1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Which formula to use to find the pH of a solution?

  1. Nov 5, 2017 #1
    • Please post this type of questions in HW section using the template.
    Hello, I just had my first exam yesterday, and I'm always tripped with this question that combines neutralization with buffer solution. The task is to find the pH. I did try to understand the general concept behind it. From my understanding, if CH3COOH reacts with NAOH and the acid has a higher amount of substance (0,005 mol vs 0,004 mol), it means that 0,005 mol of acid is used to neutralize 0,004 mol NAOH, and 0,001 mol of CH3COOH will remain together with 0,004 CH3COONA. To find the pH, one has to use the Henderson-Hasselbach equation and divide the amount of substance with the volume, with pH=pKa-(log c(CH3COOH)/c(CH3COONA). And then if NAOH has a higher amount of substance (0,0051 mol vs 0,005 mol acid), it would create 0,0001 NAOH and 0,005 CH3COONa. So to find the pH, you need to find the pOH of NAOH first. So far so good. But in the solution to the problems posted, there's suddenly an acid popping up in the solution and you have to use Henderson/Hasselbach equation. For example NH3 reacts with HCL and creates NH4CL, even if you have a bigger amount of substance for NH3, you still get an acid on the result... I'm so confused. And then there's an example with the coefficient (NA2CO3 + 2HCL --> 2NACL + H2CO3). if NA2CO3 has a higher amount of substance (let's say 0,224 mol) and HCL only 0,0816 mol, does it mean that it would result in 0,224 mol x 2 - 0,0816 NACL and 0,0816 mol H2CO3? I feel so stupid now that I can't understand this simple concept and keep making mistakes (and wasted like 8 points during the exam).
     
  2. jcsd
  3. Nov 5, 2017 #2

    Borek

    User Avatar

    Staff: Mentor

    This is quite chaotic and difficult to follow (please take your time to format the formulas correctly), some things do look apparent though.

    By acid do you mean NH4+? Yes, it is created in this reaction, but we are ignoring it just like we earlier ignored fact that CH3COO- is a base. You have solution containing an acid (NH4+) and its conjugate base (NH3), it is just another buffer (with pH described by the HH equation).

    Just follow the stoichiometry. First reaction is

    CO32- + H+ → HCO3-

    (this is in general an equilibrium reaction, but as a first approximation we can assume it proceeds to the end). Depending on the relative amounts of carbonate and hydrochloric acid you will end with some mixture of CO32- and HCO3- (again, acid and its conjugate base) or all CO32- will be protonated and the next reaction will start:

    HCO3- + H+ → H2CO3

    (again an equilibrium, again we initially ignore it). Here things get more complicated, as carbonic acid quickly decomposes and one of the products is CO2 that leaves the solution, making concentrations difficult to predict.
     
  4. Nov 5, 2017 #3
    thank you for the reply and sorry for the confusing format.. so does it mean that in the solution we have (0,224 mol x 2 minus 0,0816 mol CO3- and 0,0816 mol H2CO3? Sorry if I havent understood the full picture.
     
  5. Nov 5, 2017 #4

    Borek

    User Avatar

    Staff: Mentor

    If you have 0.224 moles of CO32- and 0.0816 moles of HCl, is there enough HCl to fully protonate CO32- (see the first reaction equation)?
     
  6. Nov 5, 2017 #5
    It's not enough I suppose.. but then why is there an acid in the solution?
     
  7. Nov 5, 2017 #6

    Borek

    User Avatar

    Staff: Mentor

    Which acid?
     
  8. Nov 7, 2017 #7
    H2CO3...
     
  9. Nov 7, 2017 #8

    Borek

    User Avatar

    Staff: Mentor

    How much of it?
     
  10. Nov 7, 2017 #9
    (0.224 x 2) - 0.0816, I suppose
     
  11. Nov 7, 2017 #10

    Borek

    User Avatar

    Staff: Mentor

    Stop guessing and just follow the stoichiometry.

    Take a look at the first equation I posted in the post #2 - what is the limiting reagent? How much HCO3- will be produced? How much HCl will be left?
     
  12. Nov 7, 2017 #11
    I'm following the stoichiometry, and that's why I still can't understand how to solve the problem.

