1 [H⁺] = [HCl] = 10⁻² ph = 2
2. ph = 8 [H⁺] = 10⁻⁸
3.. NH4⁺ + NO3⁻ ⇄ NH3 + HNO3
acid I baza II baza I acid II
cedeaza H⁺ acceptaH⁺
4. HCl + NH4OH ⇒ NH4Cl + H2O
CH3COOH + NaOH ⇒ CH3COONa + H2O
5.CH3COOH + NH4OH ⇒ CH3-COONH4 + H2O
6. Vs = 0,5 l nHCl = 7,3/36,5 = 0,2 moli cM = 0,2/0,5 = 0,4 moli/l
nr. molecule = 0,2·6,023·10²³ = 1,2046·10²³ mol2cule
7. [HO⁻] = [NaOH] = 10⁻⁵ [H⁺] = 10⁻⁹ pH = 9 (solutie bazica ; inroseste fenolftaleina)⁹
8. HCl + H2O ⇄ H3O⁺ + Cl⁻
9. H3O⁺ ; HCN
10. NH4⁺ H3O⁺
11. pOH = 5 pH = 9
12. a) nNaOH = nAl(OH)3 = 3 moli Vs = 3/0,5 = 6 l
b)NH3