Who Aced the AP Chem 2004? Answer ?s 1 and 6 Please





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Discus: SAT/ACT Tests and Test Preparation: May 2004 Archive: Who Aced the AP Chem 2004? Answer ?s 1 and 6 Please
By D231 (D231) on Monday, May 17, 2004 - 09:33 pm: Edit

Hi:
Is anyone interested in working out the answers to questions 1 and 6 from the AP Chemistry exam we all just took? I just want to make sure I am getting the right answers because our teacher is going to give us a test using these questions, or so he says. (that is the question on Ksp and the question on electrochemical cell)

http://apcentral.collegeboard.com/repository/ap04_frq_chemistry_36196.pdf

There is the link for the PDF with the questions.

Thanks to all those who take the time to help!

By Rono_G (Rono_G) on Thursday, May 20, 2004 - 02:09 am: Edit

i'm living in asia... i took the Form B... i was hoping for some Acid-Base Equilibria... but then they gave me gaseous equilibria - which i think is okay, but i just find Acid-Base much more fun... just hold on, i'll solve it and post it up...

By Rono_G (Rono_G) on Thursday, May 20, 2004 - 03:24 am: Edit

aight, i did #1... hope this helps...

1)
(a) pure solids and pure liquids are omitted from the law of mass action – thus we do not include pure solids and pure liquids in Keq, Kp, Ka, Kb, and Ksp. Therefore the Ksp for the reaction of solid Silver Chromate dissolving into the Silver ion and the Chromate ion would be represented as:

Ksp = [Ag+]2[CrO42-]

(b) to calculate the Molar concentration of the Silver ion in the solution of Silver Chromate @ 25 degrees Celsius –

known that - Ksp = [Ag+]2[CrO42-]
known that – Ag2CrO4 -> 2Ag+ + CrO42- (note, this is in equilibrium)

it can be designated that: Ag2CrO4 -> 2X+ + X2-
thereby – Ksp = [2X]^2 [X]
Ksp = 2X^3
Ksp/2 = X^3
Cuberoot(ksp/2) = X
Cuberoot (2.60E-12/2) = X
1.09 E -4 M = X

known that [Ag+] = 2X -> (1.09E-4) X 2 = 2.18 E -4 M

thus the concentration of Silver Ions in solution would be 2.18 E -4 M

(c) looking at the balanced equation of the reaction, the moles of dissolved Silver Chromate would be in a 1:1 proportion to the Chromate ion in solution. Known from the previous calculation that the value for the concentration of the chromate ions in solution is X – for which the value is 1.09 E-4 M –

now, we ware given that there is 100 mL (or 0.100 L) of water, it is known that concentration X volume = moles -> we have the concentration, we have the volume

1.09 E-4 M X 0.100 = 1.09 E -5 moles

thereby it can be seen that 1.09 E -5 moles of the Chromate ion would be present in solution, and as seen in the balanced equation, the moles of the chromate ion has a 1:1 ratio with the moles of the silver chromate; thus 1.09 E-5 silver ions would have dissolved at the maximum. To calculate the mass,

known: molar mass Silver Chromate: 331.74 g

moles X molar mass = mass
(1.09 E-5) X 331.74 = 3.62 E -3 grams

thereby, 3.62 E -3 grams of Silver Chromate would dissolve in 100 mL of water @ 25 degrees Celsius.

(d) Ag+ + NO3- + 2Ag+ + CrO42- -> Ag+ NO3- + Ag2CrO4

no new compounds are forming – the level of CrO4 2- would remain the same. I think – I’m not too confident on this part… Somebody confirm.

