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

💡 Solution & Explanation

# Solution: pH of N/1000 KOH Solution **Step 1: Determine the molarity of KOH** $$\text{Normality} = \text{Molarity} \times \text{number of OH}^- \text{ ions}$$ For $KOH$: N = 1000, and it produces 1 $OH^-$ per molecule $$\text{Molarity} = \frac{N}{1} = \frac{1}{1000} = 10^{-3} \text{ M}$$ **Step 2: Calculate $[OH^-]$ concentration** $$[OH^-] = 10^{-3} \text{ M}$$ **Step 3: Find pOH** $$pOH = -\log[OH^-] = -\log(10^{-3}) = 3$$ **Step 4: Calculate pH using the water ionization relationship** $$pH + pOH = 14 \text{ (at 25°C)}$$ $$pH = 14 - 3 = 11$$ **Answer: pH = 11** (Option D) This makes sense because KOH is a strong base, and a concentration of $10^{-3}$ M produces a basic solution with pH well above 7.