Assuming that Ba(OH)₂ is completely ionised in aqueous solution under the given conditions the concentration of H₃O⁺ ions in 0.005 M aqueous solution of Ba(OH)² at 298 K is ______________ × 10⁻¹² mol L⁻¹. (Nearest integer)

Solution

Chemistry ma’am · 3 steps

Each Ba(OH)₂ gives two OH⁻, so double the molarity first, then use K_w.

1. Count the hydroxide. Complete ionisation means Ba(OH)₂ → Ba²⁺ + 2OH⁻. One formula unit releases two OH⁻ ions, so [OH⁻] = 2 × 0.005 = 0.01 mol L⁻¹.
2. Use the water product. At 298 K, [H₃O⁺][OH⁻] = K_w = 1 × 10⁻¹⁴. Divide K_w by the hydroxide concentration to get the hydronium concentration.

   [H₃O⁺] = (1 × 10⁻¹⁴)/(1 × 10⁻²) = 1 × 10⁻¹² mol L⁻¹

3. So. [H₃O⁺] = 1 × 10⁻¹² mol L⁻¹, so the blank is 1.

The common slip: 2. Taking [OH⁻] as 0.005 M and forgetting the two OH⁻ per Ba(OH)₂ gives 2 × 10⁻¹², so 2 instead of 1.

The idea to keep. For a strong base, first find [OH⁻] using how many OH⁻ each formula unit gives. Then [H₃O⁺] = K_w/[OH⁻] at 298 K.

Answer from the NTA final answer key. Solution by PrepWiser.