You need to prepare an acetate buffer of pH 5.55 from a 0.675 M acetic acid solution and a 2.50 M KOH solution. If you have 980 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.55 ? The p K a of acetic acid is 4.76.

230 ml of 2.50M KOH

Explanation:

980ml(0.675M) HOAc + ?V(2.5M KOH) => KOAc + H₂O

=> 0.98(0.675)mol HOAc + ?V(2.5M KOH)

=> 0.6615 mol HOAc + ?V(2.5M KOH) => Bfr with pH = 5.55

For buffer solution with pH = 5.55 the [OAc⁻] : [HOAc] must be 6.46:1

That is, from Henderson-Hasselbalch Equation ...

pH = pKa + log([OAc⁻]/[HOAc]) => 5.55 = 4.74 + log([OAc⁻]/[HOAc])

=> log([OAc⁻]/[HOAc]) = 5.55 - 4.74 = 0.81

=> [OAc⁻]/[HOAc] = 10⁰°⁸¹ = 6.46/1

That is, the amount of KOH added must be such that the base concentration is 6.46x greater than the acid concentration after adding the needed volume of 2.5M KOH.

Calculation setup => For HOAc + KOH => KOAc + H₂O assume n = moles of KOH needed = moles of OAc⁻ formed.

To determine n => 6.46(0.6615 - n) = n => n = moles of KOH used and OAc⁻ formed = 0.5728 mole OAc⁻

(0.6615 - n) represents the amount of HOAc remaining after neutralization of the appropriate amount of HOAc. This acid mole value must be 6.46x less than the OAc⁻ mole value of 0.5728.

∴ M x Vol KOH = 2.5M x Vol KOH = 0.5728 mol.

=> Vol KOH needed = (0.5728/2.5)Liters = 0.23 Liters = 230 ml of 2.5M KOH.

Test Calculation:

980ml (0.675M HOAc) + 230ml (2.5M KOH)

=> 0.98(0.675) mole HOAc + 0.23(2.5) mole KOH

=> 0.6615 mole HOAc + 0.5750 mole KOH

=> (0.6615 - 0.5750) mole HOAc + 0.5750 mole OAc⁻

=> 0.0865 mole HOAc + 0.5750 mole OAc⁻  in 980 ml + 230 ml solution, or 1,210 ml solution = 1.21 Liters solution

=> (0.865mol/1.21L) HOAc + (0.5750mol/1.21L) OAc⁻

=> 0.072M HOAc + 0.4752M OAc⁻

pH = pKa + log([OAc⁻]/[HOAc]) = 4.74 = log[(0.4752)/(0.072)] = 4.74 + 0.82 = 5.55

This should be worth at least one ! Whew! :-)

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