[gilbert]

Two tritration problems...

1.  CaCO3 requires 26.7 ml of 0.0887 mol. HCl for a complete reaction.  What mass of Ca CO₃ did the table contain?

my answer...

g CaCO₃? = 0.0267 L HCl  X  0.0887 mol HCl  X  1 mol CaCO₃ =  0.001184mol CaCO₃
                                                   1L HCl                     2 mol HCl

my problem is #2

                   Read along on my reply..

[gilbert]

Problem #2 is like this..

C₆H₅COOH is used for standarization of solution of bases.  A 1.922g sample of acids reacts with 29.47 ml of an NaOH sol.  WHat is the molarity of the base solution??

                                        C₆H₅COOH + NaOH = C₆H₅COONa + H2O             
i tried to answer it in this way..

mol of NaOH ? = 1.922g C₆H₅COOH  x  1 mol C₆H₅COOH  x    1 mol NaOH        =0.01575 mol NaOH
                                                        122 g C₆H₅COOH      1mol C₆H₅COOH


                           
mol of NaOH ? =  0.02947 L NaOH  x  0.01575 mol NaOH    =   0.0004567 mol NaOH
                                                        1L NaOH


          Is this 2nd calculation correct???

          Did i do a wrong step??

[Borek]

0.0267 ml HCl

L, not mL

0.001184mol CaCO₃

OK

[Borek]

A 1.922g sample of acids

1.022g C₆H₅COOH

0 or 9 - which one is correct?

Check molar mass of C₆H₅COOH.

Not sure what you did in other places, but you have already made two mistakes.

[gilbert]

       Mr. Borek,  you were right..   it was a 9 instead of 0 and the molar mass of the acid was also wrong..  it was 122 and not werever i had before..

any way, i have made those changes on the original problem #2...

My questions was about steps..   first i converted mass of acid into moles of bases...

later i converted volume of bases into moles of bases...I'm not sure if i'm doing the right procedures for the problem.  I am confuse.  please help with the procedures and i'll take care of the details .  As you all can see, i have tried to solve this because that is one of the requirements in this forums in order to get some help ..

[Borek]

Mr. Borek

It is Borek. Just Borek.

If you really want you may call me Mr. pH

My questions was about steps

Please redo the question correcting errors pointed out. I had some doubts about what you did, but it was buried under first two errors and I didn't want to spent too much time trying to guess the what of a what.

[gilbert]

There it is Borek... or should i call you..  Mr pH.   Please, enlight me with your wissdom.. I really appreciate your help.. Sorry for the long time away, the weekend was on the middle of all this, but as always, Monday has arrived on shcedule..
 
This is the problem that i have corrected.  The problem, as you see it, has all the corrections included.

Procedures about solving this problem is what troubles me.   I see a pattern on this type of problems that i'm having difficulty with.  The book only gives one example but the  actual work has a lot of variables.  If i'm able to understand the procedures to solve this problem - than, i'll be able to solve a bunch of this tritrations 

I'll say that again...  Please help with procedures. 

Problem #2 is like this..

C₆H₅COOH is used for standarization of solution of bases.  A 1.922g sample of acids reacts with 29.47 ml of an NaOH sol.  WHat is the molarity of the base solution??

                                        C₆H₅COOH + NaOH = C₆H₅COONa + H2O             
i tried to answer it in this way..

mol of NaOH ? = 1.922g C₆H₅COOH  x  1 mol C₆H₅COOH  x    1 mol NaOH        =0.01575 mol NaOH
                                                            122 g C₆H₅COOH      1mol C₆H₅COOH


                           
mol of NaOH ? =  0.02947 L NaOH  x  0.01575 mol NaOH    =   0.0004567 mol NaOH
                                                        1L NaOH


          Is this 2nd calculation correct???

          Did i do a wrong step??

[Borek]

Is this 2nd calculation correct???

No.

To say the truth I have no idea what you did and why. Dimensional analysis is not my method of choice, especially when there are more obvious (at least for me) ways of doing such things.

You know amount of the base, you know its volume - why not use definition of molar concentration?

http://www.chembuddy.com/?left=concentration&right=molarity

[gilbert]

I will like to go the way you are suggesting but i'm to deep into this method.  (not to mention that i need to show the work done when giving an answer in the test.  For my proffessor, this is the method of choice and, weather i like it or not, mine too.)  I'm afraid i'll get more confuse if i try to learn a different method.   

I was just trying to find the moles of NaOH...   That's all..    Thanks any way.

[Borek]

But you already know number of mole of NaOH - now you have to calculate concentration.

Looks like your conversion factor for DA should be just 1/volume of NaOH solution, no idea if such things are used.

You have given value: 0.01575 mol NaOH and you are looking for unknown - which is concentration.

Looking at units:

mole NaOH   ->       mole NaOH/L

so conversion factor must have unit of 1/L.

[gilbert]

Are you saying that concentration = molarity  ?
ok..  I think i'm posting the wrong question then..
i'm asking for # of moles instead of molarity .  which is concentration.  right??
I already have the # of moles, so, in order to find "molarity" (concentration) i need to divide

0.01575 mol.NaOH  by 0.02947L NaOH ?  is that right? or is it getting worst?
 

[Borek]

Are you saying that concentration = molarity  ?

Yes.

i'm asking for # of moles instead of molarity .  which is concentration.  right??
I already have the # of moles, so, in order to find "molarity" (concentration) i need to divide

0.01575 mol. NaOH  by 0.02947L NaOH?  is that right? or is it getting worst?

That's correct.

C = n/V

n = 0.01575 mole
V = 0.02947 L

[gilbert]

Call it lack of experience...  Also lack of knowledge... Can you see the mess i was doing for not being familiar with the vocabulary?   Molarity... not the same as # of moles

[Borek]

Browse my lectures (I have already posted a link), they may straighten out some things for you.