Question

How many moles of KOH would you need to add to 1.65 L of buffer (comprised of 0.35 M CH₃COOH and 0.35 M CH₃COONa) to reach a pH of 4.849? Assume no volume change upon addition of base.

Answer 1

Initial amounts:

mol of acid = MV = 1.65*0.35 = 0.5775 mol of acid

mol of conjugate base = MV =  1.65*0.35 = 0.5775 mol of base

if we need pH = 4.849

apply buffer equation

pH = pKa + log(CH3COONa/CH3COOH )

4.849 = 4.75 + log(CH3COONa/CH3COOH )

(CH3COONa/CH3COOH ) = 10^(4.849-4.75) = 1.256

(CH3COONa/CH3COOH ) =1.256

CH3COONa = 1.256*CH3COOH

therefore... we must add KOH

mol of KOH required = "x"

after adding KOH,

CH3COOH + KOH = K + CH3COO-

CH3COOH decreaes, CH3COO- increases

mol of acid = 0.5775 - x

mol of conjugate base = 1.65*0.35 = 0.5775 + x

now.. .we know the ratio

CH3COONa = 1.256*CH3COOH

(0.5775 - x)1.256 = 0.5775 + x

0.5775 *1.256 - 1.256 x = 0.5775 + x

(0.5775 *1.256 - 0.5775 ) = 2.256 x

x = (0.5775 *1.256 - 0.5775 ) / 2.256

x = 0.0655

mol of KOH required = 0.0655