1.Tentukan pH larutan asam klorida (HCl) 0,2 M sebanyak 100 ml
~Hanya Catatan
Diketahui :
HCl = Asam Kuat
ma = 0,2 M
Ditanya :
pH .. ?
Dijawab :
[H+] = x . Ma
[H+] = 1 . 0,2
[H+] = 2 .10-¹
pH = – log [H+]
pH = – log ( 2 .10-¹ )
pH = 1 – log 2
1.Tentukan pH larutan asam klorida (HCl) 0,2 M sebanyak 100 ml
HCl = Asam Kuat
ma = 0,2 M
pH .. ?
[H+] = x . Ma
[H+] = 1 . 0,2
[H+] = 2 .10-¹
pH = – log [H+]
pH = – log ( 2 .10-¹ )
pH = 1 – log 2