Assuming that Ba(OH) is completely ionized in aqueous solution under the given conditions, then the concentration of H O ions in 0.005 M aqueous solution of Ba(OH) at 298 K is __________ × 10 mol L .

Question

Assuming that Ba(OH) is completely ionized in aqueous solution under the given conditions, then the concentration of H O ions in 0.005 M aqueous solution of Ba(OH) at 298 K is __________ × 10 mol L . (Nearest integer) 2 3 + 2 –12 –1

Prashant Kotian · Accepted Answer

Correct answer: 1.00. # Solution: Finding [H₃O⁺] in Ba(OH)₂ Solution **Step 1: Write the ionization equation for Ba(OH)₂** Ba(OH)₂ → Ba²⁺ + 2OH⁻ Since 1 mole of Ba(OH)₂ produces 2 moles of OH⁻: [OH⁻] = 2 × 0.005 M = 0.01 M = 1 × 10⁻² M **Step 2: Apply the water ionization constant** At 298 K: Kw = [H₃O⁺][OH⁻] = 1 × 10⁻¹⁴ **Step 3: Solve for [H₃O⁺]** [H₃O⁺] = Kw / [OH⁻] [H₃O⁺] = (1 × 10⁻¹⁴) / (1 × 10⁻²) **Step 4: Calculate** [H₃O⁺] = 1 × 10⁻¹² mol/L **Step 5: Express in required format** The answer is 1.00 × 10⁻¹² mol/L Therefore, the answer is 1.00.

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