larutan HNO3 0,01 molar sebanyak 100 mL mempunyai pH

Dik: mol = 0,01
= 1×10^-2
v =100 mL
=0,1 L atau 1×10^-1
Dit : pH

• M = mol / volume

[tex] \frac{1 \times 10 {}^{ - 2} }{1 \times 10 {}^{ - 1} } [/tex]
= 1×10^-1


HNO3 ---> H⁺ + NO₃⁻
valensi asamnya = 1

•[H+] = M × valensi asam
= (1×10^-1) (1)
= 1×10^-1

•pH = -log [H+]
= -log ( 1×10^-1)
= 1 - log 1
= 1