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dvolovets
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(04-16-2011, 10:27 PM)
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Hey guys, I have a chem test coming up and my instructor posted a review packet with solutions. The solutions are just the answers, though, and don't show any of the work. I worked through most of it, but I'm stumped by these problems. I'm including the answers.

3. Propanoic acid (CH3CH2COOH) has a Ka of 1.34 ´ 10-5. A 25.00 mL sample of 0.1000 mol L-1 propanoic acid (in flask) is titrated with 0.1000 mol L-1 NaOH solution, added from a buret. What is the hydroxide ion concentration after 26.00 mL of NaOH are added?

Answer: 1.96 x 10-3 mol L-1

4. What is the pH of 375 mL of solution containing 0.150 mol of propenoic acid (HA) and 0.250 mol of sodium propenoate (NaA)? (Ka for propenoic acid is 5.52 x 10-5.)

Answer: 4.48

5. Fe(NO3)3 (0.00100 mol) and KSCN (0.200 mol) are added to water to make exactly 1 liter of solution. The red complex ion FeSCN2+ is produced. Calculate the concentrations of Fe3+(aq) and FeSCN2+(aq) at equilibrium, if Kf of the FeSCN2+ is 8.9 * 10^2.

Answer: [Fe3+] = 5.6 x 10-6 mol L-1, [FeSCN2+] = 0.100 mol L-1

Thanks in advance...
Last edited by dvolovets; 04-17-2011 at 12:39 AM.
RevDM
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(04-16-2011, 10:33 PM)
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3: convert to mmol and make an ICE table

4: Henderson-Hasselbach

5: I don't remember how to do this
dvolovets
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(04-16-2011, 10:37 PM)
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Originally Posted by RevDM

3: convert to mmol and make an ICE table

4: Henderson-Hasselbach

5: I don't remember how to do this

For 3, I did the ICE table but have 0.0001 mols of NaOH AND 0.0025 mols of CH3CH2COONa left because the NaOH is in excess relative to the initial amount of CH3CH2COOH. Not really sure where to go from there, since all the problems I've done so far usually have just one species left. If, for example, only CH3CH2COONa were left, I would do an ICE table with CH3CH2COO- + H2O <-> CH3CH2COOH + OH- and calculate the hydroxide concentration that way.

For 4... wow... dumb mistake on my part. I have no idea why I tried to make an ICE table for that one.
Fallen_Hero
love has no conclusion
(04-16-2011, 10:43 PM)
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3. Propanoic acid (CH3CH2COOH) has a Ka of 1.34 ´ 10-5. A 25.00 mL sample of 0.1000 mol L-1 propanoic acid (in flask) is titrated with 0.1000 mol L-1 NaOH solution, added from a buret. What is the hydroxide ion concentration after 26.00 mL of NaOH are added?

Convert to moles of each. Take the moles of NaOH and substract that from moles of propanoic acid. Now continue on as you did before making sure you have factored in the fact that the molarity is no longer the same both due to the fewer number of moles of propanoic and the increased volume of the mixture.

See below.
Last edited by Fallen_Hero; 04-16-2011 at 10:51 PM.
dvolovets
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(04-16-2011, 10:46 PM)
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Originally Posted by Fallen_Hero

Convert to moles of each. Take the moles of NaOH and substract that from moles of propanoic acid. Now continue on as you did before making sure you have factored in the fact that the molarity is no longer the same both due to the fewer number of moles of propanoic and the increased volume of the mixture.

I did this and got the following:

CH3CH2COOH + NaOH --> CH3CH2COONa + H2O
0.0025 mols 0.0026 mol 0
-0.0025 mols -0.0025 mols +0.0025 mols
0 0.0001 mols 0.0025 mols

This is where I'm confused. I have two different species that make a solution basic. If it were just one, such as the CH3CH2COONa, I could just add that to H2O and calculate OH. What do I do in this case?

Edit - Haha, the formatting totally messed up my ICE table.
Sofo
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(04-16-2011, 10:46 PM)
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Came here expecting a failed experiment ._. Leaving disappointed.
shadowsdarknes
I'M STILL A JUNIOR
(04-16-2011, 10:48 PM)
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Does your teacher teach you Henderson?
My teacher deliberately taught us without the equation and we used a different table along with the ICE table.

I can solve these without the equation, I'm not well versed using the equation.

25 mL of .1M Acid = .0025 mols.
26 mL of .1M Strong Base = .0026 mols.

When these react,
.0001 mols of OH- will be left. All the acid will be gone. .0025 mols of conjugate base will be formed. The entire solution now contains 51mL or .051L.
Convert our moles back to Molarity.
.0001/.051= .00196M OH-. and .0025mols/.051L= .049M conjugate base.


