Destroyer General Chem Q201

Meas

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May 14, 2009
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  1. Dental Student
    HOw many ml of 0.120M KOH must be added to 60ml of 0.2M HF to produce a solution with a pH of 3.3? (pKa of HF is 3.3)

    Ans: 5ml

    - THe answer key shows that c1v1 = c2v2 eqn can be used.
    - When I first approach this question, I quickly use the pH to find out how much [H+] is needed.

    Please help! Thanks in advance!
     

    nze82

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    May 15, 2009
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    1. Dental Student
      HOw many ml of 0.120M KOH must be added to 60ml of 0.2M HF to produce a solution with a pH of 3.3? (pKa of HF is 3.3)

      Ans: 5ml

      - THe answer key shows that c1v1 = c2v2 eqn can be used.
      - When I first approach this question, I quickly use the pH to find out how much [H+] is needed.

      Please help! Thanks in advance!

      Notice how pKa = pH. This is the characteristic of the half point. Why?
      Remember that:
      pH = pKa + log [A-]/[HA]

      At half point [A-] = [HA], so log [A-]/[HA] = 1, so:
      pH = pKa + log1 = pKa

      So, in this example, you just need to find the volume of KOH added to reach the half point. How?

      Use M1V1 = M2V2

      V1(0.12M) = (60ml)(0.2M)
      V1 = 100ml

      However, this is the volume of KOH added to reach the equivalence point; whereas, we're looking for the volume at the half-point. So, just divide this volume by 2.

      Ans: 100ml/2 = 50ml
       
      Last edited:

      whawha

      Full Member
      10+ Year Member
      Apr 1, 2009
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        What nze82 did is perfectly correct, and I think the answer should be 50 mL, not 5 mL as given in the solution.

        Here is how I approached this problem before I looked at the solution.

        # of moles of [H+] in HF = 0.2 * 0.06 = 0.012

        At pKa (given to be 3.3), it will lose 50% of the protons, which means it will have 0.006 moles of [H+]

        That means, a certain amount of OH- needs to be used to get reacted with 0.006 moles of [H+] of the 0.012.

        x mL of KOH = 0.006 moles / 0.12 M = 0.05 L = 50 mL

        Hope this helps.
         
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