Problem 26 - Chemical Principles: The Quest for Insight

Question

Galculate the volume of \(0.116 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) required to neutralize (a) one-half and (b) all the hydroxide ions in \(25.0 \mathrm{~mL}\) of \(0.215 \mathrm{M} \mathrm{KOH}\) (aq). (c) What is the molarity of \(\mathrm{Cl}^{-}\)ions at the stoichiometric point? (d) Calculate the \(\mathrm{pH}\) of the solution after the addition of \(40.0 \mathrm{~mL}\) of \(0.116 \mathrm{M}\) HCl(aq) to \(25.0 \mathrm{~mL}\) of \(0.215 \mathrm{M} \mathrm{KOH}(\mathrm{aq})\).

Step-by-Step Solution

Answer

The volume of 0.116 M HCl required for half neutralization is 23.171 mL and for complete neutralization is 46.336 mL. The molarity of Cl- ions at the stoichiometric point is 0.0753 M, and the pH after adding 40.0 mL of HCl to the KOH solution is approximately 12.05.

Step 1: Determine the reaction and molar ratio

Write out the balanced chemical equation for the neutralization reaction: code{HCl(aq) + KOH(aq) -> KCl(aq) + H_2O(l)}The molar ratio of HCl to KOH is 1:1.

Step 2: Calculate moles of KOH

Find the moles of KOH present using its concentration and volume: code{moles of KOH = Molarity of KOH * Volume of KOH in liters} code{moles of KOH = 0.215 M * 0.025 L = 0.005375 moles KOH}

Step 3: Calculate volume of HCl for half neutralization

Half the moles of KOH is 0.0026875 moles. Using the molar ratio and the molarity of HCl, calculate the volume needed to neutralize half the KOH: code{Volume of HCl = moles of HCl / Molarity of HCl} code{Volume of HCl = 0.0026875 moles / 0.116 M} code{Volume of HCl ≈ 0.023171 liters ≈ 23.171 mL}

Step 4: Calculate volume of HCl for complete neutralization

Use the original moles of KOH to find the volume of HCl needed for complete neutralization: code{Volume of HCl = moles of KOH / Molarity of HCl} code{Volume of HCl = 0.005375 moles / 0.116 M} code{Volume of HCl ≈ 0.046336 liters ≈ 46.336 mL}

Step 5: Compute the molarity of Cl- ions at the stoichiometric point

At the stoichiometric point, the moles of Cl- ions equal the initial moles of KOH. The total volume is the sum of the volumes of HCl and KOH solutions: code{Total volume = 25.0 mL + 46.336 mL} code{Total volume = 71.336 mL = 0.071336 L} code{Molarity of Cl- = moles of KOH / Total volume} code{Molarity of Cl- = 0.005375 moles / 0.071336 L} code{Molarity of Cl- ≈ 0.0753 M}

Step 6: Determine the pH after addition of HCl

Determine the moles of HCl and KOH present after adding 40.0 mL of HCl. Since the molar ratio is 1:1, subtract moles of KOH from moles of HCl: code{moles of HCl added = 0.116 M * 0.04 L = 0.00464 moles} code{moles of H+ in excess = moles of HCl added - moles of KOH} code{moles of H+ in excess = 0.00464 - 0.005375} code{moles of H+ in excess = -0.000735 (Since moles cannot be negative, KOH is in excess)} code{moles of OH- in excess = 0.000735 moles} Calculate the concentration of OH- in the total volume and then the pOH: code{Total volume = 40.0 mL + 25.0 mL = 65.0 mL = 0.065 L} code{[OH-] = moles of OH- in excess / Total volume} code{[OH-] = 0.000735 moles / 0.065 L ≈ 0.0113 M} code{pOH = -log10([OH-]) ≈ 1.95} Finally, calculate the pH using the relationship pH + pOH = 14: code{pH = 14 - pOH ≈ 14 - 1.95 ≈ 12.05}

Key Concepts

Molarity Calculation

Understanding the concept of molarity is crucial for solving chemistry problems involving solutions. Molarity, denoted as M, is the measure of the concentration of a solute in a solution. It is defined as the number of moles of solute divided by the volume of the solution in liters. The formula to calculate molarity is:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
To make sense of this in a real-world context, imagine dissolving a certain amount of a substance, like salt, in water. The amount of salt is measured in moles, and the water's volume is measured in liters. By dividing the moles of salt by the volume of water, you get the concentration or molarity of the saltwater solution.
In the given exercise, molarity helps you understand the amount of HCl needed to neutralize a certain volume of KOH solution. The calculation is straightforward and involves using the known molarity of the KOH solution and its volume to find the number of moles of KOH. Then, with the molar ratio of the chemical reaction, we can find the necessary volume of HCl to achieve neutralization. Remember, getting comfortable with molarity allows us to manipulate chemical solutions confidently and is a fundamental skill in the field of chemistry.

Volume of Reactants

The volume of reactants in a chemical reaction is another critical factor when performing stoichiometric calculations. For neutralization reactions, such as the one between hydrochloric acid (HCl) and potassium hydroxide (KOH), understanding the volume relationships is essential to achieve the correct stoichiometric balance.
In the scenario of our exercise, we are tasked with finding out how much of one reactant (HCl) we need to completely or partially neutralize another (KOH). This process requires the use of the molarity of the reactants and the balanced chemical equation.
For a 1:1 molar ratio, the moles of each reactant are equal. Therefore, the volume needed for one reactant to neutralize the other is found by:
\[ \text{Volume of reactant} = \frac{\text{moles of other reactant}}{\text{molarity of reactant}} \]
Having this equation in mind is vital for students to solve problems involving finding the volume of reactants needed in a chemical reaction. Completing these types of calculations reinforces the understanding of how reactants combine and the importance of having the proper proportions to achieve the desired reaction outcome.

pH Calculation

The pH scale is an essential concept in chemistry that determines the acidity or basicity of a solution. The pH is calculated by taking the negative logarithm (base 10) of the hydrogen ion concentration ([H+]) in moles per liter. The formula for pH is written as:
\[ \text{pH} = -\log_{10}([\text{H}^+]) \]
In the case of basic solutions, we first need to find the hydroxide ion concentration ([OH-]) and then calculate the pOH, which is the negative logarithm of the hydroxide ion concentration. pOH and pH are related since:
\[ \text{pH} + \text{pOH} = 14 \]
In our exercise's part (d), after adding HCl to the KOH solution, we need to figure out which ion is in excess and then calculate the concentration of that ion in the new total volume. If KOH is in excess, it means the solution is basic, and we calculate the [OH-]. Once we have the pOH, we can easily find the pH by subtracting the pOH from 14.
Understanding pH calculations is not just a fundamental part of chemistry, but it's a crucial skill that applies to various real-life situations, such as water purification, pharmaceuticals, and agriculture. A good grasp of this concept allows for precise adjustment of solution conditions that can be critical in many scientific and industrial processes.