1. hitunglah pH larutan Al(OH)3 0,1 M apabila diketahui nilai kb= 4 × 10-5!
2. hitunglah pH larutan Al(OH)3 0,01 M yang terionisasi sebanyak 2%!
jawaban:
#. basa lemah
Penjelasan:
1.[Al(OH)3]=0,1M=1.10^-1
kb=4×10^-5
[OH-]=✓kb.basa
=√(4×10^-5).(1.10^-1)
=✓4.10^-6
=2.10^-3
pOH= -log [OH-]
= -log 2.10^-3
= 3 -log 2
= 3 – 0,3
=2,7
pH= 14 – 2,7
=11,3
1. Al(OH)3 rumus
[ OH+ ] = akar Kb × M
[ OH+ ] = akar 4 × 10 -⁵ ( 0,1 )
[ OH+] = akar 4 x 10-⁶
[ OH+] = 2 . 10^-3
pOH = – log 2. 10 ^-3
pOH = 3 – log 2
pH = 14 – 3 + log 2