You need to prepare an acetate buffer of pH 5.97 from a 0.724 M acetic acid solution and a 2.30 M KOH solution. If you have 575 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a bufer of pH 5.97? The pKa of acetic acid is 4.76.
pH goal = 5.97
the buffer:
pH = pKa + log(A-/HA)
5.97 = 4.76 + log(A-/HA)
initially
mmol of HA = MV = 0.724*575 = 416.3 mmol of HA
mmol of A- = 0
after adding MV = 2.30*Vbase
mmol of HA = 416.3 - 2.30*Vbase
mmol of A- = 0+ 2.30*Vbase
5.97 = 4.76 + log(A-/HA)
5.97 = 4.76 + log( ( 2.30*Vbase) /(416.3 - 2.30*Vbase))
10^(5.97 - 4.76) = ( 2.30*Vbase) /(416.3 - 2.30*Vbase)
16.218 * (416.3- 2.30*Vbase) = 2.30*Vbase
16.218 * (416.3) - 16.218 *2.3*Vbase = 2.3*Vbase
Vvase ( 16.218 *2.3+2.3) = 16.218 * (416.3)
Vbase = 16.218 * (416.3) / ( 16.218 *2.3+2.3)
Vbase= 170.487 mL required
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