Question

Sebanyak 3,4 gr NH3, dilarutkan dalam 1 liter air kemudian ditambahkan 3,35 gr Kb = 10^-5 garam air (NH4Cl) tentukan pH campuran tsb CAr : H = 1 , N = 14 , O = 16 , Cl = 33,5

Answer 1

Diketahui :
m NH3 = 3,4 gram
m NH4Cl = 3,35 gram
Kb = 10^ -5
Ar H = 1 , N = 14 , Cl = 35,5

Ditanya : pH larutan ?

Jawab :

• cari mol masing - masing zat :

Mr NH3 :
= 14 + 3.1
= 17 gram/mol

n NH3 :
= m / Mr
= 3,4 gram / 17 gram/mol
= 0,2 mol

Mr NH4Cl :
= Ar N + 4.Ar H + Ar Cl
= 14 + 4.1 + 35,5
= 53,5 gram/mol

n NH4Cl :
= m / Mr
= 3,35 gram / 53,5 gram/mol
= 0,063 mol

• Reaksi :
_______NH4Cl <-----> NH4+ + Cl-
awal :0,063________________
reaksi:0,063____0,063_0,063
akhir :__-________0,063_0,063

mol Asam konjugat (NH4+) = 0,063 mol

[OH-] :
= Kb × mol basa / mol asam konjugat
= 10^ -5 × 0,2 mol / 0,063 mol
= 3,18 × 10^ -5

pOH :
= - log [OH-]
= - log 3,18 × 10^ -5
= - ( log 3,18 + log 10^ -5)
= - log 3,18 - log 10^ -5
= 5 - log 3,18

pH :
= 14 - pOH
= 14 - ( 5 - log 3,18)
= 14 - 5 + log 3,18
= 9 + log 3,18