erics123 (Erics123) on Tuesday, June 01, 2004 - 06:36 pm: Edit

HSO4 really is amphoteric. In very concentrated (18 M) solution H2SO4, some H2SO4 will remain undissociated. From a dynamic standpoint, this means that some of the HS04 is accepting protons and becoming H2SO4, thus acting as a Broensted-Lowry base.

By Ware_Ru (Ware_Ru) on Tuesday, June 01, 2004 - 07:36 pm: Edit

Smazn and Erics123 are both right about HSO4- being amphoteric (it is the conjugate base of H2SO4 and the conjugate acid of SO4-), and that it is a weak acid; only the first H in H2SO4 is strong by the conventional definition of a strong acid - the H+ completely dissociates.

Sadly, Tri_N, YOU are wrong. Learn some chemistry before refuting other posters' CORRECT replies.

By Smazn (Smazn) on Tuesday, June 01, 2004 - 09:07 pm: Edit

Yes, Tri_N, YOU are wrong. BTW thanxs Ware Ru and Erics 123.

By Smazn (Smazn) on Tuesday, June 01, 2004 - 09:09 pm: Edit

Tri_N, you know that you were contradicting yourself in one post saying that HSO4 is a strong acid and in another saying that it isn't. Do you want to make a bet now?

By Kewlkiwi102 (Kewlkiwi102) on Tuesday, June 01, 2004 - 09:21 pm: Edit

Hey guys lets not turn this into a fight. Lets focus on chemistry instead. Yay chemistry!
*good vibes!*

By Smazn (Smazn) on Tuesday, June 01, 2004 - 09:30 pm: Edit

sorri Kewlkiwi102 if i made it seem like it was going to be a fight. it wasn't my intention. i just wanted to make the point that Ware Ru made, which was that Tri_N shouldn't be posting rude comments saying that people are "sadly" wrong and saying his own wrong answer is "correct". Gotta get back to studying chem.

By Tri_N (Tri_N) on Tuesday, June 01, 2004 - 10:33 pm: Edit

I'll make the wager that HSO4 isn't a weak acid. HSO4 is not a extremly strong acid but it's not a weak acid. On a scale from extremly strong to extremly weak, it would be moderately strong.

As for the Kc, it's actually [NaSO4]/[HSO4][NaOH] not [NaSO4]/[HSO4].

For the equation: HSO4 + H20 -> H30 + SO4-2
Usually, you would disregard the x, the amount of H30+, in weak acids.

In an equation with a weak acid:

[x][x]/[HSO4-x] = Ka

You would ignore the change in x of HSO4 because x is very small in comparison to the concetration fo HSO4 . Therefore, it would become:

[x][x]/[HSO4] = Ka
*This only holds true if the Ka is equal to or less than 10^-4*

In any case in which the KA is equal to or greater than 10^-2, you definitely need to account the x for the concentration of the reactants. Therfore, the full equation would be this:

[x][x]/[HSO4-x] = Ka

*This is the standard for moderately strong to strong acid and base.*

You can dispute the logic all you want but my post is from the textbook. If you want, you can look up the table of strong to weak acid and base. Sadly, HSO4 is a moderaly strong acid.

By Tri_N (Tri_N) on Tuesday, June 01, 2004 - 10:59 pm: Edit

"In terms of equilibria, HSO4 + NaOH -> NaSO4 + H20, and Kc=[NaSO4]/[HSO4] (don't count the liquids), Kc=1 (you will see why) so if NaSO4 and HSO4 are equal initially, you can remove all of the NaSO4 and HSO4 and Kc is still equal to 1. (Obviously 0/0 is not really equal to one, but . . .) "

BTW, if this is a one to one ratio, it would be .5 because Kc = [NASO4]/[HSO4][NaOH]

In this case, it would be 10^-1. The rule of thumb is that, if the Ka/Kb is equal to or greater than 10^-2, the reactant is a strong acid acid or base. Your example proves my point. Thanks.

By Tri_N (Tri_N) on Tuesday, June 01, 2004 - 11:17 pm: Edit

Spang,

The problem you give is a semi-stoichiometric problem. In this respect, you don't care about the relative strength of the acid or base. For the problem, you only care about the number of moles of acid vs. base.

By Ware_Ru (Ware_Ru) on Wednesday, June 02, 2004 - 03:16 pm: Edit

No, still, Tri_N, you are wrong. The conventional definition of a "strong acid" or a "strong base" is that it completely dissociates in solution. Your "rule of thumb" is completely wrong.

From Chemistry by Zumdahl (one of the most respected and widely used college textbooks for first year college Chemistry courses), 6th edition, p662: "The acid H2SO4 is a strong acid, virtually 100% dissociated (ionized) in water, but the HSO4- ion is a weak acid."

From wikipedia:
http://en.wikipedia.org/wiki/Strong_acid
A strong acid is an acidic compound which ionizes completely in an aqueous solution.

From about.com:
http://chemistry.about.com/library/glossary/bldef858.htm
Strong Acid
Definition: An acid that is completely dissociated in an aqueous (water) solution.

I could find more sources but I don't think that is necessary.

Now, like you said, HSO4- does not completely dissociate in water. Therefore, by the conventional definition of a strong acid, it is not a strong acid. I am positive that there is no such thing as a "not strong but not weak acid" in introductory Chemistry, and I'm sure that kind of ridiculous terminology is not used in higher-level Chemistry, either. There are two main types of acids: strong acids that completely dissociate in water (ONLY: HCl, HBr, HI, HClO4, HNO3, and the FIRST hydrogen of H2SO4); and weak acids, which are all acids that do not fall into that first category. Another property of strong acids is that their conjugate base is a weaker base than water. If HSO4- were a strong acid, SO4 2- would be a weaker base than water; but it is not.

If what you are trying to say is that HSO4- is relatively stronger acid than say... NH4+ because of its greater Ka value, then you are completely correct. However, that is not what you have been saying. Moreover, you're wrong, but obnoxious and stubborn about it, and that is even worse than the original confusion you caused. Please stop being an idiot.

By Spang (Spang) on Wednesday, June 02, 2004 - 08:16 pm: Edit

Well, it seems that books can be wrong, but thanks for all your help! HSO4- is weak I believe, but in this scenario, strength is not a matter, so thanks for clarifying that up!

By Thunder77 (Thunder77) on Wednesday, June 02, 2004 - 08:40 pm: Edit

You know you are wrong Tri_N, please stop trying to sound like you meant something else as an excuse that you were correct..
Jeesh, just admit that you were wrong. Why can't some people do that? The worse thing we'll do is laugh at you, so who cares? By repeatedly coming back and posting and insulting others thinking you are smarter, you are only making the situation worse

You either admit you're wrong or you leave and never post in this thread again. ok ?

By Erics123 (Erics123) on Wednesday, June 02, 2004 - 10:02 pm: Edit

Oops sorry, my bad. That post completely did not make sense.

Anyways, neutralization reactions do go to completion. Let's just say that K is very very big (note to self: avoid blathering useless nonsense).

How about settling that HSO4- is a moderately strong acid? It's all an issue of semantics here and it's kind of useless to argue about.

*Goes to hit himself over the head a few times with his chemistry book*


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