Because sodium hydroxide (NaOH) is a strong base and
sulfuric acid (H_{2}SO_{4}) 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.

Calculation steps are given below.

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

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

- Mixed amount of NaOH = 0.126 mol dm

- Mixed amount of NaOH = 0.00189 mol

- Mixed amount of H

- Mixed amount of H

According to the chemically balanced equation, two moles of NaOH is required to react with one mol of H_{2}SO_{4}.

According to the step 1 and step 2, for 0.001071 mol H_{2}SO_{4}, 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 H_{2}SO_{4} will remain in the solution.

Because, all NaOH is consumed, we can find how much H_{2}SO_{4} is consumed according to the stoichiometric ration of balanced equation.

- Consumed H
_{2}SO_{4}amount = 0.00189 mol * (1/2) - Consumed H
_{2}SO_{4}amount = 0.000945 mol

Now, remaining H_{2}SO_{4} amount can be found.

- remaining H
_{2}SO_{4}amount = 0.001071 - 0.000945 - remaining H
_{2}SO_{4}amount = 0.000126 mol

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

- Final volume = 15 cm
^{3}+ 21 cm^{3} - Final volume = 36 cm
^{3}

Now, we can find the concentration of H_{2}SO_{4}

- concentration of H
_{2}SO_{4}= 0.000126 mol / (36 * 10^{-3}) dm^{3} - concentration of H
_{2}SO_{4}= 0.0035 mol dm^{-3}

Because H_{2}SO_{4} remains in the solution, final solution is acidic and pH value should be below seven.

Thinking H_{2}SO_{4} is a strong acid and completely releases both hydrogen atoms as hydronium (H_{3}O^{+})
ions to the water, we can find H_{3}O^{+} ion concentration as below.

After dissociation of H_{2}SO_{4} acid, H_{3}O^{+} ion concentration is twice as remaining H_{2}SO_{4}
concentration after the reaction.

- concentration of H
_{3}O^{+}= 0.0035 mol dm^{-3}* 2 - concentration of H
_{2}SO_{4}= 0.007 mol dm^{-3}

- pH = -log(H
_{3}O^{+}_{(aq)}) - pH = -log(0.007)
- pH = 2.15

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