500 ml larutan NH4Cl mempunyai Ph = 5,5 dan kb NH4Cl = 2× 10^-5. tentukan massa NH4Cl (Mr = 33,5) yang terlarut !
Jawaban Terkonfirmasi
Diketahui :
Volume larutan = 500 ml = 0,5 L
pH = 5,5
Kb NH3 = 2 × 10^-5
Mr NH4Cl = 53,5 gram/mol
Ditanya : Massa zat terlarut NH4Cl ?
Jawab :
[H+] :
– log [H+] = pH
– log [H+] = 5,5
– log [H+] = – log 10^-5,5
[H+] = 10^-5,5
[H+] = 10^-6 . 10^0,5
[H+] = 3,16 × 10^-6
Kw = Tetapan Kesetimbangan Air = 10^-14
Molaritas garam NH4Cl :
= [H+]² . Kb / Kw
= [3,16 × 10^-6]² . 2 × 10^-5 / 10^-14
= 0,01997 mol/L
n NH4Cl :
= M . V
= 0,01997 mol/L × 0,5 L
= 0,009985 mol
massa zat terlarut NH4Cl :
= n . Mr
= 0,009985 mol × 53,5 gram/mol
= 0,534 gram