(e) Ag3PO4 @ 25 degrees Celsius, the concentration of the Ag+ ion is 5.3 E-5 M,
the balanced equation for this reaction must be:

Ag3PO4 -> 3Ag+ + PO4 3- (note this is in equilibrium)

(f) Ksp = [Ag+]^3 + [PO4 3-]

from the balanced equation, [Ag+] = 3X
[PO4 3-] = X

now we are given that @25 degrees Celsius the concentration of the Silver ion is 5.3 E-5 M, thus if 3X = 5.3 E-5 M
X = 1.77 E-5 M

Ksp = (3X)^3 (X)
= (5.3 E-5)^3(1.77E-5)
= 2.64 E -18

thus, the value for Ksp is 2.64 E-18

(g) known that [ ] = moles/volume
now it boils down to basic stoichiometry, we have to ask ourselves – what will happen to the concentration if one of the variables changes. In this case, evaporation is occurring – the moles are held constant – only the volume changes… Therefore if the volume halves (going from a 1.00 L sample to a 0.500 L sample) the concentration MUST double.

Known that @ 25 degrees Celsius the concentration of the Silver ions is 5.3 E-5 M, if the volume halves – the concentration will double thus the final concentration would be

1.06 E-4 M

By Rono_G (Rono_G) on Thursday, May 20, 2004 - 04:02 am: Edit

oh yes, one thing - about part (d) i was talking to my friend about it, and it turns out that the value of the chromate ions actually decrease, we are given AgNO3 solid and this is what happens:

it will dissociate into Ag+ and NO3-, and the Ag+ will add Ag+ ions to the products. by le chatier's principle, the system will shift to reduce the change, so it will produce more of the reactants to eliminate the Ag+. but by doing that, it must consume more of the chromate as well

By Rono_G (Rono_G) on Thursday, May 20, 2004 - 04:05 am: Edit

answers to #6 are here... please post up any corrections if you see any discrepancies in the information below - thanks

6.

(a) it is known that the mass of the Tin electrode increases, if the mass increased - then that is proof of reduction. In electrochemical cells, reduction occurs at the cathode - thus the electrode that is the cathode would be the Sn (the tin) electrode.

(b) when the switch is closed the electrons will flow from teh metal X to the Sn - thus the electron flow will be from the right to left. Cathode is the site of reduction - reduction = gain of electrons, thus electrons have to be flowing TO the site of reduction FROM the site of oxidation.

(c) E Standard Cell = E oxidation + E reduction

we know that E standard Cell = 0.60 Volts
we know that E standard reduction = - 0.14 Volts
we need to determine the standard reduction potential in volts for the X electrode...

E standard cell - E reduction cell = E oxidation
0.60 - (-0.14) = E oxidation
0.74 V = E oxidation,

this is the standard Oxidation potential of metal X, the standard reduction potential must be its opposite - thereby the standard reduction potential for this cell would be - 0.74 V

(d) looking at the chart of standard reduction potentials - the one with a standard reduction potential of: - 0.74 Volts is Chromium (Cr) thus metal X is Chromium (Cr)

(e) this one, you just have to make sure that you balance the charges & follow the law of conservation of mass, you will end up with somethign that looks like:

3Sn(aq)^2+ + 2Cr(s) -> 3Sn(s) + 2Cr^3+(aq) *note, the tin is reduced, the chromium is oxidized

(f) i'm a little unconfident on this one... help me out guys - confirm your answers... but this is what i did...

(i)

the Nernst equation is -> E cell = E standard cell - 0.0592/n X log Q

known: E standard cell = 0.60 Volts
n = 5 electrons
Q = [X3+]^2/[sn2+]^3

[sn2+] = 0.50 M
[X3+] = 0.10 M

substituting it into the nernst equation ->

E cell = 0.60 - 0.0592/5 X log (.10)^2/(0.50)^3

(ii)

now, looking at Le Chaltier's principle - if Q is bigger than K, there are more products than reactants then the reaction must shift to the left. furthermore, looking at the situation presented above... as the concentration of the tin ions are reduced, it would make Q bigger... If Q is made bigger - then according to the Nernst Equation the E cell must drop.

By Hiamerica (Hiamerica) on Thursday, May 20, 2004 - 02:31 pm: Edit

Dude, I think e cell increses, like i am getting 0.61. with a calculator. although i did not use one on the exam. maybe you are right.