    This is the original equation NA2CO3 + 2HCL --> 2NACL + H2CO3

    We have 0,244 mol NA2CO3 neutralizing 0,0816 HCL. From what I understand (and I'm following how the teacher solves the problem in the slide), you need 1 mol of NA2CO3 to neutralize 2 mol of HCL, so it's (0,244 x 2) mol - 0,0816 mol = 0,4064 of NACL left with 0,0816 mol of H2CO3.

    That's my understanding. Please correct me.
     
  13. Nov 7, 2017 #12

    Borek

    User Avatar

    Staff: Mentor

    Please reread my post #2 in this thread. This is not a single reaction, but a stepwise neutralization.
     
  14. Nov 7, 2017 #13
    ok I give up.. I really don't understand why it works differently with CH3COONA + CH3COOH where 0,005 mol CH3COONA + 0,0051 CH3COOH --> 0,0001 CH3COOH + 0,005 CH3COONA (straightforward reaction)
     
  15. Nov 7, 2017 #14

    Borek

    User Avatar

    Staff: Mentor

    Carbonic acid is diprotic and has two separate dissociation steps, each dissociation step has a different strength and a separate Ka.

    What you are probably missing is that carbonic acid can be think of as two separate acids - H2CO3 and HCO3-. First one is much stronger so it will react first, second one is much weaker and it will not react as long as there is the stronger acid present.
     
  16. Nov 7, 2017 #15
    So does the coefficient matter? Because in the original equation, there is 2HCL, but in the one you posted it's only CO3 + H --> HCO3. Does that mean that there will be 0,0816 mol of HCO3? and then 0,0816 of H2CO3? Assuming volume is 1 L, does this mean that the pH is pH = pKa - log (c(H2CO3/CO3) = pH = PKa - log (0,0816L/0.224L)? how would the pH be different in each step?
     
  17. Nov 7, 2017 #16

    Borek

    User Avatar

    Staff: Mentor

    These are different reactions, no wonder the coefficient is different.

    The overall equation

    Na2CO3 + 2HCl → 2NaCl + H2CO3

    makes sense only when the reaction goes to completion. It can't be used to predict products when the carbonate is in the excess.

    This equation:

    is wrong. Buffer contains a mixture of an acid and its conjugate base. Do you know what the conjugate base of an acid is in general? What is the conjugate base of the H2CO3? Conjugate acid of CO32-? Have you heard about Bronsted-Lowry theory of acids and bases? These are all basic things required to understand what this question is about. I strongly suggest you get back to your notes or the textbook to read on these things, otherwise I won't be able to help you further.
     
  18. Nov 7, 2017 #17
    why can't you just tell me the full solution? This is not a homework or anything (I dont know why the mod moved this thread here), I just need to understand how it works. Like in the easier example, there is 0.005 CH3COOH and reacting with 0.0051 CH3COONA. It leaves 0.0001 mol of CH3COONA and 0.005 CH3COOH, so the pH of the buffer between CH3COOH acid and its conjugate base will be pKa-(log cCH3COOH/cCH3COONA).

    Now in the previous example that confuses me, we have 0,244 Na2CO3 and 0,0816 HCL. So there is an excess of carbonate. so 0.244 Na2CO3 neutralizes 0.0816 HCL right? It should leave 0.244 minus 0.0816 mol of NA2CO3 and 0.0816 mol of H2CO3. If this reasoning is wrong, please just cut to the chase and provide the full solution. I've been reflecting over this for a week and it's driving me crazy.
     
  19. Nov 7, 2017 #18

    Borek

    User Avatar

    Staff: Mentor

    I told you several times where you are going wrong, you are just ignoring what you are being told.

    Reaction has two steps, stronger base reacts first.
     
  20. Nov 7, 2017 #19
    You only told me this is wrong, but I still do 't know how to appropriately apply the mols to the equations that you put. I just need that to understand the bigger picture of how to use the mols in the reaction.
     
  21. Nov 7, 2017 #20

    Borek

    User Avatar

    Staff: Mentor

    When I tell you to follow the equation I posted, you reverted back to your equation, which you were told is wrong. Look at my posts #2 and #10 - they contain everything you need. I am not going to repeat myself, it is a waste of time.
     
  22. Nov 7, 2017 #21
    ok
    Then forget it, I will just find someone else who is much more competent in explaining. Goodbye.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook