7.46 The ionization constant of acetic acid is 1.74 × 10–5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

Method 1

 

1.    CH3COOHCH3COO-+H+  Ka=1.74×10-52.    H2O+H2OH3O++OH-       Kw=1.0×10-14Since Ka>>Kw :        CH3COOH+H2OCH3COO-+H3O+Ci=0.05                                 0                   0        0.05α-0.5α                 0.05α           0.05αKa=.05α0.5α0.5-0.05α     =.05α0.05α0.51-α     =0.5α21-α1.74×105=0.05α21α1.74×1051.74×105α=0.05α20.05α2+1.74×105α1.74×105D=b24ac=(1.74×105)24(.05)(1.74×105)=3.02×1025+.348×105α=Kacα=1.74×105.05=34.8×105×1010=3.48×106=CH3COOHCH3COO-+H+

 

α1.86×10-3CH3COO-=0.05×1.86×10-3                      =0.93×10-31000                       =.000093Method 2Degree of dissociation,α=Kacc=0.05 MKa=1.74×10-5Then,  α=1.74×105.05α=34.8×105α=3.48×104α=1.86102CH3COOHCH3COO-+H+Thus, concentration of CH3COO-=c=.05×1.86×10-2=.093×10-2=.00093 Since OAc-=H+,H+=.00093=.093×10-2.pH=-logH+      = -log.093×-10-2pH=3.03

Hence, the concentration of acetate ion in the solution is 0.00093 M and is PH is 3.03.