Question

Consider a solution formed by mixing 50.0 $\mathrm{mL}$ of 0.100 $\mathrm{M}$ ${\mathrm{H}}_2{\mathrm{SO}}_4,30.0\mathrm{mL}$ of 0.100 $\mathrm{M}$ HOCl, 25.0 $\mathrm{mL}$ of 0.200 $\mathrm{M}$ $\mathrm{NaOH},25.0\mathrm{mL}$ of $0.100\mathrm{M}\mathrm{Ba}(\mathrm{OH}{)}_2,$ and 10.0 $\mathrm{mL}$ of 0.150 $M\mathrm{KOH}$ . Calculate the $\mathrm{pH}$ of this solution.

Short answer

The pH of the solution is 11.85.

Step-by-step solution

Calculation of moles of each reactant

First, let's convert the given volumes and concentrations to moles for each acid and base: - Moles of H2SO4: $(50.0mL)(0.100M)=5.00mmol$ - Moles of HOCL: $(30.0mL)(0.100M)=3.00mmol$ - Moles of NaOH: $(25.0mL)(0.200M)=5.00mmol$ - Moles of Ba(OH)2: $(25.0mL)(0.100M)=2.50mmol$ - Moles of KOH: $(10.0mL)(0.150M)=1.50mmol$

Water as a solvent and total volume

Water is a solvent in which all acids and bases are mixed and form the solution. The final volume of the solution can be calculated by adding the volumes of individual components: - Total Volume: $50.0+30.0+25.0+25.0+10.0=140.0mL$

Neutralization reactions

We will carry out the reactions between different acids and bases. There are three acids, $\mathrm{H}{\phantom{A}}_2\mathrm{SO}{\phantom{A}}_4$, $\mathrm{HOCl}$, and $\mathrm{H}{\phantom{A}}_2\mathrm{O}$ (since it's an aqueous solution), and three bases, $\mathrm{NaOH}$, $\mathrm{Ba}(\mathrm{OH}){\phantom{A}}_2$, and $\mathrm{KOH}$. The neutralization reactions are as follows: 1. Reaction between $\mathrm{H}{\phantom{A}}_2\mathrm{SO}{\phantom{A}}_4$ and strong bases $\mathrm{NaOH}$, $\mathrm{Ba}(\mathrm{OH}){\phantom{A}}_2$, and $\mathrm{KOH}$: Since $\mathrm{H}{\phantom{A}}_2\mathrm{SO}{\phantom{A}}_4$ is a strong diprotic acid, it will react with all three strong bases until completely neutralized. - $\mathrm{H}{\phantom{A}}_2\mathrm{SO}{\phantom{A}}_4$ reacts with $\mathrm{NaOH}$: $5.00mmol\mathrm{H}{\phantom{A}}_2\mathrm{SO}{\phantom{A}}_4+5.00mmol\mathrm{NaOH}\to 0mmol\mathrm{H}{\phantom{A}}_2\mathrm{SO}{\phantom{A}}_4$ and $0mmol\mathrm{NaOH}$ - $\mathrm{H}{\phantom{A}}_2\mathrm{SO}{\phantom{A}}_4$ is completely neutralized; there is no $\mathrm{H}{\phantom{A}}_2\mathrm{SO}{\phantom{A}}_4$ left. 2. Reaction between $\mathrm{HOCL}$ and strong bases $\mathrm{Ba}(\mathrm{OH}){\phantom{A}}_2$ and $\mathrm{KOH}$: Since $\mathrm{HOCL}$ is a weak monoprotic acid, it will react with the remaining strong bases. - $\mathrm{HOCL}$ reacts with $\mathrm{Ba}(\mathrm{OH}){\phantom{A}}_2$: $3.00mmol\mathrm{HOCL}+2.50mmol\mathrm{Ba}(\mathrm{OH}){\phantom{A}}_2\to 0.50mmol\mathrm{HOCL}$ and $0mmol\mathrm{Ba}(\mathrm{OH}){\phantom{A}}_2$ - $\mathrm{HOCL}$ reacts with $\mathrm{KOH}$: $0.50mmol\mathrm{HOCL}+1.50mmol\mathrm{KOH}\to 0mmol\mathrm{HOCL}$ and $1.00mmol\mathrm{KOH}$ - $\mathrm{HOCL}$ is completely neutralized; there is no $\mathrm{HOCL}$ left. 3. Reaction between $\mathrm{H}{\phantom{A}}_2\mathrm{O}$ and strong base $\mathrm{KOH}$: - No reaction will happen, as water will not react with strong bases under these conditions.

