plecam de la formula: pH= - log[H3O+]
a) 11,013= -log[H3O+] => [H3O+] = [tex] 10^{-11,013} [/tex]
b) 4,902= - log[H3O+] => [H3O+]= [tex] 10^{-4,902} [/tex]
c) 7,002= - log[H3O+] => [H3O+]= [tex] 10^{-7,002} [/tex]
d) 6,127= - log [H3O+] => [H3O+]= 10^{-6,127} [/tex]
e) 13,19= -log[H3O+] => [H3O+]= [tex] 10^{-13,19} [/tex]
f) 12,77= -log[H3O+] => [H3O+]= [tex] 10^{-12,77} [/tex]
sunt niste numere cam ciudate,dar asta e algoritmul corect