larutan natrium asetat ch3coona 0,1 M dengan ka ch3cooh =10-5, mempunyai harga ph sebesar ?

CH3COONa 0,1 M
Ka = 10^-5

pH ... ?

CH3COONa => CH3COO^- + Na^+
..... 0,1 M ................ 0,1 M ....... 0,1 M

Garam CH3COONa bersifat basa
[OH^-] = √( Kw/Ka × M anion )
[OH^-] = √( 10^-14/10^-5 × 0,1 )
[OH^-] = √( 10^-9 × 0,1 )
[OH^-] = √( 10^-10 )
[OH^-] = 10^-5

pOH = - log [OH^-]
pOH = - log 10^-5
pOH = 5

pH = 14 - pOH
pH = 14 - 5
pH = 9

Larutan Penyangga & Hidrolisis
Kimia XI

Terdapat garam terhidrolisis sebagian yakni CH₃COONa dengan nilai pH > 7 sebab didominasi basa kuat.
Valensi garam CH₃COONa adalah 1 sesuai jumlah anion lemah CH₃COO⁻.

[OH⁻] = √ [(Kw/Ka) x M garam x valensi]

[OH⁻] = √ [(10⁻¹⁴/10⁻⁵) x 10⁻¹ x 1]

[OH⁻] = √ [10⁻¹⁰]

[OH⁻] = 10⁻⁵

pOH = - log [OH⁻]

pOH = - log 10⁻⁵

pOH = 5

pH = 14 - pOH

Jadi nilai pH = 9