Calculation of final concentrations

After the reactions, we have the following concentrations of the remaining species: - $\mathrm{OH}{\phantom{A}}^{-}$ concentration from $\mathrm{KOH}$: ${\displaystyle \frac{1.00mmol}{140.0mL}}=0.00714M$

Calculation of pH

Now we can calculate the pH of the solution. First, let's calculate the pOH: - pOH = $-{\log }_{10}([O{H}^{-}])=-{\log }_{10}(0.00714)=2.15$ Since pH and pOH are related by the equation: $pH+pOH=14$, we can find the pH of the solution: - pH = $14-2.15=11.85$ The pH of the solution is 11.85.

Key concepts

Neutralization Reactions

In chemistry, **neutralization reactions** are processes where an acid and a base react quantitatively to produce a salt and water. These reactions are crucial in many industries and biological systems. During neutralization, the acidic hydrogen ions ($\mathrm{H}{\phantom{A}}^{+}$) and the basic hydroxide ions ($\mathrm{OH}{\phantom{A}}^{-}$) combine to form water ($\mathrm{H}{\phantom{A}}_2\mathrm{O}$).

• Strong acids and strong bases: Fully dissociate in water, leading to complete neutralization. For example, $\mathrm{H}{\phantom{A}}_2\mathrm{SO}{\phantom{A}}_4$ (sulfuric acid) neutralizes with $\mathrm{NaOH}$ or $\mathrm{Ba}(\mathrm{OH}){\phantom{A}}_2$.
• Weak acids and strong bases: The acid does not completely ionize in solution. With $\mathrm{HOCl}$ being a weak acid, it only partially neutralizes until a strong base like $\mathrm{KOH}$ completes the reaction.

In our solved exercise, each acid and base pair underwent this neutralization process, with step-by-step mole calculations ensuring complete or partial neutralization.

Acid-Base Chemistry

**Acid-base chemistry** revolves around the concepts of acids, bases, and their conjugates. Understanding this chemistry is crucial for calculating pH, a measure of the acidity or basicity of an aqueous solution.

• Acid: A substance that donates protons ($\mathrm{H}{\phantom{A}}^{+}$). $\mathrm{H}{\phantom{A}}_2\mathrm{SO}{\phantom{A}}_4$ is a strong acid that donates protons easily, making it highly reactive in neutralization reactions.
• Base: Accepts protons. Compounds like $\mathrm{NaOH}$ release hydroxide ions ($\mathrm{OH}{\phantom{A}}^{-}$) in solution. Strong bases dissociate completely in water, facilitating neutralization reactions.
• pH Scale: Ranges from 0 to 14, describing how acidic or basic a solution is:
  • pH < 7: Acidic
  • pH = 7: Neutral
  • pH > 7: Basic

In the example exercise, we calculated the pH after completing all neutralization reactions based on the stoichiometry and the properties of the acids and bases involved.

Mole Calculations

The concept of **mole calculations** is fundamental in chemistry, especially when dealing with reactions such as neutralization. This process involves calculating the moles of each reactant to predict how much product is formed.

• Moles: A mole is a standard unit in chemistry that quantifies the amount of substance. It's equivalent to Avogadro's number, $6.022\times {10}^{23}$ atoms or molecules.
• Initial Calculations: For the given exercise, we used the formula: \[molarity (M) = \frac{moles}{volume (L)}\] We calculated moles by multiplying the molarity by the volume in litres of each solution (converted from milliliters). For example, $50\text{ mL of }0.100\text{ M }\mathrm{H}{\phantom{A}}_2\mathrm{SO}{\phantom{A}}_4\Rightarrow 5000\text{ mmol}$.

These calculations helped us determine the amount of each reactant available for neutralization, crucial for determining the final pH.

Aqueous Solution Chemistry

**Aqueous solution chemistry** involves studying substances dissolved in water. Water's role as a solvent is vital in neutralization reactions and pH calculations.

• Solvent properties of water: Water is known as the "universal solvent" due to its ability to dissolve a wide range of substances, particularly ionic compounds and polar molecules.
• Dissociation: When acids or bases dissolve in water, they dissociate into ions. For example, $\mathrm{NaOH}$ dissociates into $\mathrm{Na}{\phantom{A}}^{+}$ and $\mathrm{OH}{\phantom{A}}^{-}$ ions, contributing to the basicity of the solution.
• Volume and Concentration Calculations: Given total volume (140 mL in the example), the concentration of ions post-reaction was important for pH calculations. Remaining $\mathrm{OH}{\phantom{A}}^{-}$ ions led to a basic pH, as observed in the final solution of the given problem.

Understanding these principles is crucial for accurate pH calculations because the hydronium and hydroxide ion concentrations define the solution's acid-base characteristics.