sebanyak 100 ml larutan Hbr 0.0001 m akan memiliki PH sebesar

[HBr] = 0,0001 m / 0,1 L
[HBr] = 0,001 M

[H+] = ( 1 ) . (0,001 M )
[H+] = 0,001 M
[H+] = 1 x 10^-3
[H+] = 10^-3

pH = -log[H+]
pH = 3 - log 10
pH = 3 - 1
pH = 2