Chem Question

[20090900]

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This question was taken from Top Score Test #2, Chemistry Section

#68: What volume of HCl was added if 20mL of 1M NaOH is titrated with 1M HCl to produce a pH=2

I don't really understand the solution that they give so if someone was an alternate/easier way of explaining that would be great!

 

[kc1469]

This question was taken from Top Score Test #2, Chemistry Section

#68: What volume of HCl was added if 20mL of 1M NaOH is titrated with 1M HCl to produce a pH=2

I don't really understand the solution that they give so if someone was an alternate/easier way of explaining that would be great!

Yea, this is a difficult one. The key to understanding this problem is to realize what a pH=2 actually means. pH is the -logarithmic scale of concentration. At a pH=2, the concentration of H+ ions is 1 x 10^-2.
Okay, so when we start it, we're in a basic solution right; meaning we're going to be pouring in HCl to first neutralize the NaOH. We pour it in, and of course, we need exactly 20mL since both are a 1 molar solution. We reach the equivalence point, and now we are in NEUTRAL WATER (i.e. "0" H+ concentration)--of course not really since its actually 1 x 10^-7 due to the autoionization of water, but its neglible).
So now, what do we need to do? We want a pH=2, so we want to go from 0--> 1 x 10^-2 MOLAR CONCENTRATION of H+. And remember, what are we using to accomplish that-- we're using HCl 1 MOLAR CONCENTRATION. In other words, we need to set up an equation where the moles of HCl added will equal the moles of HCl AT a concentration of 1 x 10^-2 M.
We currently have 40mL in the pot, and now we need to add an x amount of mL. This same amount (i.e. the x mL) MUST EQUAL the amount of 1M HCl added. bc this is what we are titrating with.

Therefore, our equation is

(1M)(x mL of HCl added)= (1 x 10^-2M NEEDED) (40mL already in the pot + x mL of HCl added)

Solve from there and the amount you should get is .4mL.
**The answer is in mL bc that is the value we used for the amount already in the pot, i.e. 40mL/ IF we had used .04L instead, our answer would have came out in L, i.e. .0004L

What does this mean? It means we need to add .4mL in addition to the 20mL of HCl already added to reach a pH=2, or a concentration of 1 x 10^-2M concentration. Or in total, we need 20.4mL.

 

[20090900]

Wow this is a great explanation. Thank-you!

 

[kc1469]

If anything else, just let me know!

 

[20090900]

Great thanks a lot! I have 1 topscore test left so I will most likely have more question. I actually have a topscore math question if you don't mind.

Test 2 #24: A car leaves your house at 10:00 a.m. travelling at 35mph. A second car leaves your house at 11:00 a.m. travelling the same route at 45mph. How many hours will it take for the second car to overtake the first car?
...I feel like this is an easy question but I just can't get it (even with the solution)...ya I suck at QR

 

[kc1469]

Great thanks a lot! I have 1 topscore test left so I will most likely have more question. I actually have a topscore math question if you don't mind.

Test 2 #24: A car leaves your house at 10:00 a.m. travelling at 35mph. A second car leaves your house at 11:00 a.m. travelling the same route at 45mph. How many hours will it take for the second car to overtake the first car?
...I feel like this is an easy question but I just can't get it (even with the solution)...ya I suck at QR

Of course I don't!

The trick to this question is to realize that the only way for one car to overtake another is for the distance of both cars to be the same. Lets call the car that leaves at 10 AM car A and the car that leaves at 11 AM car B.
The only way for car B to overtake car A is for their distances to be equal, so we have dB=dA. They gave us both rA and rB, and also tA and tB. More specifically, they gave us rA and rB literally (i.e. 35mph and 45mph, respectively) and they gave us tA and tB in terms of x.
In other words, since car B has been traveling one less hour, it has been travelling for x-1 hours, while car A has been travelling for x hours. However, they want the time of car B, so it makes more sense to define x in terms of car B.
So instead of defining X as previously stated, we'll define car B as x. If we do that, we have to redefine car A. Car A has now been travelling for x+1 hours, rather than x, since we changed what x stands for.
Okay, so now we know dA=dB and we also know the rates/hour and times of both cars. Using the distance formula-- d=rt, we can exchange the d in the equation for rt. In other words, we used to have:
dA=dB
But now we have
rAtA=rBtB, or, more specifically, we have...
(35mph)(x+1hr)=(45mph)(xhr)

The only way to get good at these type of equations is to do a ton of them, and MATH DESTROYER is amazing with that. When I first started with these, i was so flustered and found them so difficult. If you go through MATH DESTROYER twice, once in the beginning and once at the end before your DAT, you'll see how much you improved. (My breakdown explains it better)

Anything else, let me know.

 

[chax1]

EDIT:

^^^^^ better explanation.

 

[aqz]

Great thanks a lot! I have 1 topscore test left so I will most likely have more question. I actually have a topscore math question if you don't mind.

Test 2 #24: A car leaves your house at 10:00 a.m. travelling at 35mph. A second car leaves your house at 11:00 a.m. travelling the same route at 45mph. How many hours will it take for the second car to overtake the first car?
...I feel like this is an easy question but I just can't get it (even with the solution)...ya I suck at QR

This question could be solved in a much simpler method. It's conceptually the same as kc's but without all the algebra.

35 mph car = Car A, 45 mph car = Car B

At 11 am, Car A has driven 35 miles (because 10 am - 11 am is 35 mph for 1 hr) and Car B has driven 0 miles.

