Question

3. Determine the pH of the following solutions:

a. A 25.00 mL aqueous solution containing 0.05000 M sodium acetate (CH3COONa), given that the Ka of acetic acid (CH3COOH) is 1.75 x 10-5. Kw = 1.00*10-14.

b. The solution in (a) after 5.50 mL of 0.1000 M HCl is added.

Answer 1

Ka for acetic acid = 1.75*10^-5
Kb of sodium acetate = 10^-14 / ( 1.75*10^-5 )
                                             = 5.71*10^-10

a)
CH3COONa + H2O <-----> CH3COOH + NaOH
0.05 M                                           0                   0     (initial)
0.05-x                                             x                   x   (final)

Kb = [CH3COOH] [NaOH] / [CH3COONa]
5.71*10^-10 = x*x / (0.05-x)
since x is very small, it can be ignored as compared 0.05
5.71*10^-10 = x*x / 0.05
x = 5.34*10^-6 M

[OH-]=[NaOH]=x = 5.34*10^-6 M
[H+] = 10^-14 / [OH-]
          =10^-14 / (5.34*10^-6)
          =1.87*10^-9 M

pH = -log [H+]
       = -log (1.87*10^-9 )
       = 8.73
Answer: 8.73

b)
moles of acetate = M*V = 0.05 M *25 mL = 1.25 mmol
moles of HCL = M*V = 0.1 M *5.5 mL = 0.55 mmol

CH3COONa + HCl <-----> CH3COOH + NaCl
1.25                     0.55                0                   0     (initial moles)
1.25-0.55           0.55-0.55      0.55             0.55   (final)
0.7                              0                0.55             0.55    (final)

after reacion:
moles of CH3COOH = 0.55 mmol
moles of CH3COONa = 0.7 mmol
total volume= 5.5 mL + 25 mL = 30.5 mL

[CH3COOH] = number of moles / volume
                         =0.55/30.5
                          = 0.018 M

[CH3COONa] = number of moles / volume
                         =0.7/30.5
                          = 0.023 M

CH3COOH and CH3COONa forms buffer

pKa = -log Ka
        = -log (1.75*10^-5)
        = 4.757
usE:
pH = pKa + log {[CH3COONa] / [CH3COOH]}
      = 4.757 + log (0.023/0.018)
      = 4.863
Answer: 4.863