# 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) + H2SO4(aq) → Na2SO4(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+ + SO42-

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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