You need to prepare an acetate buffer of pH 5.50 from a 0.872 M acetic acid solution and a 2.41 M KOH solution. If you have 580 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.50? The p? a of acetic acid is 4.76.

Need to add 178ml of 2.41M KOH into the 580ml of 0.872M HOAc = > Buffer Solution with pH = 5.50.

Explanation:

1. Determine the Base to Acid Ratio using the Henderson-Hasselbalch Equation

Given pH (Bfr) = 5.50; pKa (HOAc) = - logKa = - log (1.8x10ˉ⁵) = 4.76

pH (Bfr) = pKa (HOAc) + log ([OAcˉ]/[HOAc]) = > 5.5 = 4.76 + log ([OAcˉ]/[HOAc])

=> log ([OAcˉ]/[HOAc]) = 5.50 - 4.76 = 0.74 = > [OAcˉ]/[HOAc] = 10^0.74 = 5.495*

*For an HOAc/OAcˉ Buffer to have a pH = 5.50 the [OAcˉ] concentration must be 5.495 times greater than the [HOAc] concentration.

2. Determine moles of KOH need be added to 580ml of 0.872M HOAc such that the moles of OAc⁻ is 5.495 times greater than moles of HOAc. The Acid/Base Rxn is HOAc + KOH = > KOAc + H₂O.ˉ

That is, given 580ml (0.872M HOAc) + V (L) ∙2.41M KOH = > 5.495 x moles HOAc

=> 0.58 (0.872) mole HOAc + X moles KOH = > 5.495 x 0.58 (0.872) mole KOAc (=OAcˉ)

=> 0.506 mole HOAc + X moles KOH * = > 2.7805 mole KOAc ( = 2.7805 mole OAcˉ)

*note = > KOH is the limiting reactant and will be consumed when added into HOAc solution leaving HOAc with OAcˉ that constitutes the buffer solution. The moles of KOH added is equal to the moles of OAcˉ produced. [HOAc] will decrease and [OAcˉ] will increase until the OAcˉ concentration is 5.495 times grater than the HOAc concentration.

3. After adding X moles of KOH, the following solution results ...

=> (0.506 mole - X mole) HOAc + X mole OAcˉ

=> Since the moles OAcˉ is 5.495 x moles HOAc, the following linear expression in one unknown is generated ...

=> moles OAcˉ produced/moles HOAc remaining = X / 0.506 - X = 5.495 where X = OAcˉ produced from rxn and 0.506 - X is the HOAc remaining. The moles of OAcˉ must be 5.495 times greater than moles of HOAc.

Solving for X = > X = 5.495 (0.506 - X) = 2.7805 - 5.495X = > X = 2.7805/6.495 = 0.428 mole OAcˉ produced. The 0.428 mole OAcˉ is 5.495 time greater than (0.506 - 0.428) mole HOAc remaining.

4. This means that 0.428 mole KOH is needed for reaction with 580ml of 0.872M HOAc to give 0.428 mole OAcˉ with (0.506 - 0.482) mole HOAc remaining after mixing solution.

The volume of 2.41M KOH needed to deliver 0.428 mole KOH is V (KOH) ∙2.41M = 0.428 mole = > V (KOH) = (0.428/2.41) Liters = 0.178 Liter = 178ml of 2.41M KOH.

5. Verification of Results ...

580ml (0.872M HOAc) + 178ml (2.41M KOH)

=> 0.58 (0.872) mole HOAc + 0.178 (2.41) mole KOH

=> 0.506 mole HOAc + 0.428 mole KOH

=> (0.506 - 0.428) mole HOAc + 0.428mole OAc⁻

=> 0.078 mole HOAc + 0.428mole OAcˉ

=> (0.078mol HOAc) / (0.58 + 0.178) Liter Soln + (0.428 mole OAcˉ) / (0.58 + 0.178) Liter Soln

=> (0.078mole/0.758L) HOAc + (0.428mole/0.872L) OAcˉ

=> 0.1029M HOAc + 0.5646M OAcˉ (Buffer Solution with pH = 5.50)

Checking pH of this buffer solution ...

HOAc ⇄ H⁺ + OAcˉ

C (eq) 0.1029M [H⁺] 0.5646M

Ka = [H⁺][OAcˉ]/[HOAc] = > [H⁺] = Ka[HOAc]/[OAcˉ]

= [ (1.8 x 10ˉ⁵) (0.1029) / (0.5646) ]M = 3.28 x 10ˉ⁶M

pH = - log[H⁺] - log (3.28 x 10ˉ⁶) = 5.50