Hitunglah ph dari 25mL larutan NH3 0,005 M, Kb=1,8x10
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![M=5.10^{-3} \\ \\ .[OH^-]= \sqrt{Kb.M} \\ \\ = \sqrt{1,8.10^{-5}.5.10^{-3}} \\ \\ = \sqrt{9.10^{-8}} \\ \\ =3.10^{-4} \\ \\ pOH=-log3.10^{-4}=4-log3 \\ \\ pH=14-pOH \\ \\ =14-(4-log3) \\ \\ =10+log3](https://tex.z-dn.net/?f=M%3D5.10%5E%7B-3%7D+%5C%5C++%5C%5C+.%5BOH%5E-%5D%3D+%5Csqrt%7BKb.M%7D+%5C%5C++%5C%5C+%3D+%5Csqrt%7B1%2C8.10%5E%7B-5%7D.5.10%5E%7B-3%7D%7D+%5C%5C++%5C%5C+%3D+%5Csqrt%7B9.10%5E%7B-8%7D%7D+%5C%5C++%5C%5C+%3D3.10%5E%7B-4%7D+%5C%5C++%5C%5C+pOH%3D-log3.10%5E%7B-4%7D%3D4-log3+%5C%5C++%5C%5C+pH%3D14-pOH+%5C%5C++%5C%5C+%3D14-%284-log3%29+%5C%5C++%5C%5C+%3D10%2Blog3+++ "M=5.10^{-3} \\ \\ .[OH^-]= \sqrt{Kb.M} \\ \\ = \sqrt{1,8.10^{-5}.5.10^{-3}} \\ \\ = \sqrt{9.10^{-8}} \\ \\ =3.10^{-4} \\ \\ pOH=-log3.10^{-4}=4-log3 \\ \\ pH=14-pOH \\ \\ =14-(4-log3) \\ \\ =10+log3")
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