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Creating a buffer solution

  1. May 18, 2009 #1

    I have encountered a problem relating to buffer solutions. The question states that I am to make a Sodium Acetate buffer solution of 0.25mol l-1, at pH 6.8. The reagents used to make the buffer solution however, is Acetic acid and Sodium Hydroxide. The pKa of Acetic acid is given as 4.75. I need to calculate the amounts (in moles) of acid/salt present in the buffer at the desired pH.

    So far, all I got is the balanced reaction:
    CH3COOH + NaOH <--> H2O + CH3COONa.

    Where do I go from there?

    Edit: I am aware of the Henderson-Hasselbalch equation but I do not know how the values fit in save for the pH and pKa. The final concentration is know (sodium acetate) but there are no values for the acid or the base, also, the Henderson-Hasselbalch equation is probably more useful for finding the pH of known solution concs.

    So, I do not know how to proceed!
  2. jcsd
  3. May 18, 2009 #2


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    From the Henderson-Hasselbalch equation you should be able to figure out the ratio of [CH3COOH]/[CH3COONa] required for the buffer to be pH 6.8. Since you know that [CH3COONa] + [CH3COOH] = 0.25 mol L-1, you have two equations and two unknowns.
  4. May 18, 2009 #3
    Well using the Henderson-Hasselbalch equation,

    6.8 = 4.75 + log [CH3COOH]/[CH3COONa];

    log [CH3COOH]/[CH3COONa] =2.05

    After which, I approach a problem!
  5. May 18, 2009 #4


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    log [CH3COOH]/[CH3COONa] =2.05

    means that

    [CH3COOH]/[CH3COONa] = 102.05

    Because of the definition of a logarithm.

    But, you've got another problem, you've confused the signs on the HH equation. It should be:

    pH = pKa - log [CH3COOH]/[CH3COONa]


    pH = pKa + log [CH3COONa]/[CH3COOH]

    The way to remember this is to check whether the equation makes sense with what you know about buffer systems. If [CH3COOH] is greater than [CH3COONa] then the pH of the buffer will be below the pKa of the acid and vice versa. The original way you had set up the equation, the former statement would not have held.
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