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Finding volume needed for buffer

  1. Apr 8, 2010 #1
    Finding volume needed for buffer (URGENT!!)

    1. The problem statement, all variables and given/known data
    What volume of 0.50M NaClO must be added to 0.20M HClO in order to prepare 500mL of a NaClO/HClO buffer with pH=7.00. Ka(HClO)= 3.5e-8


    2. Relevant equations

    I'm using the H-H equation pH=pKa + log ([A]/[HA])
    NaClO being A and HClO being HA

    3. The attempt at a solution

    I tried doing it by method mentioned in another thread but I still did not get the right answer (answer is supposed to be NaClO = 61mL)

    pKa= -log(3.5e-8) = 7.46

    Using H-H found that [A]/[HA] = .35

    mols A = mols HA (.35)

    (mols A + mols HA)/.5 = .2
    (.35 mols HA + mols HA)/.5 = .2
    1.35 mols HA / .5 =.2
    mols of HA = 7.4e-2 mols

    mols A = (7.4e-2)(.35) = 2.58e-2 mols

    .5= 2.58e-2/ L
    L= 5.16e-2
    mL = 51.6

    Close but no cigar. What am I doing wrong?
    I have a midterm tomorrow and I've been trying to solve this problem for 5 hours now and I feel like crying I am so frustrated. Please help!
     
    Last edited: Apr 8, 2010
  2. jcsd
  3. Apr 8, 2010 #2
    I can't figure out where you got [A]/[HA] = .35, that's wrong.

    To find the actual concentrations of [A-] and [HA] in the buffer solution, you need to find the moles of each and divide by the volume of solution, 0.500L. Since the final solution has a volume of 500mL, the volumes of the two solutions must add to 500mL or 0.500L. I solved it by letting x be the volume of NaClO solution and (0.500 - x) be the volume of HClO solution, then dividing each by 0.500mL solution to get the concentrations of [A-] and [HA]. Then plug everything into pH=pKa + log ([A]/[HA]) and solve for x, the volume of NaClO solution.
     
  4. Apr 8, 2010 #3
    Thank you. At first I didn't understand but now I get it. I appreciate the help.
     
    Last edited: Apr 8, 2010
  5. Apr 8, 2010 #4
    The concentration of NaClO in the final solution is the moles of NaClO divided by the volume of the final solution. Volume × molarity = moles of the solute in the original NaClO solution, but we don't know that volume; let that unknown volume be x. So x(0.50M NaClO) will give you moles of NaClO as I have in the numerator below.

    The concentration of NaClO has changed in the new solution because you're adding HClO solution, but you know the final volume is 0.500L. Dividing the moles of NaClO by 0.500L gives you the concentration of NaClO, or [A]:

    [tex][A] = \frac{\overbrace{x}^{volume~of~NaClO}\overbrace{(0.50M~NaClO)}^{molarity~of~NaClO}}{0.500L~solution} \rightarrow\frac{moles~of~NaClO}{divided~by~volume} = concentration~of~salt~or~[A][/tex]

    We will do something similar for the concentration of HClO. Volume × molarity = moles, so we need to express the volume somehow. The two volumes of solution must add to 0.500L and we already let x be one of the volumes, so the volume of the other solution has to be (0.500L - x).
    [tex][HA] = \frac{\overbrace{(0.500L - x)}^{volume~of~HClO}\overbrace{(0.20M~HClO)}^{molarity~of~HClO}}{0.500L~solution}\rightarrow\frac{moles~of~HClO}{divided~by~volume} = concentration~of~acid~or~[HA][/tex]

    Now to get [A]/[HA]:
    [tex]\frac{[A]}{[HA]} = \frac{\frac{x(0.50)}{0.500}}{\frac{(x - 0.500)(0.20)}{0.500}} = \frac{0.50x}{0.20(x - 0.500)}[/tex] since the 0.500 in the denominators cancel each other.
    Now that you have [A] and [HA], plug them and pH and pKa into the equation and solve for x.

    [tex]7.00 = 7.46 + \log\left(\frac{0.50x}{0.20(x - 0.500)}\right)[/tex]
     
    Last edited: Apr 8, 2010
  6. Apr 9, 2010 #5

    Borek

    User Avatar

    Staff: Mentor

    Strong words :wink:

    After some calculations you got to the point where you have shown that

    and then you put it into Henderson-Hasselbalch equation:

    it follows that

    [tex]\frac{[A]}{[HA]} = \frac{0.50x}{0.20(x - 0.500)} = 10^{7.00-7.46} = 0.35[/tex]

    so initial part of the Leila's calculations was correct.

    --
     
  7. Apr 9, 2010 #6
    I know what I did wrong, I was using another Ka value for HClO I found online that was different; I thought at first that might have been the problem. No wonder I didn't get 0.35 :blushing:
     
  8. Apr 9, 2010 #7

    Borek

    User Avatar

    Staff: Mentor

    Happens all the time :wink:
     
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