Answer:
Ba(OH)₂ > NaOH > HCl = HOCl > N₂H₂
Explanation:
The pH = -log[H⁺] and pOH = -log[OH⁻], and by the water equilibrium, pH + pOH = 14. The value of pH determines the acid-base solutions: for pH< 7 the solution is an acid, for pH > 7 the solution is a base, and for pH = 7 the solution is neutral.
To determine the [H⁺] and the [OH⁻], let's do the dissolution reactions.
a. N₂H₂ → 2N⁻ + 2H⁺
For the stoichiometry (1:2:2), [H⁺] = 2*0.10 = 0.20 M
pH = -log(0.20) = 0.70
b. Ba(OH)₂ → Ba⁺ + 2OH⁻
For the stoichiometry (1:1:2), [OH⁻] = 2*0.10 = 0.20 M
pOH = -log(0.20) = 0.70
pH = 14 - 0.70 = 13.30
c. HOCl → H⁺ + OCl⁻
For the stoichiometry (1:1:1), [H⁺] = 0.10 M
pH = -log(0.10) = 1.00
d. NaOH → Na⁺ + OH⁻
For the stoichiometry (1:1:1), [OH⁻] = 0.10 M
pOH = -log(0.10) = 1
pH = 14 - 1 = 13.00
e. HCl → H⁺ + Cl⁻
For the stoichiometry (1:1:1), [H⁺] = 0.10 M
pH = -log(0.10) = 1.00
Ba(OH)₂ > NaOH > HCl = HOCl > N₂H₂
