What is the pH value of N/1000 KOH solution — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

# pH of N/1000 KOH Solution **Step 1: Determine the molarity** - N/1000 means normality = 1/1000 = $10^{-3}$ N - For $KOH$ (monoprotic base): Molarity = Normality = $10^{-3}$ M **Step 2: Calculate $[OH^-]$ concentration** - $KOH$ is a strong base, completely ionizes: $KOH \rightarrow K^+ + OH^-$ - $[OH^-] = 10^{-3}$ M **Step 3: Calculate pOH** $$pOH = -\log[OH^-] = -\log(10^{-3}) = 3$$ **Step 4: Calculate pH** $$pH = 14 - pOH = 14 - 3 = 11$$ **Answer: pH = 11** (Option D) The solution is basic because $KOH$ is a strong base, producing $OH^-$ ions. With $[OH^-] = 10^{-3}$ M, the pOH is 3, giving a pH of 11.