Question

Describe two ways [one starting with sodium acetate (ie, no acetic acid available) and the other starting with only acetic acid] of preparing 1 liter of 0.1 N acetate buffer at pH 4.5. Note you have plenty of HCl and NaOH. You must clearly show your calculations using the Henderson-Hasselbalch equation.

Answer 1

pH of acidic buffer

1) pH = pka + loga(base(or)salt/acid)

    4.5 = 4.76+log(x)

   x = salt/acid = 0.549

total no of equivalents of buffer = 0.1*1 = 0.1 equiv

buffer = acid+salt

from ratio, salt = 0.549 acid

   if acid = x equiv

buffer = 0.59*x+x = 0.1

x = no of equivalents of acid(CH3COOH) = 0.063 equiv

CH3COONa(aq)+HCl(aq) ----> CH3COOH(aq) + NaCl(aq)

   1equiv HCl = 1 equiv

no of equivalents of HCl must add = 0.063 equiv

   volume of lab reagent HCl = n/M = 0.063/11.6   ( lab reagent HCl = 11.6 M)

                                   = 5.43 ml
  
no of equivalents of CH3COONa = 0.1-0.063 = 0.037 equiv

as some ch3coona converts into acetic acid.we must take 0.1 equiv of salt

    amount of CH3COONa must take = 0.1*82 = 8.2 g

so that,

take 8.2 grams of CH3COONa , and add 5.43 ml of lab reagent HCl and diluted in limited water and finally add water up to 1 L.

path 2)

pH = pka + loga(base(or)salt/acid)

    4.5 = 4.76+log(x)

   x = base/acid = 0.549

total no of equivalents of buffer = 0.1*1 = 0.1 equiv

buffer = acid+base

from ratio, base = 0.549 acid

   if acid = x equiv

buffer = 0.59*x+x = 0.1

x = no of equivalents of acid(CH3COOH) present in buffer = 0.063 equiv

no of equivalents of NaOH = 0.1-0.063 = 0.037 equiv

amount of NaOH must take = 0.037*40 = 1.48 g

CH3COOH(aq)+NaOH(aq) ----> CH3COONa(aq) + H2O(aq)

   1equiv NaOH = 1 equiv CH3COONa

as some ch3cooH converts into CH3COONa.we must take 0.1 equiv of ch3cooh.

no of equivalents of cH3COOH must take initially = 0.063+0.037 = 0.1 equiv

    amount of CH3COOH must take = 0.1*60 = 6 g
  
       so that,

take 6 grams of CH3COOH , and add 1.48 g of NaOH and disolved in limited water and finally add water up to 1 L.