General chem questions!!!!

[allday0109]

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The question is A solution of NaOH and HF are titrated to the end point of a titration. THe final solution will contain?
a. Na+ and F-
b. H2O, H+, OH-, HF, Na+ and F-
c. H2O, F-, and Na+
d. Na+ and F-, NaOH
e. others
the answer is b. what I wonder is if NaOH can be contained in the final solution?
Hope that somebody answer to this.!!

 

[fifi1314]

No, there is impossible NaOH contained in the solution. Cuz NaOH will immediately dissolved in H2O and become Na+ and OH- ions.

 

[doc toothache]

The question is A solution of NaOH and HF are titrated to the end point of a titration. THe final solution will contain?
a. Na+ and F-
b. H2O, H+, OH-, HF, Na+ and F-
c. H2O, F-, and Na+
d. Na+ and F-, NaOH
e. others
the answer is b. what I wonder is if NaOH can be contained in the final solution?
Hope that somebody answer to this.!!

No, there is impossible NaOH contained in the solution. Cuz NaOH will immediately dissolved in H2O and become Na+ and OH- ions.

NaOH is not present in the final solution because all of it was used up in the titration with the acid and not because this was a trick question as suggested by fifi. Why there is an excess of HF as indicated in answer b is not clear, unless the titration was not potentiometric and the end point was determined using an indicator such as methyl red whose transition pH range is on the acid side (4.4-6.2). In such a case, a small concentration (<0.001M) of HF would be expected to have negligible effect on the final pH since, unlike HCl, HF is not fully ionized (Ka of 6.8 x 10^4). www.chemicalforums.com/index.php?page=miscellaneous

 

[sublimegd]

NaOH is not present in the final solution because all of it was used up in the titration with the acid and not because this was a trick question as suggested by fifi. Why there is an excess of HF as indicated in answer b is not clear, unless the titration was not potentiometric and the end point was determined using an indicator such as methyl red whose transition pH range is on the acid side (4.4-6.2). In such a case, a small concentration (<0.001M) of HF would be expected to have negligible effect on the final pH since, unlike HCl, HF is not fully ionized (Ka of 6.8 x 10^4). www.chemicalforums.com/index.php?page=miscellaneous

I think your over thinking this... This question reminds me of a question I got wrong on a gen chem 2 test that I'm still mad about. The question was asked about a titration curve of NaOH and HCl, with concentration of titrant graphed with pH. It asked whether pH started low and ended high or started low and ended neutral, or started low and ended slightly acidic... (I figured testing acid/base salt pH knowledge). I put that it started low (all HCl) and ended in the middle (I figured the titration was complete at the equivalence point, and NaCl is neutral)

Apparently the answer was pH starts low and ends high. My prof told me that "titration" in of itself doesn't really indicate how much is added. Therefore, the answer to the question could be whats in the solution before the equivalence point, after the equivalence point, or even after the titrant is saturated and can no longer ionize. You have to look at the potential answers and look for one that is even possible, which is B. HF is only partially soluble. NaOH could possibly be an answer I think, once the solution is saturated with it. All the confusion with this question stems from the poorly defined term titration.

Then again, I may have no clue what I'm talking about.

 

[sublimegd]

The question is A solution of NaOH and HF are titrated to the end point of a titration. THe final solution will contain?
a. Na+ and F-
b. H2O, H+, OH-, HF, Na+ and F-
c. H2O, F-, and Na+
d. Na+ and F-, NaOH
e. others
the answer is b. what I wonder is if NaOH can be contained in the final solution?
Hope that somebody answer to this.!!

Okay, I didn't read that it said end point of a titration. The dentist is still over thinking it though. I wouldn't assume that an indicator was used that wouldn't give the true endpoint for a standardized test question, maybe for a real life question. The final solution doesn't have to be acidic to have H+ ions because of the self ionization of water. The F- ions will continue to form HF acid with the H+ ions from the self ionization of water, so a small concentration of HF will be present at the end point. The Na ions won't form any NaOH with the OH ions however, so it won't be in the final solution. I hope I did more harm than good trying to answer your question. haha jk

 

[doc toothache]

I think your over thinking this...

HF is only partially soluble. NaOH could possibly be an answer I think, once the solution is saturated with it (????????). All the confusion with this question stems from the poorly defined term titration.

Then again, I may have no clue what I'm talking about.

