Solve
NaAc---->(Na+)+(Ac-)
1mol/L 1mol/L 1mol/L
HAc<==>(H+)+(Ac-)
C(before reaction) 1 0 1
C(when balance) 1-x x 1+x
K[HAc]= C(H+).C(Ac-)/C(HAc)
Look up the table,find K[HAc]=1.8*0.00001
0.000018=x(1+x)/(1-x)
x is tiny, so we can regard 1+x & 1-x as 1
therefore C(H+)=x=0.000018mol/L
pH=-lg[C(H+)]=-lg0.000018=4.74
|