AI(OH3)0,1M jika kb=2,5Ã10pangkat min6

[OH-]=
[tex] \sqrt{kb \times molar} [/tex]
[tex] \sqrt{2.5 \times {10}^{ - 6} } \times 0.1[/tex]
[tex] \sqrt{25 \times {10}^{ - 7 - 1} } [/tex]
[tex]5 \times {10}^{ - 4} [/tex]
p[OH-]=
[tex] - log( {oh}^{ - } ) [/tex]
[tex] - log(5 \times {10}^{ - 4} ) [/tex]
p[OH-] = 4 - log 5
p[H+] = 14 - 4 - log 5
= 10 - log 5