diketahui 100 mL larutan  CH3COONa 0,1 M dan Ka CH3COOH = 1 x 10-5. TENTUKAN: a. reaksi hidrolisis' b. pH larutan!

CH3COONa _ CH3COO+Na
(H) = akar dari ka dikali M
        akar dari 1x 10-5
pH= hasil tadi dimasukan ke rumus ini 
     = - log (H)

V CH3COONa = 100ml
M CH3COONa = 0,1M
MxV= 0,1 x 0,1 = 0,01 = 1x10^-2
CH3COOH = 1x10^-5
Ka = 10^-5

[H^+] = Ka.mol asam/mol garam
         = 10^-5 . 1x10^-5/1x10^-2
         = 10^-5 . 1x10^-3
         = 1x10^-8

pH = -log . H^+
     = -log . 1x10^-8
     = 8 -log 1 atw 8
reaksi hidrolisis == CH3COO + Na ----> CH3COONa