You need to prepare an acetate buffer of pH 6.22 from a 0.662 M acetic acid...

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!

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Expert Solution

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

queen_honey_blossom answered 1 year ago

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