I know this question is from the homework before the exam, but I am so confused about 14.32. My calculator always says "error" every time I try to put in 9.8 x 10^128. I'm not sure how to attack this homework problem.

14.32

You need to set up the problem and then factor to put the 10^128 and 1(10^-52)^2 portions separate. Do this with the math you learned to handle exponents. Then do the rest of the math on your calculator.

I honestly don't even know how to start/set up this problem.

Which problem?

Please explain how to solve problems 5 & 6 on chapter 14 online problems without a graphing calculator. These deal with x^4 and above, are we supposed to know how to do these by hand??

14.54 part B

Is the answer none because when we write the equilibrium constant we do not include solids?

Thank you!

This type of problem will be done in class tomorrow. Don't miss it! Critical skills will be covered! No graphing calculators needed.

14.54 part B

Yes, solids are not part of the equilibrium expression so adding or removing them does not affect the equilibrium concentrations of the remaining soluble reactants and products.

I'm not sure how to start/set up 14.32.

Dr. Patty,

For number 5 on the additional Ch. 14 problems, what do I do after I've set up Kc= ([0.30+3x)^3][x])/[0.6-2x] ? At that point, even after expanding, there's a polynomial of x>2. Do we have to use successive approximations?

Thanks!

14.32.

This is just like problem 13 on Exam 2,and uses the concepts discussed on page 5 of the Chapter 14 outlines. The basic approach is to use the two given equations and make them add up to the third equation. This involves flipping one (taking the inverse of the K), and multiplying one equation by 2 (then squaring the K). Once added the resulting Ks for the two equations need to be multiplied. This concept shows up again at the end of Chapter 15.

LeChatlier - Heat

Could you please clarify problem 7b of Chapter 13 addition problems. I thought that if a reaction was endothermic, heat would be considered a reactant. If heat is raised, then shouldn't the forward reaction increase more than the reverse?
I'd like to clear up my logic.

Thank you!

If pure liquids and solids are not included in the equilibrium constant, why does increasing water push a weak-acid reaction to the right?

If pure liquids and solids are not included in the equilibrium constant, why does increasing water push a weak-acid reaction to the right?


Another way to look at this is the following.

HA <===> H+ + A-

For simplicity, assume that at a particular equilibrium, the concentrations of HA, H+, and A- are all 2 M. This would mean that K = 2, i.e, (2)^2/2.

Now dilute the solution by adding enough water to double the volume. Since the system is now perturbed, it is necessary to calculate Q to see how the system will react to this change. At double the volume, the concentration of HA, H+, and A- are all changed to 1. Q will then equal (1)^2/1 = 1. Q is less than K, so the system responds by making more products, i.e. H+ and A-.

7b of Chapter 13

Once again, only some of the questions posted are resulting in an immediate or timely message sent to my email.

This is actually a question still under discussion among several of us on the faculty. We'll get back to you.

Can you please explain how to do number 5 on the chapter 15 additional problems?

Number 5 on the chapter 15 additional problems

Do you remember in class how I said decomposition increases at higher temperature? So when you heat up water, more water ionizes (splits) into H+ and OH-. When you multiply the concentrations of these two ions, the product is 1 x 10 ^-12 instead of 1 x 10^-14. So using the same rationale as used in the notes, or a simple ICE table where at equilibrium there is x of H+ and x of OH-, the [H+] concentration can be determined.

What would you use as the initial concentration of the water?

Dear 1234,

Let me ask you a question before I answer. I think it might help you make a better decision.

What would you use for the concentration of water in an aqueous reaction?

For the benefit of everyone, please post your answer so I can respond if it leads to another question.

Thanks,
Dr. P.

Would you do the number of moles of water dissolved in 1 L?

1234,

You need to do the following:

Read pages 670 - 672 in the book
Read your notes, especially the top of page 2 in Chapter 15
THINK! REMEMBER CHAPTER 14, page 4 of notes

15.100

How would we approach this problem? I tried using PV=nRT, but my answer came out to be 0.124, rather than 0.106. I used the grams given and the mole ratio to find the moles of CO2(g), but my answer came out too big.

