In: Chemistry
You need to prepare an acetate buffer of pH 6.22 from a 0.662 M acetic acid solution and a 2.71 M KOH solution. If you have 775 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 6.22? The pKa of acetic acid is 4.76.
Answer in mL
And please explain, Thanks!
This is a buffer preparation, since acetate ion and acetic acid are present
The henderson hasselbach equation
pH = pKa + log(A-/HA)
pH goal = 6.22
pKa = 4.76
6.22 = 4.76 + log(A-/HA)
initially:
mol of HA = Macid*Vacid = 775*0.662 = 513.05 mmol
mol of A- = 0
after adding:
mmol of base = Mbase*Vbase = 2.71*Vbase
then, base reacts with HA and A- is formed
mol of HA = 513.05 - 2.71*Vbase
mol of A- = 0 + 2.71*Vbase
substitute in ratio
6.22 = 4.76 + log(A-/HA)
6.22 = 4.76 + log((2.71*Vbase) / (513.05 - 2.71*Vbase))
10^(6.22 -4.76 ) = (2.71*Vbase) / (513.05 - 2.71*Vbase))
28.8403 * (513.05 - 2.71*Vbase) = 2.71*Vbase
513.05 - 2.71*Vbase = 0.093*Vbase
(0.093+ 2.71)*Vbase = 513.05
Vbase = 513.05/(0.093+ 2.71)
Vbase = 183.0360 mL of KOH must be added