The pH of 0.1 M - N2H4 solution is (For N2H4, Kb1 = 3.6 * 10^-6, Kb2 = 6.4 * 10^-12, log 2 = 0.3, lo — Ionic Equilibrium Chemistry Question
Question
The pH of 0.1 M - N2H4 solution is (For N2H4, Kb1 = 3.6 * 10^-6, Kb2 = 6.4 * 10^-12, log 2 = 0.3, log 3 = 0.48)
Answer: C
💡 Solution & Explanation
Since Kb1 >> Kb2, [OH-] = √(Kb1*C) = √(3.6*10^-6 * 0.1) = 6.0*10^-4 M. pOH = 4 - log6 = 4 - 0.78 = 3.22. pH = 14 - 3.22 = 10.78.
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