Two solutions, A and B, each of 100L was made by dissolving 4g of NaOH and 9.8 g of H₂SO₄ in water, respectively. The pH of the resultant solution obtained from mixing 40L of solution A and 10L of solution B is ________. (Round off to the nearest integer)

Solution

Chemistry ma’am · 4 steps

Count moles, not molarity: H₂SO₄ gives two H⁺ each, and the leftover OH⁻ sets the pH.

1. Moles in each part. Solution A has 4/40 = 0.1 mol NaOH in 100 L, so 40 L of it holds 0.04 mol OH⁻. Solution B has 9.8/98 = 0.1 mol H₂SO₄ in 100 L, so 10 L holds 0.01 mol H₂SO₄.
2. Diprotic acid. Each H₂SO₄ releases two H⁺, so the 10 L portion supplies 0.02 mol H⁺. This neutralises 0.02 mol of the OH⁻ and leaves 0.04 - 0.02 = 0.02 mol OH⁻ in excess.

   [OH⁻] = (0.02 mol)/(40 L + 10 L) = 4 × 10⁻⁴ M

3. pOH to pH. pOH = 4 - log 4 = 4 - 0.602 = 3.398. Then pH = 14 - 3.398 = 10.602.
4. So. The pH of the mixture is 10.6, which rounds to 11.

The common slip: 3. Stopping at the pOH of the mixture gives 3. The question asks for pH, so subtract the pOH from 14.

The idea to keep. For acid and base mixtures, work in moles: find the H⁺ and OH⁻ supplied, cancel them, then divide the excess by the total volume.

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