Calculate the molar concentration of OH - ions in an 1.42 M solution of a weak base, hydrazine, N2H4.?
(Kb = 4.0 x 10-6).
What is the pH of this solution?
- BIILv 79 years agoFavorite Answer
N2H4 + H2O ⇋ [N2H5]^+ + OH^-
Kb = [N2H5^+][OH^]-/[N2H4] = 4.0×10-6
At equilibrium let [N2H5^+] = [OH^]- = x; [N2H4] = 1.42 - x ≈ 1.42M (looking at √Kb this will be fine)
(x)(x)/1.42 = 4.06×10-6
x = √(1.42×4.06×10-6) = 2.40 ×10-3M pOH = -log[OH^]- = 2.62; pH = 14.0 -pOh = 11.38
- Anonymous5 years ago
BrO- acts as a base by the following equation: BrO- + H2O <---> HBrO + OH- Kb = [HBrO][OH-] / [HBrO] Recognize that [HBrO] = [OH-] = x, and [BrO-] = 0.545-x. Assume that x is small compared to 0.545, so, Kb = 4X10^-6 = x^2 / 0.545 x = [OH-] = 1.5X10^-3 M