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bufi bufi
wrote...
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11 years ago
I have to calculate how much 0.05 M HCl I need to add too my puffer (pka=7.21).
nB=0.0503
pH= the start pH (I know that one from my experiment)
pH=pka+log(nB-x/nA-x)  <=> 7.21+log(0.0503-x/0.05-x)=6.8,x  = x= 0.0504

Is this correct?
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wrote...
11 years ago
I don't think it is, but I'm not certain what you're trying to do. You appear to be preparing a buffer at pH=6.8 by adding HCl to the conjugate base of a weak acid. It's not clear though whether nB is a number of moles, or concentration, or something else.

If nB is the number of moles of base, then by adding a volume x of acid you add (0.05 M)x moles of acid. Since base and acid are dissolved in the same total volume V,
[base]/[acid] = { (0.0503 mol)/V } / { (0.05x mol) / V } = (0.0503) / (0.05x)

and x = 2.6 L (!).

If nB is the concentration of base and not the moles of base, then the amount of base is
nB * Vb where Vb is the volume of base you start with, and now
[base]/[acid] = (nB * Vb) / 0.05x

Now you can solve for (x / Vb), the volume of acid over the volume of base:
x / Vb = 2.6 mL/mL

which is much more reasonable.
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