[NewtoAtoms]

Calculate the H⁺ of 25 mL of 2.50 x 10^-3 M nicotine (for which kb= 1.05 x 10^-6) and 50 mL of 1.25 x 10^-3 M HCl

I have done  most of the leg work, but I could also be out in left field. Would someone be so kind to review my work and tell me if I am correct, or if I need to re-hit the books...

Step #1

      # of moles nicotine                  
                        
          X        =        2.50 x 10^-3    =  6.25 X 10^-5 moles   
        -------           ---------------      
         0.025 L               1L            
                        
      # of moles HCl               
                        
          X               =     1.25 x 10^-3   = 6.25 x 10^-5 moles      
         ---------           ------------
           0.050 L                1L            

IN THIS MIXTURE THE NICOTINE COMPLETELY NEUTRALIZES THE HCL! But what is the remaining H+???
                        
                  HCl              Nicotine   ---------->    ‘salt’                             H20
   Initiation   5.00 x 10^-3      5.00 X 10^-3                         O                        O
   Change   5.00 X 10^-3      5.00 X 10^-3                 5.00 X 10^-3                      O
-----------------------------------------------------------------------------------
   Final                  O            O                          5.00 X 10^-3                   O            
                        
   salt =    5.00 x 10^-5                  = 7.14 x 10^-4 M   
              -------------            
                0.070 L     (total volume)               
                        
   Kb =       [Nicotine] [HCl]   
               ----------------               
                       [Salt]      
            
   1.05 x 10^-6 =   [x) (x]                  
                      -------------
                  [7.14 x 10^-4 - x]      
            
    7.50 x 10^-10 = X²                     
   2.74 x 10^-5   = x                     
   Therefore the concentration of H+ is 2.74 x 10^-5   

Is this at all correct.... OH I THINK either I have it or I am TOTALLY lost. 

Pleeeeeeeeeeeeease help me!

Thank you for your time
New to atoms                  

[Borek]

You are twice right. First time you are right when you write about completele neutralization. Second time you are right when you write "I am completely lost" 

Salt consist of protonated nicotine (weak conjugate acid of nicotine) and Cl^- (base so weak, that its presence doesn't change pH).

You have to find out Ka of protonated nicotine (hint: pKa + pKb = .), then you have just a solution of weak acid.

7.14x10^-4 is not salt concentration, check your math.

[NewtoAtoms]

Thank you for the reply Borek,

You're right that was a bad math error, I apologize,

salt =    5.00 x 10^-3                  = 7.14 x 10^-2 M 
              -------------           
                0.070 L     (total volume)               
                       
   Kb =       [Nicotine] [HCl]   
               ----------------               
                       [Salt]     
           
   1.05 x 10^-6 =   [x) (x]                 
                      -------------
                  [7.14 x 10^-2 - x]     
           
    7.50 x 10^-8 = X²                     
   2.74 x 10^-4   = x                     
   Therefore the concentration of H+ is 2.74 x 10^-4

Now the part I am confused about is the part where you said "you have to find out Ka of protonated nicotine"  I do not have a great teacher, so this might be completely wrong. However, she said that when the acid and base completely ionize to leave nothing but a salt (nicotine salt in this case) the kb is used and no longer the Ka. 
Therefore I am torn between what she said and what you suggest.
Can you at all tell me why I would calculate the Ka???  Is Kb = [nicotine][acid]/[salt] not the correct equation to use in this situation???
I apologize if my question seems elementary, however I am a tad lost in the acidity of chemistry.

But thank you for your time

New to atoms

[NewtoAtoms]

Okay wait a minute.... wait a minute....

2.74 x 10^-4   = x      this would represent the OH-

Therefore the following must proceed as:

pOH = -log (2.74 x 10^-4)
pOH = 3.56
pH + 3.56 = 14
pH = 10.44

pH = - log (H+)
10.44 = - log (H+)
H+ = 10^-10.44
THEREFORE H+ concentration would be 3.63 x 10^-11

That's better right?!?!?!?  Oh I think I might be getting closer..

I hope anyways..

Thank you for your time.

