Besar pH larutan HCL 1×10–9 M adalah

H+=1×10^-9
=M×H
=1×10^-9×1
=1×10^-9

PH= -log[H+]
=-log[1×10^-9]
=9-log

Menentukan konsentrasi asam sangat lemah 

HCl ==> H+ + Cl-
H2O ==> H+ + OH -


pH = - Log ( H+ total )

H+ Total = H+ H2O + H+ HCl
H+ Total = 10^-7 + 10^-9
H+ Total = 1,01 x 10^-7

pH = - Log ( 1,01 x 10^-7 )
pH = 7 - log 1,01 

Semoga dapat membantu !!!