Dik :v ch3cooh=10ml ,ph =3 diencerkan 90ml air ,Ka =1 × 10 min 5,ditanya ph yang berbentuk?

10 mL CH3COOH pH = 3
+ 90 mL air
Ka = 1 × 10^-5

pH yang terbentuk ... ?

pH = 3
- log [H^+] = 3
[H^+] = 10^-3

[H^+] = √( Ka × Ma )
10^-3 = √( 10^-5 × Ma )
(10^-3)^2 = (√( 10^-5 × Ma ) )^2
10^-6 = 10^-5 × Ma
Ma = 10^-6/10^-5
Ma = 10^-1
Ma = 0,1

Konsentrasi awal larutan adalah 0,1 M

V1 = 10 mL
M1 = 0,1 M
V2 = 10 + 90 = 100 mL
M2 = belum diketahui

V1 × M1 = V2 × M2
10 × 0,1 = 100 × M2
.......... 1 = 100 × M2
.. 1/100 = M2
.... 0,01 = M2
...... M2 = 0,01

[H^+] = √( Ka × Ma )
[H^+] = √( 10^-5 × 0,01 )
[H^+] = √( 10^-5 × 10^-2 )
[H^+] = √( 10^-7 )
[H^+] = 10^-3,5

pH = - log [H^+]
pH = - log 10^-3,5
pH = 3,5