No ice table is needed. If you run the ICE table, you will find that your X value will be some tiny negligible number and there's no sense in adding it to your molarity of OH-.
Hope this helps!
Last edited by shadowsdarknes; 04-16-2011 at 10:58 PM.
RevDM
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(04-16-2011, 10:48 PM)
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3. Propanoic acid (CH3CH2COOH) has a Ka of 1.34 ´ 10-5. A 25.00 mL sample of 0.1000 mol L-1 propanoic acid (in flask) is titrated with 0.1000 mol L-1 NaOH solution, added from a buret. What is the hydroxide ion concentration after 26.00 mL of NaOH are added?

Answer: 1.96 x 10-3 mol L-1

All of the prop acid will be removed because it is the limiting reagent, you really have to just determine the amount of NaOH in 1mL of .1M = .1mmol

and, .1mmol/51mL = .00196
dvolovets
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(04-16-2011, 10:49 PM)
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Originally Posted by shadowsdarknes

Does your teacher teach you Henderson?
My teacher deliberately taught us without the equation and we used a different table along with the ICE table.

I can solve these without the equation, I'm not well versed using the equation.

We only use the Henderson equation in isolated cases -- notably stuff like problem 4. Most of the acid/base and titration problems are done using ICE tables in my class.
Lebron
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(04-16-2011, 10:49 PM)
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for 5 you have to do a BRA table then an ICE table.
salva
Más perro que Dios y Jesús combinados, más machín que blue demon y más famoso que el santo
(04-16-2011, 10:52 PM)
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Ugh, i hate acids and bases. I had to tutor it for two weeks and it was dreadful. At least i have to tutor the more fun electrochemistry and nuclear chemistry now!
dvolovets
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(04-16-2011, 10:52 PM)
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Originally Posted by RevDM

3. Propanoic acid (CH3CH2COOH) has a Ka of 1.34 ´ 10-5. A 25.00 mL sample of 0.1000 mol L-1 propanoic acid (in flask) is titrated with 0.1000 mol L-1 NaOH solution, added from a buret. What is the hydroxide ion concentration after 26.00 mL of NaOH are added?

Answer: 1.96 x 10-3 mol L-1

All of the prop acid will be removed because it is the limiting reagent, you really have to just determine the amount of NaOH in 1mL of .1M = .1mmol

and, .1mmol/51mL = .00196

Interesting. Why is the presence of CH3CH2COONa not taken into account when doing this calculation? I had a similar problem where the pH had to be calculated when mols of base and mols of acid were equal. In that case, the left side of the ICE table was 0 and the only thing left in the solution was the conjugate base of the acid. You had to use that in another ICE table to get concentration/pH... I guess I'm confused because it's not taken into account here.

@ salva - No kidding. This is, strangely enough, the most difficult part of chemistry for me so far. You'd think acids and bases would be more simple than this, haha.
salva
Más perro que Dios y Jesús combinados, más machín que blue demon y más famoso que el santo
(04-16-2011, 10:54 PM)
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Originally Posted by dvolovets

Interesting. Why is the presence of CH3CH2COONa not taken into account when doing this calculation? I had a similar problem where the pH had to be calculated when mols of base and mols of acid were equal. In that case, the left side of the ICE table was 0 and the only thing left in the solution was the conjugate base of the acid. You had to use that in another ICE table to get concentration/pH... I guess I'm confused because it's not taken into account here.

All the acid is used up at that point.
RevDM
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(04-16-2011, 10:55 PM)
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Originally Posted by dvolovets

Interesting. Why is the presence of CH3CH2COONa not taken into account when doing this calculation? I had a similar problem where the pH had to be calculated when mols of base and mols of acid were equal. In that case, the left side of the ICE table was 0 and the only thing left in the solution was the conjugate base of the acid. You had to use that in another ICE table to get concentration/pH... I guess I'm confused because it's not taken into account here.

There are only 4 possible types of acid base problems

1. strong acid, strong base
2. strong acid, weak base
3. weak acid, strong base
4. weak acid, weak base

whenever you are dealing with a Strong and a Weak, and all of the weak is consumed, the pH is determined by the remaining Strong (always). You can calculate the affect of the conjugate salt of the Weak but it is so negligible the numbers get lost in significant figures.
dvolovets
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(04-16-2011, 10:55 PM)
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Originally Posted by salva

All the acid is used up at that point.

Right, but doesn't the presence of CH3CH2COONa (0.0025 mols in this case) affect the acidity/basicity of the solution?

@ RevDM - Thank you! This makes a hell of a lot more sense now.
shadowsdarknes
I'M STILL A JUNIOR
(04-16-2011, 11:01 PM)
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Originally Posted by dvolovets

Right, but doesn't the presence of CH3CH2COONa (0.0025 mols in this case) affect the acidity/basicity of the solution?

