larutan CH3COONH4 0,1M mempunyai pH sebesar....(Kb:NH4OH=2x10^-5 dan Ka CH3COONH=2x10^-5)

[H+] = akar Kw×Ka/kb
[H+] = akar 10^-14×2.10^-5/2.10^-5
[H+] = akar 2.10^-19/2.10^-5
[H+] = akar 1.10^-14
[H+] = 1.10^-7

pH= -log [H+]
pH= -log 1.10^-7
pH = 7-log1
pH= 7

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