Ph NH4OH 10L 0,2 M terionisasi 1%

[tex] { \alpha }^{2} = \frac{kb}{ma} \\ ({1.10}^{ - 2} {)}^{2} \times {2.10}^{ - 1} = kb \\ {2.10}^{ - 5} = kb \\ (o {h}^{ - } ) = \sqrt{kb.mb} \\ = \sqrt{2. {10}^{ - 5} }.1.10^-1 = 1.4108 \times {10}^{ - 3} \\ poh = 3 - log \: 1.4108 \\ ph = 14 - (3 - log \: 1.4108) = 11 + log \: 1.4108[/tex]
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