Tentang hidrolisis garam Hitunglah ph CH3COONA 0,1 M (Ka =2*10-5). Tolong ya

[tex][OH-] = \sqrt{ \frac{kw}{ka} \times M garam } \\ = \sqrt{ \frac{ {10}^{ - 14} }{2. {10}^{ - 5} } \times 0.1} \\ = \sqrt{ {10}^{ - 10} \times 0.5 } \\ = \sqrt{5 \times {10}^{ - 11} } \\ = {10}^{ - 5.5} \sqrt{5} [/tex]
pOH= -log [OH-]
= -log 10^-5,5 √5
= 5,5 - log√5

pH= 14 - pOH
= 14-(5,5-log√5)
= 8,5 + log√5

#semoga membantu :)