What is the pH of 4 * 10^-3 M - Y(OH)2 solution assuming the first dissociation to be 100% and secon — Ionic Equilibrium Chemistry Question
Question
What is the pH of 4 * 10^-3 M - Y(OH)2 solution assuming the first dissociation to be 100% and second dissociation to be 50%, where Y represents a metal cation? (log 2 = 0.3, log 3 = 0.48)
Answer: A
💡 Solution & Explanation
[OH-]1 = 4*10^-3. [OH-]2 = 2*10^-3. Total [OH-] = 6*10^-3. pOH = 3 - log6 = 3 - 0.78 = 2.22. pH = 11.78.
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