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What happens if an acid is added to this buffer?

  1. Jul 13, 2014 #1
    1. The problem statement, all variables and given/known data
    In grapes, there is an equilibrium between tartaric acid and hydrogen tartrate and hydrogen ions:

    (1) H2T (aq) <=> HT- (aq) + H+

    NOTE: "T" stands for the tartrate ion C4H4O6.

    There is also a buffer system in grapes, involving a solubility equilibrium of potassium hydrogen tartrate and hydrogen tartrate and potassium ions:

    (2) K+ (aq) + HT- (aq) <=> KHT (s)

    What happens if you add H2T to this system?


    2. Relevant equations
    N/A


    3. The attempt at a solution
    If I added an acid to this system, the buffer will oppose the decrease in pH and so the pH will stay the same, right?

    Suppose I added H2T to this system: then I have increased the concentration of H2T, and due to Le Chatelier's Principle, equilibrium (1) will shift to produce more HT- and H+. This will increase the concentration of HT- however, so equilibrium (2) will shift so that K+ reacts with HT- to produce KHT, which will precipitate. But hold on, the concentration of H+ has increased, meaning the pH has gone down. The buffer has not opposed a decrease in pH!

    What's going on?
     
    Last edited: Jul 13, 2014
  2. jcsd
  3. Jul 14, 2014 #2

    Borek

    User Avatar

    Staff: Mentor

    Assuming you managed to prepare solution that behaves the way you describe, you add the acid AND remove the conjugate base at the same time, so it is not as simple as just adding acid to the buffer.

    Besides, when you add acid (base) to the buffer pH always goes down (up) - just slower than if the buffer wasn't present.
     
  4. Jul 26, 2014 #3
    The reason for this is because while the acid is neutralized, remember that base was used to neutralize it. You lose hydroxide ions from solution and the volume of acid added will decrease the concentration of all entities involved.
     
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