By Mkjwoa40 (Mkjwoa40) on Thursday, May 20, 2004 - 05:11 pm: Edit

I got that the voltage increases because the log is negative and then multiplied by another negative number making the change positive

By Kewlkiwi102 (Kewlkiwi102) on Thursday, May 20, 2004 - 06:14 pm: Edit

for 1G, I think that the molarity stays constant. It was saturated. It can only hold so much. If you decrease the amount of water some will come out of solution so you get more solid on the bottom of the container. So molarity would stay constant becuase it would still be saturated.

By Hiamerica (Hiamerica) on Thursday, May 20, 2004 - 09:03 pm: Edit

Kewlkiwi dude! That's exactly what i put. The ratio remains the same, no matter what. that's what i put. but who know's, i have no idea what the correct answer is. But glad that someone put what i put. Booyah! And Mkjwoa40, Your explaination is what i put.

By Kewlkiwi102 (Kewlkiwi102) on Thursday, May 20, 2004 - 09:18 pm: Edit

Im glad also! And I too concur with Mkjwoa40's explanation for that question on electro :)
yay

By Wrathofgod64 (Wrathofgod64) on Friday, May 21, 2004 - 03:08 am: Edit

the molarity stays the same until the system is disturbed. a simple touch of the glass can cause the reverse reaction to occur and the molarity can decrease. that's wat i learned at least

By 1212 (1212) on Friday, May 21, 2004 - 06:41 pm: Edit

Rono G, i believe your answer to 1.ii. is incorrect, the ksp = (2x)^2 * (x) = 4x^3, so ksp needs to be divided by 4 and then cube rooted

By Scion (Scion) on Friday, May 21, 2004 - 07:13 pm: Edit

(i)

known: E standard cell = 0.60 Volts
n = 5 electrons ISNT IT 6 ELECTRONS?
Q = [X3+]^2/[sn2+]^3

E cell = 0.60 - 0.0592/5 X log (.10)^2/(0.50)^3

and i got that it DECREASES too, because if the concentration of the reactant is decreased, then according to le chatelier, the reaction should shift to the left, decreasing the E. o well, maybe u guys r right... yeah i think u r.

By Hiamerica (Hiamerica) on Friday, May 21, 2004 - 07:36 pm: Edit

Dude you get log 0.08 which is -1.0969 blah blah. which when multiplied by -0.0592/6(yeah! you are right!) we get a positive value namely 0.0108, and addiding that to 0.60 will give 0.61, which implies its decrease. Dude! i seriously need to know who is right! I am getting nervous, someone mediate, correct, etc.Please

By Hiamerica (Hiamerica) on Friday, May 21, 2004 - 08:47 pm: Edit

anyone! pleaseeee

By Zik (Zik) on Friday, May 21, 2004 - 10:46 pm: Edit

"addiding that to 0.60 will give 0.61, which implies its decrease...."

i don't see how you ADD and then suddenly declare that it DECREASES.

???

By Hiamerica (Hiamerica) on Friday, May 21, 2004 - 11:47 pm: Edit

Oh i got so tensed that i put it decrease. sorry dude. it increses. Haha! But i get the credit for working the math on this board

By Mattlord (Mattlord) on Saturday, May 22, 2004 - 01:05 am: Edit

it definitely INCREASES; .6 -.0592/6 *logQ

In this case, logQ (based on whatever the original problem said for the concentrations, too lazy to plug in for Q) is NEGATIVE (Q is less than 1); therefore you ADD something to .6.

By Wrathofgod64 (Wrathofgod64) on Saturday, May 22, 2004 - 01:18 am: Edit

do u get credit if u wrote it increases but didn't really state y (in my case, i completely forgot wat the nerst equation was)

By Scion (Scion) on Saturday, May 22, 2004 - 01:18 pm: Edit

will i get credit even though i said it DECREASES? i did set up the problem...

