Question

Suatu larutan mengandung 0,525 gram HOCl tiap liternya. jika harga Ka= 3x10^-8 maka pH larutan adalah.........

Answer 1

Diketahui :
asam lemah HOCl (Ka = 3 x 10⁻⁸) 
M = (m/Mr) x (1000/V) 
M = (0,525/52,5) x (1000/1000) 
M = 0,01 M = 10⁻² M 

[H⁺] = √(Ka . M) 
[H⁺] = √(9 x 10⁻⁸ . 10⁻²) 
[H⁺] = √(9 x 10⁻¹ᴼ) 
[H⁺] = 3 x 10⁻⁵ M 

pH = -log [H⁺] 
pH = -log (3 x 10⁻⁵) 
pH = 5 – log 3

Answer 2

[OH-]=√kw/ka . [G]
       =√1.10^-14/3.10^-8 . 525.10^-3
       =√1575.10^-9
       =39,6.10^-3
pOH= - log [OH-]
       = - log 39,6.10^-3
       = 3-log 39,6
pH=14-(3-log39,6)
     = 11 - log 39,6