Can you go over how to tell whether a salt solution will be acidic, basic, or neutral?? (Problems 15.78 & 84) Thanks!!

Can you go over 15. 107 please?

Dear Chem

15.100

Can you tell me how many moles of CO2 you calculated from the reaction stoichiometry?

Dear T,

Why don't we start with you telling me how much you already know about salt solutions. That will tell me where to start.

Name one salt that is acidic, one that is basic, and one that is neutral. You may use the book and your notes.

Once I hear back, I'll fill in the gaps.

Dear L,

15.107.

Start by writing out he reaction for the dissociation of formic acid, HCOOH.

Then write out reaction for the dissociation of H2O. Put the K's beside each equilibrium reaction.

Now rearrange (may reverse one) the reactions and treat the Ks appropriately just like you did on the last exam.

If you don't get it, then please type out the reactions and Ks, and I'll give you the next hint.

I'm having trouble with 15.129 and 15.130. If you could explain the first one I could probably figure out the second since they are similar. Thanks

Dear sf.

Try the problem at the top of page 13 of the Chapter 15 outlines.

Start by writing out the equilibrium reactions. Then fill in the part of the ICE table that is given in the problem. That might not be in the I part of the table!

Just to make sure I did number 5 on the Chapter 15 additional problems correctly...

According to the Chapter 14 notes, in pure water
[H+]=[OH-]=x^2=Kw

So, x^2=1.0 x 10^-12
and x= 1x 10^-6

Then to find pH: -log (1x 10^-6)= 6

Did I reason through that correctly?

Thank you!

Number 10 on Chapter 15 additional problems

I got everything but the concentration of C2O4 2- correct.
I set up my second ICE table like this:

HC2O4- + H20 <-> H30+ + C204 2-
I 0.315 0 0
C -x +x +x
-----------------------------------------------
E (0.315-x) x x


Then I set up my second equation as

K2= 6.4 x 10^-4= (x^2)/(0.315-x)

I tried assuming and using quadratic equation and neither got me to the correct answer. Did I miss something important??

Chapter 14 additional problems.

Hmm. Seems you might not have reviewed the problem we did in class on diprotic acids. Why is there no H+ in your equation above. I thought there was a bunch of it from the first dissociation step. Where did it go? H+ is not a gas. It does not bubble out of the solution. It has to be somewhere in that beaker.



Hmm, remember the shortcut. Tell me Ka2 and I will tell you A^2- in a flash. But prove it like in the notes.

For number 15.130 I got the right answer by following the procedure in the notes, but for some reason I did not get number 15.129 correct. I found [HCN]=5.0x10^-4 which should also be equal to [NaCN] and then did stochiometry and molar masses but I got 6.13x10-3 and not your answer of 7.4x10^-3 g

15.129.

Send me the reaction you are using. Give it some serious thought. There IS NOT AT RULE that says that HCN = NaCN !!!!!!!!!!!!!!!!!!

Write the reaction. Then I'll know where to start. I suspect you still have not looked at your notes and the exact same problem. You cannot fall into the "x" trap here.

number 5 on the Chapter 15 additional problems


Looks correct, except for what is probably a typo. Instead of
[H+]=[OH-]=x^2=Kw

it is

[H+][OH-]=x^2=Kw

SMU is not allowing me access to the website at the moment, so here is some important information.

On the exams posted from last year (2012), Exam 2 and problems #2, 5, 6, and 7 apply to YOUR 2013 Exam 3 (March 20).

14.90

Would we just solve this problem using the quadratic rule, or could we also assume that it has undergone completion and only a trace of y is left?

This is the equation that I got:
54.3=[0.369-2x]^2/[0.230-x][0.410-x]
Would that be correct?

15.100

I'm a bit puzzled by why stoichiometry also works here. I'm assuming it's because the ICE table set up makes a bunch of things cancel out and for it to appear that the antacid reacts completely? If this is the case... would we use the "Ka" of HCO3- <=> CO3(2-).

Thanks!

14.90

First you have to decide which way the equilibrium will shift since it's clearly not at equilibrium. Calculate Q which shows that Q is too small. There needs to be more product (hence 0.369 + 2x, not -2x as you typed).

The concentration for H2 should be 0.298 and not 0.230.