By Catalinakc (Catalinakc) on Saturday, May 22, 2004 - 05:17 pm: Edit

even if you forgot the nerst equation, i'm pretty sure they gave it to you on the green sheets

By Hiamerica (Hiamerica) on Monday, May 24, 2004 - 09:13 pm: Edit

Hi guys,
Just wondering about Question 2. Lets not write pages of solutions, but a quick recap. Answers, bam bam bam. alright. here goes.
Question 2
1.i, ii is o.k.
But What is the limiting reactant. I have no idea. What is it guys? our whole class has a debate on this. Please anyone. No solutions required, just the answers. And what drives the reaction, enthalpy or entropy. Anyone?

By Jshifton (Jshifton) on Monday, May 24, 2004 - 09:37 pm: Edit

I didn't even take the AP exam, but the limiting reactant would be Fe, since 1.342/2 is less than 1.250/1.5.

By Hiamerica (Hiamerica) on Monday, May 24, 2004 - 09:40 pm: Edit

Dude! You made my day! These kids in my class said it is O2. Thanks man. Any idea about other answers, guys. anyone.

By Hiamerica (Hiamerica) on Monday, May 24, 2004 - 10:05 pm: Edit

Anyone bro!

By Hiamerica (Hiamerica) on Tuesday, May 25, 2004 - 12:51 am: Edit

Please guys! I just need the answers not even the explanations.

By Hiamerica (Hiamerica) on Tuesday, May 25, 2004 - 07:04 pm: Edit

Bro dudes! Please!

By K0005 (K0005) on Wednesday, May 26, 2004 - 12:10 am: Edit

(b) to calculate the Molar concentration of the Silver ion in the solution of Silver Chromate @ 25 degrees Celsius –

known that - Ksp = [Ag+]2[CrO42-]
known that – Ag2CrO4 -> 2Ag+ + CrO42- (note, this is in equilibrium)

it can be designated that: Ag2CrO4 -> 2X+ + X2-
thereby – Ksp = [2X]^2 [X]
Ksp = 2X^3
Ksp/2 = X^3
Cuberoot(ksp/2) = X
Cuberoot (2.60E-12/2) = X
1.09 E -4 M = X

known that [Ag+] = 2X -> (1.09E-4) X 2 = 2.18 E -4 M

thus the concentration of Silver Ions in solution would be 2.18 E -4 M

--------------------------------------------

actually, wouldn't it be
Ksp = [2x]^2[x]
= [2x][2x] . [x]
Ksp = 4x^2 . x
Ksp = 4x^3

so the value would be diff?

By Hiamerica (Hiamerica) on Wednesday, May 26, 2004 - 10:09 am: Edit

so what drove the reaction. enthalpy or entropy? i have a test coming tomorrow guys,

By Miscanon (Miscanon) on Wednesday, May 26, 2004 - 01:47 pm: Edit

enthalpy; the entropy was positive (it 'reduced' the spontaenity of the reaction)

By Hiamerica (Hiamerica) on Wednesday, May 26, 2004 - 11:38 pm: Edit

I thought it was entropy! Oh well! Anyone else! Please

By Scion (Scion) on Wednesday, May 26, 2004 - 11:51 pm: Edit

hey hiamerica, i think ur right. the entropy is SUPPOSED to be positive.

By Hiamerica (Hiamerica) on Thursday, May 27, 2004 - 02:14 am: Edit

Yeah! Scion. I got a - value for the entropy. Moreover. It is like 3.5 moles (reactants) to 2 moles of (products). which implies that it is moving towards the products, and the entropy is negative, and thus drives the reaction. Oh well! i am really confused about this one. Someone please answer our question, and give us evidence to prove that answer. I have a test coming in a few days. Quick! Thanks scion.

By Hiamerica (Hiamerica) on Thursday, May 27, 2004 - 01:59 pm: Edit

cOME ON GUYS


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