Jawaban:

Diketahui :

Volume NH3 = 500 ml = 0,5 L

Molaritas NH3 = 0,1 M = 0,1 mol/L

pH = 9

Kb = 10^-5

Ar H = 1 , N = 14 , O = 16 , S = 32

Ditanya : Berapa gram (NH4)2SO4 yang harus ditambahkan ?

Jawab :

pOH :

= 14 - pH

= 14 - 9

= 5

[OH-] :

- log [OH-] = pOH

- log [OH-] = 5

- log [OH-] = 10^-5

[OH-] = 10^-5

mol NH3 :

= M . V

= 0,1 mol/L . 0,5 L

= 0,05 mol

Indeks ion garam :

(NH4)2SO4 <----> 2NH4+ + SO4²-

Indeks ion = 2

mol (NH4)2SO4 :

= Kb × mol basa / [OH-] . indeks ion

= 10^-5 × 0,05 mol / 10^-5 . 3

= 0,025 mol

Mr (NH4)2SO4 :

= 2.Ar N + 8. Ar H + Ar S + 4.Ar O

= 2.14 + 8.1 + 32 + 4.16

= 132 gram/mol

Massa (NH4)2SO4 :

= n . Mr

= 0,025 mol . 132 gram/mol

= 3,3 gram

Penjelasan:

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