Hitunglah ph larutan fe(oh)3 yg memuliki derajat ionisasi 0,3%dan kb=9×10-5!

Diketahui :
Derajat ionisasi (q) = 0,3% = 0,003
Kb = 9 × 10^-5

Ditanya : pH larutan Fe(OH)3 ?

Jawab :

Molaritas Fe(OH)3 :
= Kb / q²
= 9 × 10^-5 / (0,003)²
= 9 × 10^-5 / 9 × 10^-6
= 10 M

[OH-] :
= √ Kb . M
= √ 9 × 10^-5 . 10
= 3 × 10^-2 M

pOH :
= – log [OH-]
= – log 3 × 10^-2
= – ( log 3 + log 10^-2)
= – log 3 – log 10^-2
= 2 – log 3

pH :
= 14 – pOH
= 14 – ( 2 – log 3)
= 14 – 2 + log 3
= 12 + log 3