7. pH = -lg[H(+)] = -lg 1*10^(-9) = 9
pH>7 => caracter bazic
8. [H+] = 10^(-4) moli/L
pH = -lg[H+] = -lg 10^(-4) = 4
9. Aplicând aceeași formulă:
a. pH = 6
b. pH=4
c. pH = 7
d. pH = 1
10. pOH = -lg [OH-] = -lg 10^(-3) = 3
pH + pOH = 14 => pH = 14-3 = 11
sau
Kw = [OH-][H3O+] = 10^(-14) (moli/L)^2
=> [H3O+] = [10^(-14)]/[10^(-3)] = 10^(-11) moli/L
pH = -lg[H3O+] = -lg 10^(-11) = 11
