Tentukan pH dari larutan 100 ml NH4OH 0,01 M ​

Jawaban:

100 mL NH4OH 0,1 M

Kb = 1,8 × 10^-5

pH … ?

[OH^-] = √( Kb × Mb )

[OH^-] = √( 1,8 × 10^-5 × 0,1 )

[OH^-] = √( 1,8 × 10^-6 )

[OH^-] = 10^-3 × √1,8

pOH = – log [OH^-]

pOH = – log 10^-3 × √1,8

pOH = 3 – log √1,8

pH = 14 – pOH

pH = 14 – (3 – log √1,8)

pH = 14 – 3 + log √1,8

pH = 11 + log √1,8

pH = 11 + log 1,34