How many ml of 0.8M KOH should be added to 5.02g of 1,5 pentatonic acid (H2A) Mwt= 132.11 to give a pH of 4.4 when diluted to 250ml before the first eq point pKa1= 4.345

If that IS correct then I need help b/c I can't get that answer...

I've got it but I didn't get the 25.2ml but rather 25.1ml...which is correct?

(my error was in the H-H eqzn..I divided the pH by the pKa instead of subtracting) I wasted a whole 2hrs on this Q..shakes head*

hm..I don't understand why it doesn't work out when I don't use the H-H eqzn...=(

It works using the one you like to use.
I used that and obtained 25.25 mL.

(H^+))(HP^-)/(H2P) = Ka
(HP^-)/(H2P) = Ka/(H^+)
(HP-)/(H2P) = 4.5186/3.981E-5 = 1.135
(HP-)=1.135(H2P). Since the volume is the same for both salt and acid, we can let the molarities = mols. Let x = mols HP-
x = 1.135 x (mols H2P - x)
x = 1.135 x (0.03800 - x)
x = 0.04313 - 1.135x
2.135x = 0.04323
x = 0.0202 mols KOH to form 0.0202 mols HP-
volume KOH needed = 0.0202mols/0.8 M = 0.02525 L = 25.25 mL.

Check my arithmetic and thinking.

But isn't the base supposed to be on the top of the acid for the buffer eqzn??
For the H-H eqzn it's the other way around with the base on the top...

Withe the H-H equation, that happens because when you change H to pH, it gets that negative log in and we don't like negative signs so what WAS pH = pK - log (acid/base) becomes pH = pK + log (base/acid). Either equation will work but most texts prefer to leave equations with plus signs if that is easily done. In the case of the H-H equation, it is easily done simply by inverting the log term.

Okay,
I get it now

Thanks Dr.Bob

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