Okay, I didn't read that it said end point of a titration. The dentist is still over thinking it though. I wouldn't assume that an indicator was used that wouldn't give the true endpoint for a standardized test question, maybe for a real life question. The final solution doesn't have to be acidic to have H+ ions because of the self ionization of water. The F- ions will continue to form HF acid with the H+ ions from the self ionization of water, so a small concentration of HF will be present at the end point. The Na ions won't form any NaOH with the OH ions however, so it won't be in the final solution. I hope I did more harm than good trying to answer your question. haha jk

Let's see if we can over analyze this problem one more time.
1. NaOH + HF= NaF + H2O

2. Neutralization reactions are irreversible reactions.

3. There are 3 possible outcomes of the titration.
a. the end point is at pH 7- The reaction is stoichiometric and the reaction products are NaF and H2O. The titration is followed potentiometrically.(no NaOH or HF present)
b. the end point has an acidic pH- resulting from an excess of HF if using an indicator whose pH range is acidic (methyl red, methyl orange, bromcresol green and others)-presence of HF
c. the end point has a basic pH-resulting from an excess of NaOH if using an indicator whose pH range is basic (phenolphthalein with a pH range of 8-10)-presence of NaOH

Incidentally:
HF is infinitely soluble in water.
The poorly defined term is end point not titration.
The hydronium ions do not react with F- unless you are suggesting that the reaction is reversible. If it does,as you suggest, what happens to the OH- and Na+? Are they just spectator ions?

 

[sublimegd]

Let's see if we can over analyze this problem one more time.

Okay, my turn.
I didn't mean to come of as condescending by stating that I think you over analyzed the question. I even stated that I may have no clue what I'm talking about. But now you have sparked my interest and I want to make sure I understand this question. Don't take what I'm about to say as me telling you the correct way to do the problem, just my thought process of working through the problem(which may be wrong).

First off, HF is infinitely soluble in water, what I mean was it only partially ionizes. This is my bad. As you pointed out this is a neutralization reaction, so the HF will react with NaOH as you stated...

"1. NaOH + HF= NaF + H2O"

You state that there are 3 possible outcomes, which is why I said you are over analyzing the problem. In a lab, you are right, but for a textbook test question you have to assume the reaction is stoichiometric...

"There are 3 possible outcomes of the titration.
a. the end point is at pH 7- The reaction is stoichiometric and the reaction products are NaF and H2O."

This is where I believe you may be wrong. The end point pH here is not 7, but is slightly basic. The reason for this is because NaF is a basic salt. As my old chem textbook states, "a salt of a strong base and a weak acid- the anion of the salt is the conjugate of the weak acid. It (F-) hydrolyzes to give a basic solution."

"The hydronium ions do not react with F- unless you are suggesting that the reaction is reversible. If it does,as you suggest, what happens to the OH- and Na+? Are they just spectator ions?"

Snap, your right. My bad yet again. BUT, the F- ions do react with water. I am suggesting this reaction occurs at the endpoint,

F- + H20 <-> HF + OH-

the Kb for this reaction is (1x10^-14 / the Ka for HF)

While on the other hand, Na+ does not hydrolyze

Na+ + H20 -> no reaction

So yes, Na+ is a spectator ion i guess? as it doesn't do anything after the endpoint.

I understand you being skeptical of my answer because my reasoning was way off. I also said quite a few incorrect things (solubility, the stupid titration definition rant, others). However, I am 99.9% sure that this final answer is correct, as I am staring at my textbook that explains this. None of your reasoning is incorrect, you just forgot about the conjugate acid/base salt detail. To solve these problems you do the neutralization reaction as it is irreversible, then go about finding the pH of the resulting salt as if the neutralization reaction never occured.


Example from my textbook: Calculate the pH of the solution at the equivalence point when 25 mL of .2 M nicotinic acid is titrated by .1 M NaOH. The Ka for nicotinic acid is 1.4 x 10^-5.

Solution: Assume the neutralization reaction is complete, so equal molar amounts of acid and base react, leaving the salt. Use basic stoichiometry to find the molar concentration of the nicotinate ion directly following the neutralization reaction, 0.05 M, Then find the pH of this solution, which is 8.78.

Really, thanks for calling me out on this one. I realize now I have to go back and check up on my gen chem.

 

[sublimegd]

adfdafd

 

[doc toothache]

Okay, my turn.


"1. NaOH + HF= NaF + H2O"

You state that there are 3 possible outcomes, which is why I said you are over analyzing the problem. In a lab, you are right, but for a textbook test question you have to assume the reaction is stoichiometric...

"There are 3 possible outcomes of the titration.
a. the end point is at pH 7- The reaction is stoichiometric and the reaction products are NaF and H2O."

This is where I believe you may be wrong. The end point pH here is not 7, but is slightly basic. The reason for this is because NaF is a basic salt. As my old chem textbook states, "a salt of a strong base and a weak acid- the anion of the salt is the conjugate of the weak acid. It (F-) hydrolyzes to give a basic solution."

Snap, your right. My bad yet again. BUT, the F- ions do react with water. I am suggesting this reaction occurs at the endpoint,

F- + H20 <-> HF + OH-

the Kb for this reaction is (1x10^-14 / the Ka for HF)

You are right. NaF is a salt of a strong base and a weak acid and therefore a solution of the salt would be basic. The OH- as you pointed out comes from the base hydrolysis rather than NaOH.