pH and pOH of mixture of NaOH and H₂SO₄ Solutions

Because sodium hydroxide (NaOH) is a strong base and sulfuric acid (H₂SO₄) is a strong acid, they immediately react with each other and give sodium sulfate and water as products. According to the mixed amount of reactants, final solution can be acidic or basic depending on which reactant remains. pH and pOH of final solution can be calculated by considering remaining chemical after the reaction.

It is good to learn this kind of lessons with examples.

pH of mixture of 15.0ml of 0.126M NaOH and 21.0ml of 0.051M H₂SO₄

Calculation steps are given below.

1. Step 1: Find amount of mixing NaOH and H₂SO₄ solutions [equation: n = CV]
2. Step 2: Write chemically balanced equation between NaOH and H₂SO₄
3. Step 3: Decide Limiting Reagent - Finding reacting amount of each reactant
4. Step 4: Calculate the concentration of remaining reactant
5. Step 5: Calculate the pH or pOH according to the concentration of remaining reactant

Step 1: Find amount of mixing NaOH and H₂SO₄ solutions [equation: n = CV]

Amount (mol) = concentration (mol dm^-3) * volume (dm³)

Mixed amount of NaOH

Mixed amount of NaOH = 0.126 mol dm^-3 * 15 * 10^-3 dm³

Mixed amount of NaOH = 0.00189 mol

Mixed amount of H₂SO₄

Mixed amount of H₂SO₄ = 0.051 mol dm^-3 * 21 * 10^-3 dm³

Mixed amount of H₂SO₄ = 0.001071 mol

Step 2: Chemically balanced equation between NaOH and H₂SO₄

2NaOH_(aq) + H₂SO_4(aq) → Na₂SO_4(aq) + 2H₂O_(l)

According to the chemically balanced equation, two moles of NaOH is required to react with one mol of H₂SO₄.

Step 3: Decide Limiting Reagent - Finding reacting amount of each reactant (one reactant will remain and one reaction will be completely consumed)

According to the step 1 and step 2, for 0.001071 mol H₂SO₄, 0.001071 * 2 (= 0.002142 mol) mol of NaOH is required for complete reaction. But, there is only 0.00189 mol NaOH.

Therefore, all NaOH will be consumed and become the limiting reagent. Some H₂SO₄ will remain in the solution.

Because, all NaOH is consumed, we can find how much H₂SO₄ is consumed according to the stoichiometric ration of balanced equation.

• Consumed H₂SO₄ amount = 0.00189 mol * (1/2)
• Consumed H₂SO₄ amount = 0.000945 mol

Now, remaining H₂SO₄ amount can be found.

• remaining H₂SO₄ amount = 0.001071 - 0.000945
• remaining H₂SO₄ amount = 0.000126 mol

Step 4: Calculate the concentration of remaining reactant

Because two solutions are mixed, we can assume that final volume will be summation of initial volumes of two solutions.

• Final volume = 15 cm³ + 21 cm³
• Final volume = 36 cm³

Now, we can find the concentration of H₂SO₄

• concentration of H₂SO₄ = 0.000126 mol / (36 * 10^-3) dm³
• concentration of H₂SO₄ = 0.0035 mol dm^-3

Step 5: Calculate the pH or pOH according to the concentration of remaining reactant

Because H₂SO₄ remains in the solution, final solution is acidic and pH value should be below seven.

Thinking H₂SO₄ is a strong acid and completely releases both hydrogen atoms as hydronium (H₃O⁺) ions to the water, we can find H₃O⁺ ion concentration as below.

H₂SO₄ + 2H₂O → 2H₃O⁺ + SO₄^2-

After dissociation of H₂SO₄ acid, H₃O⁺ ion concentration is twice as remaining H₂SO₄ concentration after the reaction.

• concentration of H₃O⁺ = 0.0035 mol dm^-3 * 2
• concentration of H₂SO₄ = 0.007 mol dm^-3

Now, we can calculate the pH value.

• pH = -log(H₃O⁺_(aq))
• pH = -log(0.007)
• pH = 2.15

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