Question

3. A buffer is made by adding 0.3 moles of CH₃COOH and 0.3 moles of NaCH₃COO to 1 liter of water.

A. Calculate the pH of the solution

  

B. Calculate the change in the pH of the solution when:

5 ml of 1 M NaOH are added.

5 ml of 1 M HCl are added.

Answer 1

Ka of CH3COOH = 1.8*10^-5 M
pKa= -log Ka
= -log (1.8*10^-5)
= 4.745

CH3COOH   ----> CH3COO-   + H+
0.3                                 0.3                 0 (initial)
0.3-x                          0.3+ x                x(at equilibrium)
Ka =(0.3+x)*x / (0.3-x)
1.8*10^-5 =(0.3+x)*x / (0.3-x)
since Ka is very small, x will be very small:
1.8*10^-5 =(0.3)*x / 0.3
x = 1.8*10^-5 M
[H+] = 1.8*10^-5 M
pH =-log [H+]
    =-log ( 1.8*10^-5)
   = 4.745
Answer: 4.745

B)
When 5 ml of 1 M NaOH are added.    :
number of moles of NaOH added = 1 M * 0.005 L = 0.005 mol
CH3COOH reacts with this NaOh to form CH4COONa
So,
[CH3COONa] = 1.005/ (1+0.005) = 1 M
[CH3COOH] = (1-0.005)/ (1+0.005) = 0.99 M

Use:
pH = PKa + log [CH3COO-] /[CH3COOH]
pH = 4.745 + log (1/0.99)
     = 4.749

when 5 ml of 1 M HCl are added.    :
number of moles of HCl added = 1 M * 0.005 L = 0.005 mol
CH3COONa + HCl will form 0.005 mol of CH3COOH
So,
[CH3COOH] = 1.005/ (1+0.005) = 1 M
[CH3COONa] = (1-0.005)/ (1+0.005) = 0.99 M

Use:
pH = PKa + log [CH3COO-] /[CH3COOH]
pH = 4.745 + log (0.99/1)
     = 4.741