Since Car B drives 10 mph faster, for every hour that goes by, Car B gains 10 miles on Car A.

Car B needs to gain 35 miles on Car A, and it gains 10 miles for every hour so then if Car B drives 3.5 hours, it will have gained 35 miles.

I feel as if your solution's method is like this one. If so, you should really try to understand this method because QR is all about speed and this method took me about 20 seconds mentally.

 

[kc1469]

This question could be solved in a much simpler method. It's conceptually the same as kc's but without all the algebra.

35 mph car = Car A, 45 mph car = Car B

At 11 am, Car A has driven 35 miles (because 10 am - 11 am is 35 mph for 1 hr) and Car B has driven 0 miles.

Since Car B drives 10 mph faster, for every hour that goes by, Car B gains 10 miles on Car A.

Car B needs to gain 35 miles on Car A, and it gains 10 miles for every hour so then if Car B drives 3.5 hours, it will have gained 35 miles.

I feel as if your solution's method is like this one. If so, you should really try to understand this method because QR is all about speed and this method took me about 20 seconds mentally.

Genius! lol

 

[stpele01]

How did you come up with 40 ml?

 

[kc1469]

How did you come up with 40 ml?

I did not notice, but you answered your own question, lol. Yes, you are right. The key here is to reach neutrality, that way we can pretend as if we are at "0M" H+ concentration. Once we reach neutrality, i.e. the equivalence point for the titration, we are technically pretending as if we never added the NaOH prior. (In other words, it does not matter because the salt produced, NaCl, is neutral and if it wasn't, would make the question so muchhhh harder; definitly beyond the scope of the DAT).
From here on out, we can pretend as if we are actually starting with 40mL of neutral water, which is in fact what we have; the question now should be, starting from 40mL of NEUTRAL WATER, how much HCl should I add TO THOSE 40ML OF NEUTRAL WATER (the result of the titration, as you realized and answered in your questoin) to reach a pH=2 (i.e. 1 x 10^-2M concentration).

If still not sure, let me know.

 

[20090900]

Haha. K a bio question: how many ATP are produced from OXIDATIVE PHOSPHORYLATION in glycolysos? I'm pretty sure it's 32 but topscore says 34? Here's why I think it's 32.

2NADH only produce 2 ATP each instead of 3 each b/c it requires 1ATP/NADH for the pyruvate shunt/shuttle or whatever. Therefore...
2NADH --> 4 ATP
8NADH (remaining NADH give you 3ATP/NADH) --> 8x3 = 24 ATP
2FADH2 (each FADH gives 2ATP) --> 2x2 = 4 ATP
Total for oxidative phosphorylation = 4 + 24 + 4 = 32 ATP

 

[20090900]

*sorry for all the spelling mistakes 🙂

 

[kc1469]

I just want to clarify; is that EXACTLY how the question is worded in TOPSCORE?

 

[kc1469]

Your right for Eukaryotic cells. This is why I asked is that how the question is worded. If it had been in prokaryotes, there is no mitochondria, so that "shuttle" system is non-existent and actually, net ATP from oxidative phosphorylation is 34. In Eukaroyotes, its 32 ATP.

 

[20090900]

It wasn't actually a question but as I was reading the solution for a question and I came across it. It's from the solution of question #7, topscore test #3. But ya, I definitely did not know that, thank-you!...Here is the question and part of the solution...

Question: " The maximum ATP yield from the oxidation of one molecule of glucose to six molecules of carbon dioxide is"

This is what topscore says in the solution "Two ATP from substrate-level phosphorylation, minus two ATP for shuttling electrons from NADH in the cytosol, plus two ATP again from substrate-level phosphorylation via Kreb's cycle, THEN PLUS 34 ATP BY OXIDATIVE PHOSPHORYLATION totaling 36 ATP".

I love how the question doesn't specify if it's a eukaryotic or prokaryotic cell

 

[kc1469]

Yea, dont wry, on the DAT, they would be more specific.

 

[kc1469]

Its always 4 from substrate level.
32 or 34 depends on if its a Eurkaryotic vs a Prokaryotic Cel.

 

[20090900]

Ok cool thank-you!

 

[20090900]

Random bio question: for the purposes of the DAT, what should we know about erythroblastosis fetalis? More specifically, it is known that it can be caused by incompatibility of blood type (aka ABO system) as well as the Rh antigen system or whatever...but Rh incompatibility is more severe and happens more often...but bottom erythroblastosis CAN occur due to blood type incompatibility. Kaplan says it cannot occurs b/c anti-A and anti-B bodies cannot cross the placenta...but they definitely can. So which one should we know b/c I know sometimes knowing "too much" can actually hurt you on the test. Thanks!

 

[20090900]

Also very basic chem question that is going to make me look pretty dumb. What is the difference between a sub-shell, orbital, and sub-orbital?

 

[kc1469]

FOr the purpose of the DAT, know that erythrobastosis fetalis is due to the Rh antibodies.

Orbitals are locations of electrons. (We cannot define an electron in terms of its EXACT speed and position, but we can define it in terms of a region of space, which is defined by the orbital.

A subshell is composed of specific types of orbitals (i.e. based on shape). So the s subshell will have ONE circular orbital.The p subshell will have three orbitals which are DUMBBELL SHAPED.
The # of orbitals in a given shell (i.e. first quantum number N=1,2,3,4,5, etc.) is n^2, with 2n^2 being the number of electrons.
Each subshell has 2l+1 orbitals, and 2(2l+1) electrons.

Anything else, let me know.