Larutan garam 
0,1 M mengalami hidrolisis menurut reaksi :
+ 
---> 
+ 
Jika tetapan hidrolisis, Kh =
, maka pH larutan garam adalah.....
Jika tetapan hidrolisis, Kh =
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![[H^+]= \sqrt{Kh.[NH_4^+]} \\ \\ = \sqrt{10^{-9}.10^{-1}} = \sqrt{10^{-10}}=10^{-5} \\ \\ pH=-log[H^+]=-log10^{-5}=5](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D+%5Csqrt%7BKh.%5BNH_4%5E%2B%5D%7D+%5C%5C++%5C%5C+%3D+%5Csqrt%7B10%5E%7B-9%7D.10%5E%7B-1%7D%7D+%3D+%5Csqrt%7B10%5E%7B-10%7D%7D%3D10%5E%7B-5%7D+%5C%5C++%5C%5C+pH%3D-log%5BH%5E%2B%5D%3D-log10%5E%7B-5%7D%3D5++ "[H^+]= \sqrt{Kh.[NH_4^+]} \\ \\ = \sqrt{10^{-9}.10^{-1}} = \sqrt{10^{-10}}=10^{-5} \\ \\ pH=-log[H^+]=-log10^{-5}=5")
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