diketahui konsentrasi larutan CH3COOH=0,03M. hitunglah ph larutan jika diketahui harga ka=3x10~4

[tex]\displaystyle pH \ Larutan \ [XI] \\ \\ H^+ = \sqrt{Ka.M} = \sqrt{3.10^{-4}.0.03} = 3.10^{-3} \\ pH = -logH^+ = -log3.10^{-3} = 3 - log 3 = 2.53 \ \xrightarrow \ Answer [/tex]

Asam Lemah
CH3COOH = 0,03 M
Ka = 3 × 10^-4

[H+] = √ ( Ka × M )
= √ ( 3 × 10^-4 × 0,03 M )
= √ 9 × 10^-6
= 3 × 10^-3

pH = - log [H+]
= - log ( 3 × 10^-3 )
= 3 - log 3