jika larutan garam KX 0,01 M mempunyai pH: 9 harga Ka asam HX sebesar

pH=9
pOH=14-9=5
[OH⁻]= 10⁻⁵
[OH⁻⁻]=√(Ka*M)
10⁻⁵=√(Ka*10⁻²)
10⁻¹⁰=Ka*10⁻²
Ka=10⁻⁸

KX 0,01 M
pH = 9

Ka HX ... ?

pH garam KX = 9
pOH = 14 - pH
pOH = 14 - 9
pOH = 5
- log [OH^-] = 5
[OH^-] = 10^-5

....... [OH^-] = √( Kw/Ka × M )
......... 10^-5 = √( 10^-14/Ka × 0,01 )
......... 10^-5 = √( 10^-16/Ka )
... (10^-5)^2 = (√( 10^-16/Ka ) )^2
........ 10^-10 = 10^-16/Ka
Ka × 10^-10 = 10^-16
.............. Ka = 10^-16/10^-10
.............. Ka = 10^-6

Jadi, Ka asam HX = 10^-6.