Suatu garam (Mr = 53) ditambahkan ke dalam 200 ml larutan NH4OH 0,2 M (kb. = 10^-5) shingga pH larutan mnjdi 9. Massa garam tsb adalah…gram?

Diketahui :
Mr garam = 53,5 gram/mol
Volume larutan = 200 ml = 0,2 L
Molaritas NH4OH = 0,2 M = 0,2 mol/L
Kb NH4OH = 10^-5
pH = 9

Ditanya : massa garam ?

Jawab :

n basa NH4OH :
= M . V
= 0,2 mol/L × 0,2 L
= 0,04 mol

pOH :
= 14 – pH
= 14 – 9
= 5

[OH-] :
– log [OH-] = pOH
– log [OH-] = 5
– log [OH-] = – log 10^-5
[OH-] = 10^-5

n garam :
= Kb. n basa / [OH-]
= 10^-5 × 0,04 mol / 10^-5
= 0,04 mol

massa garam :
= n . Mr
= 0,04 mol × 53,5 gram/mol
= 2,14 gram