Question

You need to prepare an acetate buffer of pH 5.64 from a 0.747 M acetic acid solution and a 2.93 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 5.64? The pKa of acetic acid is 4.76. Already tried 1498.88 and 1499 and they are wrong. Thank you!

Answer 1

174.6ml

Explanation

Henderson-Hasselbalch equation is

pH = pKa + log([A^-]/[HA])

where,

[A^-] = concentration of conjucate base i.e concentration of CH3COO^-

[HA] = concentration of weak acid i.e concentration of weak acid CH3COOH

pKa of acetic acid = 4.76

required pH = 5.64

5.64 = 4.76 + log([CH3COO^-]/[CH3COOH])

log([CH3COO^-]/[CH3COOH]) = 0.88

[CH3COO^-]/[CH3COOH] = 7.586

[CH3COO^-] = 7.586[CH3COOH]

moles of CH3COO^- = 7.586×moles of CH3COOH

Total no of moles = (0.747mol/1000ml)×775ml = 0.57893 mol

moles of CH3COOH + 7.586moles of CH3COOH = 0.57893mol

8.586molea of CH3COOH = 0.57893mol

moles of CH3COOH = 0.06743 mol

moles of CH3COO^- = 0.57893mol - 0.06743mol = 0.5115

0.5115 moles of KOH react with 0.5115moles of CH3COOH to produce 0.5115 moles of CH3COO^-

Volume of 2.93 M KOH solution containing 0.5115 moles of KOH =(1000ml /2.93mol)× 0.5115mol = 174.6ml