[Borek]

salt =    5.00 x 10^-3                  = 7.14 x 10^-2 M 
              -------------           
                0.070 L     (total volume)

It is still wrong. Please check what was number of moles of nicotine (your original post contains the correct number) and what was the final volume. Neither 5.00 x 10^-3 nor 0.070 is correct.

Now the part I am confused about is the part where you said "you have to find out Ka of protonated nicotine"  I do not have a great teacher, so this might be completely wrong. However, she said that when the acid and base completely ionize to leave nothing but a salt (nicotine salt in this case) the kb is used and no longer the Ka. 
Therefore I am torn between what she said and what you suggest.

When you have solution of acid you will use Ka, when you have solution of base - you will Kb.

Nicotine is a base, so when you have solution of JUST nicotine, you should use Kb.

However, when nictotine gets protonated (in salt) it becomes its conjugate acid (do you know Brønsted-Lowry theory of acids and bases?). As it is an acid now, you should use Ka.

Unfortunately your other approach (the one with pH 10.44) is wrong. You should use ICE table.

Besides, it is obvious that something is wrong even without checking calculations. pH 10.44 means solution is basic, and you have solution of protonated nicotine, that is weak acid, its pH must be slightly acidic, below 7.0.

[shanmac]

Calculate the H+ of 25 mL of 2.50 x 10^-3 M nicotine (for which kb= 1.05 x 10^-6) and 50 mL of 1.25 x 10^-3 M HCl

I did find this question posted from a few years ago, but the final answer never came up.  I've done the problem here and am hoping that I'm on the right track.  Any feedback would be appreciated as I'm still pretty confused!

Here is what I have done:

25mL ÷ 1000 mL = 0.025 L nicotine
0.025 L x 2.50 x 10^-3 moles = 6.25 x 10^-5 M nicotine

50.0 mL of 1.25 x 10^-3 M HCl = 0.050 L x 1.25 x 10^-3 M = 6.25 x 10^-5 M HCl.

Since HCl is a strong acid and disassociates completely, it has 6.25 x 10^-5 M [H+].

Kb nicotine= 1.05 x 10^-6

pKb nicotine= -log 1.05 x 10^-6 = 5.98

14 - 5.98 = 8.02 = pKa nicotine

10^-8.02 = 9.55 x 10^-9= Ka

              Nicotine             Nicotine +   OH-
Initial (M)   6.25 x 10^-5           0             0
Change   -x                         +x           +x
Equilibrium   6.25 x 10^-5 - x           x             x

Ka = 9.55 x 10^-9 = x^2 / (6.25 x 10^-5)


x = 7.73 x 10^-7 = [H+].

6.25 x 10^-5 M HCl - 7.73 x 10^-7 M = 6.17 x 10^-5 M

6.17 x 10^-5 M ÷ 0.075 L = 8.23 x 10^-4 moles H+ per 75 mL.

Please let me know if this makes sense.

[Borek]

Please give the final answer as pH, even if that's not exactly what the question asks for.

25mL ÷ 1000 mL = 0.025 L nicotine
0.025 L x 2.50 x 10^-3 moles = 6.25 x 10^-5 M nicotine

You are already doing some strange things - I guess you have tried to calculate number of moles, but the units you have listed suggest you did something completely different.

You have a mixture of base (nicotine) and acid - they react. You can't use ICE table for nicotine ignoring presence of H⁺.

[shanmac]

OK, yes it is evident that I am very confused. 
If I change the ICE chart to this does it make more sense?

                 Nicotine                 H+          OH-
Initial (M)   6.25 x 10^-5           0             0
Change           -x                    +x           +x
Equilibrium   6.25 x 10^-5 - x     x             x

Ka = 9.55 x 10^-9 = x^2 / (6.25 x 10^-5)

x = 7.73 x 10^-7 = [H+].

6.25 x 10^-5 M HCl - 7.73 x 10^-7 M = 6.17 x 10^-5 M (my logic here is that I am taking the difference between the [H+] of HCl and the [H+] of protonated nicotine because all of the HCl will have reacted.)

