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Calculating the molar concentration of H3O+ and pH of solutions

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  1. Apr 29, 2016 #1

    NYK

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    1. The problem statement, all variables and given/known data
    Calculate the the molar concentration of H3O+ ions and the pH of the following solutions:
    a) 25.0 cm3 of 0.144 M HCl(aq) was added to 25.0 cm3 of 0.125 M NaOH(aq)
    b) 25.0 cm3 of 0.15 M HCl(aq) was added to 35.0 cm3 of 0.15 M KOH(aq)
    c) 21.2 cm3 of 0.22 M HNO3(aq) was added to 10.0 cm3 of 0.30 M NaOH(aq)

    2. Relevant equations
    pH = -log[H3O+] Handerson - Hasselbach eqn.

    3. The attempt at a solution

    I have only been able to solve part a)

    I did that by multiplying the molairty of the strong acid/base by the total volume

    then: [H3O+] = [HCl] - [NaOH] = 9.5 x 10-3mol

    then took the negative log of that number to find the pH = 2.0

    which are the correct answers, but when I do that process for part b) and c) which are also strong acid/base combos, the answers are no where near correct.

    Any help will be appreciated.
     
  2. jcsd
  3. Apr 30, 2016 #2

    Borek

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    Staff: Mentor

    In the second case - what is the limiting reagent? What is left in the solution after the neutralization reaction took place?
     
  4. Apr 30, 2016 #3

    NYK

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    In the second case the LR is the acid. When i do that calculation I find that the [H3O+] = .04875 mol/L

    the answer is 21 mmol/L and a pH = 12.3
     
  5. Apr 30, 2016 #4

    Borek

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    Staff: Mentor

    Good.

    I am afraid neither of these numbers is correct.

    First of all - you said acid was the limiting reagent. If so, how come there is mo much H3O+ left?

    21 mmol/L of what?

    pH of 12.3 is quite close - but it is possible to easily give a better answer.
     
  6. Apr 30, 2016 #5

    epenguin

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    Homework Helper
    Gold Member

    Strong acids and bases, so Henderson - Hasselbach eqn. never came into it.
     
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