Question

Calculate the pH of 0.100 L of the buffer 0.110 M CH3COONa/0.130 M CH3COOH before and after the addition of the following species. (Assume there is no change in volume.)

(a) pH of the starting buffer:

(b) pH after addition of 0.0030 mol HCl:

(c) pH after addition of 0.0040 mol NaOH (added to a fresh solution of the starting buffer):

Answer 1

no of moles of CH3COONa = molarity * volume in L
                           = 0.11*0.1 = 0.011moles
no of moles of CH3COOH   = molarity * volume in L
                           = 0.13*0.1 = 0.013 moles
   PKa = 4.75 for CH3COOH
    PH = PKa + log[CH3COONa]/[CH3COOH]
        = 4.75+ log0.011/0.013
         = 4.75-0.0725 = 4.6775
b. no of moles of CH3COONa after addition of 0.003 moles of HCl = 0.011-0.003 = 0.008mole
   no of moles of CH3COOH after addtion of 0.003 moles of HCl   = 0.013+ 0.003 = 0.016moles
PH = Pka + log[CH3COONa]/[CH3COOH]
      = 4.75+ log0.008/0.016
      = 4.75-0.3010 = 4.449
c. no of moles of CH3COONa after addition of 0.004 moles of NaOH = 0.011+0.004 = 0.015mole
   no of moles of CH3COOH after addtion of 0.004 moles of NaOH   = 0.013- 0.004 = 0.009moles
PH = Pka + log[CH3COONa]/[CH3COOH]
      = 4.75+ log0.015/0.009
      = 4.75+0.2218 = 4.9718