6.17 x 10^-5 M ÷ 0.075 L = 8.23 x 10^-4 moles H+ per 75 mL.  (This takes the final concentration  of 75 mL into account).

This would make the final pH 3.08-- which I suspect is a bit too strong!

[AWK]

Ka = 9.55 x 10^-9 = x^2 / (6.25 x 10^-5)

this should be concentration

[Borek]

Apart from AWK wrote - that you should use concentrations in ICE table - your ICE table is still wrong. ICE table should refer to the reaction taking place. Write the reaction equation.

[shanmac]

Thanks for the tips.
OK.  I hope that I have the reaction equation right here. I'm having trouble envisioning what is happening, which is why I am so confused with which equations/reactions to work with.  Am I right in saying that the HCl completely reacts with the nicotine, forming a salt?
Then what I need to figure out is how much of that salt disassociates in the solution to reach equilibrium?  (This is my first attempt at Chem).

           HCl              Nicotine   ------------>       salt             + H20
   Initiation   6.25 x 10^-5      6.25 x 10^-5                         O                       
   Change    -6.25 x 10^-5      -6.25 x 10^-5                +6.25 x 10^-5                      
-----------------------------------------------------------------------------------
   Final                  O            O                          6.25 x 10^-5                              
                        
   salt =    6.25 x 10^-5                  = 8.33 x 10^-4 M   
              -------------            
                0.075 L     (total volume) 

So, if that is right.... I have a salt of 8.33 x 10^-4 M sitting in a 0.075L solution.  All of the HCl has been used up, and all of the nicotine has been used up.  I must be wrong somehow.  I don't understand why I need to then calculate the pKa of the nicotine.  Could you explain that further?

[Borek]

I have a salt of 8.33 x 10^-4 M sitting in a 0.075L solution.  All of the HCl has been used up, and all of the nicotine has been used up.

That's correct. As amounts of both substances was exactly identical, what you have in the solution now is HNicotine⁺ - protonated nicotine, and identical amount of the counterion, Cl^-. HNicotine⁺ is a weak acid (that's Bronsted-Lowry theory at work), that dissociates lowering pH.

Your ICE table is slightly wrong, as in fact final concentrations of nicotine and H⁺ won't be exactly zero, but it apparently pushed you in the right direction

[shanmac]

Whew!  So I got that far... but I still have no final answer.  Nor have I taken the Kb of nicotine into account at all.  Carrying on from there, is this right?

[salt] = 6.25 x 10^-5 ÷ 0.075L = 8.33 x 10^-4 .

Kb nicotine= 1.05 x 10^-6

pKb nicotine= -log 1.05 x 10^-6 = 5.98

14 - 5.98 = 8.02 = pKa nicotine

pH = pKa + log [salt]/[acid]

pH = 8.02 + log (8.33 x 10^-4) ÷ (6.25 x 10^-5)
pH = 8.02
H = 10^-8.02

[H+] = 9.5 x 10^-9

[Borek]

Whew!  So I got that far... but I still have no final answer.  Nor have I taken the Kb of nicotine into account at all.  Carrying on from there, is this right?

[salt] = 6.25 x 10^-5 ÷ 0.075L = 8.33 x 10^-4 .

Kb nicotine= 1.05 x 10^-6

pKb nicotine= -log 1.05 x 10^-6 = 5.98

14 - 5.98 = 8.02 = pKa nicotine

OK so far.

pH = pKa + log [salt]/[acid]

No. Just calculate pH of the acid solution. You can't use concentrations of acid and conjugate base, as you don't know them yet.

But I see you are still confused. Yes, you have a salt in the solution (nicotine hydrochloride), but it is this salt that is the source of the acid - HNicotine⁺.

[shanmac]

I really appreciate your help Borek.  I am trying to teach this stuff to myself and it's not easy.  Have I finally got it now....?

Ka nicotine= 10^-8.02 = 9.5 x 10^-9

9.5 x 10^-9 = x2 ÷ (6.25 x 10^-5 -x)
x= 7.7 x 10^-7 = [H+]

pH = 6.11 (slightly acidic, so should be on track....)