Problem 10 - Analytical Chemistry

Question

Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of a solution obtained by mixing equal volumes of \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) and \(0.30 \mathrm{M} \mathrm{NaOH}\).

Step-by-Step Solution

Answer

The pH is 12.70 and the pOH is 1.30.

Step 1: Write the balanced chemical equation

The reaction between sulfuric acid (\(\mathrm{H}_2\mathrm{SO}_4\)) and sodium hydroxide (\(\mathrm{NaOH}\)) is a neutralization reaction. The balanced chemical equation is: \[ \mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + 2 \mathrm{H}_2\mathrm{O} \] This means one mole of \(\mathrm{H}_2\mathrm{SO}_4\) reacts with two moles of \(\mathrm{NaOH}\).

Step 2: Determine moles of reactants present initially

Assume you mix 1 L of each solution (since volumes cancel out when mixed in equal amounts). Thus, for \(0.10 \mathrm{M} \mathrm{H}_2\mathrm{SO}_4\), the moles present are: \(0.10\,\text{mol/L} \times 1\,\text{L} = 0.10\,\text{mol}\). For \(0.30 \mathrm{M} \mathrm{NaOH}\), the moles present are: \(0.30\,\text{mol/L} \times 1\,\text{L} = 0.30\,\text{mol}\).

Step 3: Calculate the limiting reactant and amount remaining

From the balanced equation, we need 2 moles of \(\mathrm{NaOH}\) for each mole of \(\mathrm{H}_2\mathrm{SO}_4\). Thus, 0.10 mol of \(\mathrm{H}_2\mathrm{SO}_4\) requires 0.20 mol of \(\mathrm{NaOH}\). Since there are 0.30 mol \(\mathrm{NaOH}\) available, \(\mathrm{H}_2\mathrm{SO}_4\) is the limiting reactant. \(0.30 \text{ mol NaOH} - 0.20 \text{ mol NaOH} = 0.10 \text{ mol NaOH}\) remain.

Step 4: Determine concentration of excess \(\mathrm{NaOH}\)

Since we have mixed equal volumes of the solutions (total volume becomes 2 liters), the concentration of excess \(\mathrm{NaOH}\) is: \[ \frac{0.10\,\text{mol}}{2\,\text{L}} = 0.05\,\mathrm{M} \]

Step 5: Calculate \(\mathrm{pOH}\) of the solution

The concentration of hydroxide ions, \(\mathrm{OH}^-\), is 0.05 M, since \(\mathrm{NaOH}\) is a strong base that fully dissociates. The \(\mathrm{pOH}\) is given by: \[ \mathrm{pOH} = -\log[\mathrm{OH}^-] = -\log(0.05) \approx 1.30 \]

Step 6: Convert \(\mathrm{pOH}\) to \(\mathrm{pH}\)

Use the relationship between \(\mathrm{pH}\) and \(\mathrm{pOH}\): \(\mathrm{pH} + \mathrm{pOH} = 14\). Thus, \(\mathrm{pH} = 14 - 1.30 = 12.70\).

Key Concepts

pH calculation

The \(\mathrm{pH}\) of a solution is a measure of its acidity or basicity. It is calculated using the concentration of hydrogen ions (\(\mathrm{H}^+\)) in the solution. The formula is: \[\mathrm{pH} = -\log[\mathrm{H}^+]\]Calculating \(\mathrm{pH}\) is crucial in chemistry as it gives insights into the chemical nature of a solution.
To find the \(\mathrm{pH}\) of a solution with a known \(\mathrm{pOH}\), one can use the relationship: \[\mathrm{pH} + \mathrm{pOH} = 14\]This equation comes in handy for reactions involving strong acids and bases since they dissociate completely, creating clear opportunities to use either \(\mathrm{pH}\) or \(\mathrm{pOH}\) in our calculations.
In our exercise, after determining the \(\mathrm{pOH}\) to be 1.30, we use the above equation to calculate the \(\mathrm{pH}\) as \(12.70\), indicating a basic solution.

Neutralization reaction

In solution chemistry, a neutralization reaction is between an acid and a base. The acid donates protons (\mathrm{H}^+ o o the base, which accepts them, forming water and a salt. Our example involves sulfuric acid (\(\mathrm{H}_2\mathrm{SO}_4\) and sodium hydroxide (\(\mathrm{NaOH}\).
The formula representing neutralization is:\[\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + 2 \mathrm{H}_2\mathrm{O}\]This reaction is an example of a complete neutralization because it shows the conversion of all reactants into products without any leftover acid or base.
Remember:

• Neutralization reactions are exothermic, releasing heat.
• They typically result in a pH closer to 7 if the acid and base have equal strength.
• These reactions are important in various practical applications like wastewater treatment and digestion processes.

Understanding neutralization helps synthesize ideas on how different chemical compounds interact to form new substances.

Chemical equilibrium

Chemical equilibrium refers to a state in a chemical reaction where the rate of conversion of reactants to products and the rate of conversion of products back to reactants are equal. In simpler words, the reaction seems to stop changing because forward and backward reactions occur at the same rate.
In a neutralization reaction between a strong acid and a strong base, the equilibrium concept underscores that these reactions generally proceed to completion, meaning they do not show typical equilibrium behavior as weaker acids or bases might.
Key points to remember about equilibrium are:

• Equilibrium doesn't mean reactants and products are equal in concentration, but that their rates are.
• It can be shifted by changing pressure, temperature, or concentration (Le Chatelier's Principle).
• Not all reactions achieve equilibrium—the completion of a reaction depends on the strength of acids and bases.

Gaining insights into chemical equilibrium is essential for understanding reaction dynamics and the conditions needed to achieve desired results.

Concentration determination

The concentration of a solution provides information about how much solute is present in a given volume of solvent. It is generally expressed in molarity (M), moles per liter.
In the original exercise, determining the concentration of remaining \(\mathrm{NaOH}\) was crucial as it allowed us to calculate \(\mathrm{pOH}\) and therefore the \(\mathrm{pH}\). After the neutralization reaction, the leftover \(\mathrm{NaOH}\) concentration was calculated using the formula: \[\text{Concentration} = \frac{\text{Moles of solute}}{\text{Total volume of solution}}\]Where 0.10 moles of \(\mathrm{NaOH}\) remained in a 2-liter mixture.
Understanding concentration is fundamental because:

• Most chemical properties depend on concentration including reaction rates and yields.
• Concentration informs how solutions should be prepared in laboratory and industrial settings.
• It is pivotal in dosing for pharmaceuticals and nutrient solutions in agriculture.

Recognizing how to determine concentration allows chemists to predict and control the outcome of chemical reactions efficiently.