54.3=[0.369+2x]^2/[0.298-x][0.410-x]

Yes, the quadratic would need to be solved. Just make sure you understand the setup.

15.100

This turns out to be a simple acid plus base reaction. HCO3^- is a base. H+ will completely transfer to HCO3^- making H2CO3 which will from CO2 and HCl. So there are no Ks to worry about. The Ka of HCl is essentially infinity and overpowers anything that HCO3^- might do alone.

Can you help with 7 on the chapter 15 additional problems?

Dear 713 (and other aliases I believe you might be using)

This is a very unusual question, given that the questions before it are usually deemed to be more difficult. I think you have misinterpreted the question. It's simply asking for the the H+ concentration of weak acids with very simple Kas. If you can solve this for acetic acid where Ka is 1.8 x 10^-5, then it is much easier for these hypothetical acids where Kas are 1 x 10^-x.

If this does not get you started, please respond by posting again, but please type the equilibrium reaction in your responding post and include the Initial concentrations to be used in the ICE table.

15.76

To find the salts that undergo hydrolysis, is it just the salts that do not come from a strong acid and strong base?

Number 2 part e on the chapter 15 additional problems.

I don't get why the product has an H+ if (CH3)3N is a base?

# 6 on the additional problems for chapter 15,

I set the whole problem and got 1.46 for the pH?

#8 part f on the additional problems, I assumed that x<1.0X10^-4 and got 1X10-5 for x and then took the negative log of that to find the ph and got 5. But the answer online is 6

Could you explain the reasoning behind #112 on Ch. 15 homework problems? (This is purely a conceptual question.) What makes a strong base strong? And why is "F is more electronegative than H" given as a hint?

Chapter 15 Additional Problems- Number 5

I can't seem to figure this out

15.76

At least one ion of the salt must be from a weak acid or a weak base.

We will do some of these in class on Monday, before beginning Chapter 16.

2e on Chapter 15 additional problems:

I corrected the omission of water on the email just sent to all students in the class. The equation was not balanced without the water. This should clarify your question. It is analogous to putting ammonia, NH3, into water.

Dear 1234,

# 6, Chapt 15 additional problems.

You calculated an acidic pH, but no acid was added to the water. A SALT was added.

Write the equilibrium for what was put into water, as was done on page 13 and page 20 of the notes. i.e.,

A- + H2O <===> HA + OH-

Thus the solution is, therefore, basic and Kb must be used (calculated from Ka).

# 8, Chapter 15 online

Turne out that we should not assume x < 1 - 10^-6 here. The H+ concentration from water becomes important in this very very dilute solution. Hence the ICE table must show the original H+ concentration as 1 x 10^-7 from water.

So the quadratic must be solved:

10^-4 = [10^-7 +x][x] / [10^-6-x]

The solution just gave me 6.0044 for the pH but with sig figs this is 6.00.

#112 on Ch. 15 homework problems


The electron pair in NF3 is pulled toward N by the three very electronegative F atoms and is, therefore, not available to donate the pair (Lewis definition) or to accept an H+ (Bronsted definition)

Chapter 15 Additional Problems- Number 5

As posted above:


Number 5 on the chapter 15 additional problems

Do you remember in class how I said decomposition increases at higher temperature? So when you heat up water, more water ionizes (splits) into H+ and OH-. When you multiply the concentrations of these two ions, the product is 1 x 10 ^-12 instead of 1 x 10^-14. So using the same rationale as used in the notes, or a simple ICE table where at equilibrium there is x of H+ and x of OH-, the [H+] concentration can be determined.

old test # 2 problem 5

I'm having trouble finding pH of a 0.25 M solution of Ba(OH)2

I can figure out that since 2 mols of OH are produced per 1 of Ba, that the M of OH- is 0.125 but log of that is 0.9, I find the ph would be 13.1?
please guide me to my issue.

Missing from email message:

#2, 5, 6, and 7 from Exam 3 last year apply to this exam (Exam 3) as well as all of Exam 2.

Additional chapter 5 problems number 6

I did it like you said using kb and got 8.5 instead of 7.5?

I got kb to be 1X10^-8

and x=3.16X10^-6 which was the OH- and then found the pOH to be 5.5.

