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I need help solving this question -- aqueous solution of sodium hydroxide + sulfuric acid...

  1. Dec 10, 2016 #1
    1. The problem statement, all variables and given/known data
    A 50.0 mL aqueous solution of sodium hydroxide has a pH of 12.50. If 36.00 ml of 0.0200 mol/L sulfuric acid is added to this sodium hydroxide solution, what will be the new pH of the resulting solution? Assume that the temperature stays constant at 25C, and that the volumes are perfectly additive.

    2. The attempt at a solution
    I tried looking at a similar post -->https://www.physicsforums.com/threads/quick-ph-calculation-please-verify.753708/#post-5639696, (I have no idea how he/she did it)

    I've tried doing 10^-12.50 to get [OH-] but I'm not sure it's right.

    3. I have no idea what to do to solve this question. Please help, and thank you. :D
     
  2. jcsd
  3. Dec 10, 2016 #2

    Borek

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    Staff: Mentor

    What you did was more or less a correct first step - that should give you the concentration of NaOH.

    Now write the reaction equation.
     
  4. Dec 10, 2016 #3
    So,
    2NaOH(aq) + H2SO4(aq) ---> Na2SO4(aq) +2H2O(l) ?
     
  5. Dec 10, 2016 #4

    Borek

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    How many moles of the NaOH were present? What is the limiting reagent?

    Please try to apply what you know to the problem, if you are going to just do whatever you are told you will never learn how to move ahead.
     
  6. Dec 13, 2016 #5

    epenguin

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    Will there is a way of being more sure one way or the other. In a neutral pure water [OH-] is 10-7. So for an NaOH solution does 10-12.5 sound reasonable?

    It would be nice and helpful also to alingy him/herself if he/she came back and explained
     
  7. Dec 13, 2016 #6
    Are you familiar with the fact that $$log[H^+]+log[OH^-]=-14$$ and that $$pH=-log[H^+]$$
     
  8. Dec 13, 2016 #7
    Yes, sorry, I had my exam yesterday and was really tired. But I did find the concentration by doing the following.

    14-12.50 =1.50

    then 10^-1.50 to get 3.16x10^-2

    Then I divided that by the 50.00ml of water to find initial moles.

    After doing that I made an IRF table..I also found the moles of H2SO4 by using the given information. I pretty much just solved the IRF table. (I'd have to look at my notes again to see what else I did.) But I think I undertand how to solve the problem. Thanks and sorry.
     
  9. Dec 13, 2016 #8
    You should multiply by 0.05 liters of water to find the initial moles. The 0.0316 has units of moles/liter.
     
  10. Dec 13, 2016 #9


    Ya that's pretty much what I did. I converted ml to liters by dividing by 1000. After that I multiplied the [OH-] by the L to get the moles of [OH-]
     
  11. Dec 13, 2016 #10

    epenguin

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    Checking for reasonableness and orders of magnitude needs to become a second-nature habit for chemical calculations. Even if you were partly along the right lines humourless Profs etc. may give you no credit at all or even less if your answer is out by 10 orders of magnitude! :oldsmile:
     
  12. Dec 13, 2016 #11
    What do you mean by orders of magnitude?
     
  13. Dec 13, 2016 #12

    epenguin

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    Factors of 10
     
  14. Dec 13, 2016 #13
    I see. Thanks.
     
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