Al(OH)3 0,000025 M (kb=10^-5) . tentukan PH

[OH^-] =
[tex] \sqrt{mb.kb} [/tex]
[OH^-] =
[tex] \sqrt{25 . {10}^{ - 5}. {10}^{ - 5} } [/tex]
[OH^-] = 5 × 10^-5

pOH = -log [OH^-]
= -log 5 × 10^-5
= 5 - log5

pH = 14 - pOH
= 14 - (5 - log5)
= 9 + log5

Dik : Al(OH)3 = 0.000025 M
Kb = 10 pangkat -5
Dit : PH ?
Jawab :
Rumus = akar kb kali basa lemah
= akar 10 pangkat -5 kali 25 kali 10 pangkat -5
= akar 25 kali 10 pangkat -10
= 5 kali 10 pangkat -5
POH = -log 5kali 10pangkat -5
= 5 - log 5
PH = 14 -(5-log 5)
=9+log 5
=9+0.69
=9.69

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