Consider solution prepared from 0.10 mol HAc/0.10 mol NaAc and diluted to 1.00 L:

HAc  H⁺¹ +Ac^-1          K_a = 1.8x10^-5
Ac^-1  HAc  +  OH^-1     K_b = 5.6x10^-10
Solution acidic (K_a>K_b) but Ac^-1 has ability to accept H⁺¹
Called common-ion effect because Ac^-1 present in both HAc & NaAc
 

To Calculate pH of solution:
In:    0.1M            0         0.1M
       HAc  H⁺¹ + Ac^-1
Eq:  0.1-X            X         0.1+X
Ka = 1.8x10^-5  = (X)(.1+X)/(.1-X) = .1X/.1
X = [H⁺¹] = 1.8x10^-5      pH = 4.74

^{How does this compare to 0.10 M HAc?}
In:    0.1M            0           0
       HAc   H⁺¹ + Ac^-1
Eq:   0.1-X             X          X
Ka = 1.8x10^-5  = (X)(X)/(.1) = X²/.1
X = [H⁺¹] = 0.0013      pH = 2.89
 

Why does adding NaAc result in higher pH?
By increasing [Ac^-1], [H⁺¹] reduced in equilibrium expression!