Question

Sebanyak 400 mL KOH 1 M direaksikan denɡan 400 mL asam lemah (Hx) 1 M. Tentukan nilai Ka asam tersebut jika pH garam yang terbentuk adalah 10!

Answer 1

400 mL KOH 1 M + 400 mL HX 1 M
pH = 10

Ka HX ... ?

n KOH = M × V
n KOH = 1 × 400
n KOH = 400 mmol

n HX = M × V
n HX = 1 × 400
n HX = 400 mmol

....... HX + KOH ➡KX + H2O
M : 400 .. 400
R : 400 .. 400
___________---
S : . --- ..... --- ..... 400 ... 400

M = n / V total
M = 400 / (400+400)
M = 400 / 800
M = 0,5

pH = 10
pOH = 14 - pH
pOH = 14 - 10
pOH = 4
- log [OH^-] = 4
[OH^-] = 10^-4

[OH^-] = √( Kw/Ka × M )
10^-4 = √( 10^-14 / Ka × 0,5 )
10^-4 = √( 5 × 10^-15 / Ka )
(10^-4)^2 = (√( 5 × 10^-15 / Ka ) )^2
10^-8 = 5 × 10^-15 / Ka
10^-8 × Ka = 5 × 10^-15
Ka = 5 × 10^-15 / 10^-8
Ka = 5 × 10^-7

Jadi, nilai Ka asam tersebut adalah 5 × 10^-7