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A pH problem

  1. Oct 7, 2005 #1
    Hi,

    My Question:

    Upon evisceration, it was discovered that the frog had died due to acid ingestion. 10mL of gastric juice was found in its stomach required 7.2mL of 0.1 M NaOh for neutralization. Calculate the pH of its gastric juice.

    I'm confused, how would I approach this problem.

    Thank You:smile:
     
  2. jcsd
  3. Oct 7, 2005 #2
    Well first you might want to find the pH of the NaOH solution. Also, what have you tried?

    Alex
     
  4. Oct 7, 2005 #3
    So it would be like this

    poH=log[H]
    poh=-log[0.1]
    poH=1

    and

    ph+poH=14
    ph+1=14
    ph=13

    Is this correct:confused:

    thank you
     
  5. Oct 7, 2005 #4
    That looks good for the pH of the NaOH solution. Now, you want to find the pH of the gastric acid. Do you know how to do so from here?

    Alex
     
  6. Oct 7, 2005 #5
    No:cry: , I still don't know what to do. Also what do I do with the mL's

    Thank You:smile:
     
  7. Oct 7, 2005 #6

    Borek

    User Avatar

    Staff: Mentor

    No need for pOH.

    Calculate from simple stoichiometry amount of HCl in the gastric juice, then its concentration - once you know concentration of HCl pH calculation should be a breeze.
     
  8. Oct 8, 2005 #7
    Sorry to reply so late, so I would first find the concentration like so,

    M1V1=M2V2

    and then use the concentration to find the pH of the acid using this equation

    ph=-log(H+)

    Is that the correct method?

    Thank You
     
  9. Oct 9, 2005 #8

    Borek

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    Staff: Mentor

    Show what you did as your answer about using M1V1=M2V2 is ambiguous.
     
  10. Oct 9, 2005 #9
    Wouldn't it help if you knew the formula of the gastric juice?
     
  11. Oct 10, 2005 #10

    Borek

    User Avatar

    Staff: Mentor

    P-man - gastric juice contains mostly HCl.
     
  12. Oct 10, 2005 #11
    Oh, sorry. I guess that's the stuff in your stomach?
     
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