@ RevDM - Thank you! This makes a hell of a lot more sense now.

Nope, that's just an extra product. Only the [OH-] and [H+] matters for pH.
You can run the ice table again using your Ka value, but it's going to be so small that the OH- concentration will only change by maybe half a percent.
Clemsontigers35
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(04-17-2011, 12:36 AM)
dvolovets, I'm a little confused by the answer to question #5. It doesn't make sense that you'd get .1 mol of a product from .001 mol of a reactant if they react at a one to one ratio. When I calculated it out, I got .00099 mol of the product, and 5.6E-6 moles of reactant iron. The method I used was to set up an ICE table, where the product was x at equilibrium, and the reactants were .2-x, and .001-x. I set up the law of mass action, then just solved for x. If you have figured this one out, please let me know.
dvolovets
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(04-17-2011, 07:17 AM)
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Originally Posted by Clemsontigers35

dvolovets, I'm a little confused by the answer to question #5. It doesn't make sense that you'd get .1 mol of a product from .001 mol of a reactant if they react at a one to one ratio. When I calculated it out, I got .00099 mol of the product, and 5.6E-6 moles of reactant iron. The method I used was to set up an ICE table, where the product was x at equilibrium, and the reactants were .2-x, and .001-x. I set up the law of mass action, then just solved for x. If you have figured this one out, please let me know.

How did you set up the problem in the first place? I honed my buffer problem solving skills all day today and feel very good about those, but problem #5 is confusing the hell out of me. If you could please post a solution (for 5.6e-6 mols of Fe), I would appreciate it. Not sure re: discrepancy between .1 mols and .001 mols...
thetechkid
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(04-17-2011, 07:30 AM)
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Well since Chemistry help is being delt, someone want to explain the principal quantum number and magnetic quantum number? I get how to go from Principal > Azimuthal > Magnetic > Spin but I don't get what those two mean, and how they relate to an electron.
dvolovets
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(04-17-2011, 07:36 AM)
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Originally Posted by thetechkid

Well since Chemistry help is being delt, someone want to explain the principal quantum number and magnetic quantum number? I get how to go from Principal > Azimuthal > Magnetic > Spin but I don't get what those two mean, and how they relate to an electron.

Principal quantum number indicates size. The higher it is, the higher in energy the orbital is. Magnetic quantum number indicates orbital orientation. These numbers, along with l (angular momentum, defines shape) identify one orbital.
thetechkid
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(04-17-2011, 07:42 AM)
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Originally Posted by dvolovets

Principal quantum number indicates size. The higher it is, the higher in energy the orbital is. Magnetic quantum number indicates orbital orientation. These numbers, along with l (angular momentum, defines shape) identify one orbital.

Thanks, my textbook doesn't understand how to state this stuff simply. Say you're given only the principal number and need to find the number of electrons, how would you go about doing that?
dvolovets
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(04-17-2011, 07:47 AM)
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Originally Posted by thetechkid

Thanks, my textbook doesn't understand how to state this stuff simply. Say you're given only the principal number and need to find the number of electrons, how would you go about doing that?

Principal number = n
l = 0 to n - 1
ml = -l to +l

The number of orbitals will be the number of ml values. So for example, say your n = 3.

n = 3
l = 0 (ml = 0)
l = 1 (ml = -1, 0, 1)
l = 2 (ml = -2, -1, 0, 1, 2)

Adding up total ml values gives you 9 orbitals. For some reason I'm drawing a blank as to how you'd calculate the number of electrons from orbital number, though...

Edit - Oh, wait, I'm an idiot... each orbital can hold 2 electrons. So your electron number is 9 * 2 = 18.

Edit 2 - You can also use 2n^2 to find # of electrons, bypassing finding l and ml.
Last edited by dvolovets; 04-17-2011 at 07:51 AM.
Dead Man
I got d 2 tha eepdicked
d-e-e-p-d-i-c-k-e-d
(04-17-2011, 07:50 AM)
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I do not miss Chemistry classes. At all. God, I sucked at this stuff.
thetechkid
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(04-17-2011, 07:51 AM)
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Yeah I get how to set those up, I just can't figure out the # of electrons.

Originally Posted by dvolovets

Edit - Oh, wait, I'm an idiot... each orbital can hold 2 electrons. So your electron number is 9 * 2 = 18.

Edit 2 - You can also use 2n^2 to find # of electrons, bypassing finding l and ml.

Alright that is pretty easy, thanks.
dvolovets
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(04-17-2011, 07:52 AM)
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Originally Posted by thetechkid

Yeah I get how to set those up, I just can't figure out the # of electrons.

See my edit above -- each orbital can only hold two electrons, so for n = 3, it's gonna be 9 orbital * 2 = 18 electrons. 2n^2 will also give you the same number.

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