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Chemistry - Adding Strong Acid to Buffer

  1. Nov 18, 2007 #1
    1. The problem statement, all variables and given/known data
    A beaker with 135 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1 M. A student adds 7.80 mL of a 0.400 M solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.76.


    2. Relevant equations

    Henderson-Hasselbach equation: pH = pKa + log([base]/[acid])

    3. The attempt at a solution

    hat I did was find the total number of moles of acid + base = 0.0135. Then using the pH = pKa + log([base]/[acid]) I found the ratio of [base]/[acid] to be 1.74. Then using these two equations I found the moles of acid to be 4.93*10^-3 and base = 5.45*10^-3. Next I found the moles of HCl = 0.00312 mol and substracted that from the base and added it to the acid so: acid = 8.05*10^-3 and base = 5.45*10^-3. Now I put these values into pH = pKa + log([base]/[acid]) and got the pH to be 4.59...therefore the change would be -0.40...But I think this value is too high...I think it's supposed to be lower. Can anyone tell me where I went wrong? Any help is appreciated.
     
  2. jcsd
  3. Nov 20, 2007 #2

    chemisttree

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    7.80 mL of a 0.400 M solution of what? Possibly a strong acid? I'll assume it is a strong acid (HCl).
    No. What does your question say about the total concentration of the acid [HA] and conjugate base[A-]? It isn't 0.0135.

    This part is right.

    Your logic is sound here but you used the wrong value for the ratio of [A-]/[HA]. Can you calculate initial [A-] given only the initial pH? You should also keep in mind that [A-](initial) + [HA](initial) = 0.1 and that total doesn't change throughout the problem, only the individual values of [A-] and [HA].

    Hope it helps.
     
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