I then did pH= 14-5.5= 8.5?

15.50 part B

I did 10^-1 to find the H+ which was found to be 0.1 Then I divided 7.75X10^-3(from part A) by 0.1 and multiplied by 100 to get 7.75% ionization. My answer is way off.

Additional chapter 5 problems number 6

Kb is not 1X10^-8 . It is (1 x 10^-14) / (1 x 10^-4) or 1 x 10^-10.

Dear Justin,

What a horrible choice for an alias! Not even very creative!

old test # 2 problem 5

The molality of the packaged deal, i.e, Ba(OH)2 is 0.25 M. When the package is opened up (dissolves) all the pieces come apart (ions) giving one Ba2+ and 2 OH- ions per package. That means the MOLALITY of the OH- ions is 0.50 M.

15.50 part B

Set up a new ICE table with the HA (aspirin) as 0.20, the H+ as 0.1, and zero for the A- as the Initial concentrations. Then take it from there. This is the common ion effect and we will see MUCH MUCH MUCH more of this.

15.100

How should i approach this? I feel like im missing something really simple

15.100

See answer to person who asked about this previously. It's somewhere above.

I am a little confused on when we can assume a reaction as undergone completion as we did in additional ch.14 problems #4 and 5. I am aware that we can assume that the reaction has undergone completion when Ka is very large, but is there any specific number in which we can consider Ka to be large enough to assume completion like when Ka is above 10^3 or something?

Dear Ka,
Assuming is all relative. There is no magic number that says assume one way or the other. It depends on concentrations as well as the size of the number. If you think about it logically, it's not that hard.

15.34

Why would CN- be a weak base. Isn't it the conjugate base of a weak acid a strong base?

And also 15.104

Why would it take the same volume to neutralize both HA and HB?

15.34

Why is NH2- a strong base?

15.34

What is the Kb for the hydrolysis of CN-? The fact that I ask, indicates there is one, and since K is not infinity, it is not labeled strong. It is "stronger" than e.g., Cl-, but not as strong as OH-.

15.104

HA + NaOH ---> NaA + H2O
HB + NaOH ---> NaB + H2O

Both go to completion, i.e, acid + base gives salt.

15.34

NH2_ is a STRONG STRONG STRONG BASE, because it reacts with water in a flash (literally) to give OH- (and NH3). There is no reverse reaction. It's 100 %.

15.85

Why would CO and NO be neutral instead of acidic?

15,85

This is a poor question and it should not have been included in the book. These two are exceptions to the trend.

14.80

I am not getting the correct answer and I think it's becauase my ICE table is off. I think I am misinterpreting the information given. Could you please help?

The ICE table I have is:

SO2Cl2 <-> SO2 + Cl2

0.0252 0 0
-0.0173 +0.0173 +0.0173
0.0079 0.0173 0.0173

With these values I get kp = 0.0389 and kc = 7.3 * 10^-4.

Hi Dr. Patty,

Never mind about 14.80. My mistake (which I repeated several times on other problems--lesson learned) was treating concentrations as partial pressures because the chemicals are gases. (So, I tried to find kp using molarities and then find kc from that.)

Thank you!

I came to your office hours today about #5 on the chapter 14 additional problems and you said to do 5.8X10^6=(0.3-y)(1.2-3y)^3/(y)^2.

Why wouldn't you do (2y)2??

14.46

I was doing my homework problems again and seemed to read this question wrong. And now I have no idea how to do this

Dar AZ,

Go ahead and do 2y squared. Once you have solved for y, you will need to multiply by 2 to get the concentration of NH3.

Your answer will be the same.

14.46

First determine K from the equlibrium concentrations given. Just plug them into the K expression.

For part b, three of the concentrations stay the same as they were, but the fourth, CO2 changes to 0.50 M which means the reaction shifts to the right making more CO and H2O (call this change x for each of them). At the same time x H2 and x CO2 is also used up.

So the K = (0.086 + x)(0.040+ x)/ (0.5 - x)(0/045 -x). No assumptions can be made so this leads to a lot of algebra and solving a quadratic equation --- which you don't have time to do tonight!