Page 32 - E-Modul Larutan Penyangga Kelas XI
P. 32
- mol basa
[OH ] = K x mol asam konjugasi
b
= 2 x 10 x NH OH l
−5
4
NH Cl
4
−5 15 mmol
= 2 x 10 x 5 mmol
−5
=6 x 10
[pOH] = -log [OH]
−5
= -log 6 x 10
=5 – log 6
pH = 14 – pOH
= 14 – (5 – log 6)
= 9 log 6
32
E-Modul Kimia Berbasis Scientific Critical Thinking
