Problem 32 - Chemical Principles: The Quest for Insight

Question

Calculate the \(\mathrm{pH}\) at each stage in the titration in which \(0.116 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) is added to \(25.0 \mathrm{~mL}\) of \(0.215 \mathrm{M} \mathrm{KOH}(\mathrm{aq})(\mathrm{a})\) initially; (b) after the addition of \(5.0 \mathrm{~mL}\) of acid; (c) after the addition of a further \(5.0 \mathrm{~mL}\); (d) at the stoichiometric point; (e) after the addition of \(5.0 \mathrm{~mL}\) of acid beyond the stoichiometric point; \((f)\) after the addition of \(10 \mathrm{~mL}\) of acid beyond the stoichiometric point.

Step-by-Step Solution

Answer

Initially, the pH depends on the concentration of KOH. After adding 5.0 mL and a further 5.0 mL of HCl, the pH gradually lowers due to the neutralization. At the stoichiometric point, pH is 7. Beyond that, the pH depends on the excess HCl's concentration, becoming more acidic with each addition of HCl.

Step 1 - Calculate Initial Moles of KOH

First, find the initial moles of KOH by multiplying its concentration (\(0.215 \text{ M}\)) by the volume in liters (\(25.0 \text{ mL} = 0.0250 \text{ L}\)). Moles of KOH = concentration \times volume = \(0.215 \text{ M} \times 0.0250 \text{ L} = 0.005375 \text{ moles}\).

Step 2 - Determine the Change for Each Stage

For each stage, calculate the moles of HCl added. Since the concentration of HCl is stated as \(0.116 \text{ M}\), conversion of added volume to liters allows moles of HCl to be determined.

Step 2a - Initially

No HCl is added initially, so \( \text{pH} \) is determined only by the KOH present. KOH is a strong base which fully dissociates in water. Calculate \( \text{pOH} \) using \( \text{pOH} = -\text{log}([OH^-]) \) and \( \text{pH} = 14 - \text{pOH} \) to find the pH.

Step 2b - After Addition of 5.0 mL of HCl

Calculate moles of HCl added by multiplying its concentration by the volume in liters added (5.0 mL = 0.0050 L). Then, subtract this from the initial moles of KOH to find the moles of KOH remaining. Convert back to concentration by dividing by the new total volume and calculate the pH using the \( \text{pOH} \) then \( \text{pH} \) formulas.

Step 2c - After Further 5.0 mL of HCl

Repeat the process done in Step 2b but add this volume to the previous 5.0 mL of HCl already accounted for.

Step 2d - At the Stoichiometric Point

At the stoichiometric point, the number of moles of HCl added equals the number of moles of KOH initially present. Use the moles of HCl to get the volume required to reach the endpoint, and then determine the pH of the neutral solution (pH=7).

Step 2e - After 5.0 mL of Acid Beyond the Stoichiometric Point

Calculate the excess moles of HCl after considering the moles required to neutralize the KOH. Find the concentration of excess HCl, then calculate the pH directly since HCl is a strong acid.

Step 2f - After 10.0 mL of Acid Beyond the Stoichiometric Point

Calculate the total excess moles of HCl (additional to those required for neutralization) and use this to determine the new pH.

Key Concepts

Acid-Base Titration

Acid-base titration is a process used to determine the concentration of an acid or a base by reacting it with a substance of known concentration. The principle behind titration lies in the reaction between an acid and a base, known as neutralization, where the acid donates protons (H+) to the base. The endpoint of the titration is typically indicated by a color change of an indicator or by reaching a pH that is characteristic of the equivalence point, where moles of acid equal moles of base.

During the titration, careful measurements of the volume added allow for precise calculations of concentration. In the given exercise, the process involves titrating a potassium hydroxide (KOH) solution with hydrochloric acid (HCl), which are both strong electrolytes and completely dissociate in solution. The pH at different stages reflects how the reaction progresses and eventually reaches the stoichiometric point, where equivalent amounts of acid and base have been mixed.

Stoichiometry

Stoichiometry is the area of chemistry that pertains to the calculation of reactants and products in chemical reactions. It helps us understand the proportional relationships in a chemical equation and predict the amounts of substances consumed and produced.

In the context of this titration problem, stoichiometry is employed to calculate the initial moles of KOH, the moles of HCl added at each stage, and the amounts remaining or in excess after neutralization. Having a solid grasp of stoichiometry is crucial for accurately determining the pH changes during the various stages of titration, as it defines the molar ratio between the acid and the base in the neutralization reaction.

pH and pOH

pH and pOH are measures of the acidity and basicity of a solution. The pH scale, ranging from 0 to 14, denotes the concentration of hydrogen ions \(H^+\) in a solution, with lower values being more acidic and higher values more basic or alkaline. pOH, on the other hand, measures the concentration of hydroxide ions \(OH^-\).

In an aqueous solution, pH and pOH are inversely related and always add up to 14. This relationship is crucial for solving the titration problem provided. For instance, after calculating the concentration of \(OH^-\) ions from remaining KOH, we convert it to pOH and subsequently to pH. When dealing with strong acids like HCl, we directly determine the pH from the \(H^+\) concentration. The ability to interchangeably use pH and pOH via the equation \( pH + pOH = 14 \) is integral to solving acid-base problems.

Neutralization Reaction

A neutralization reaction occurs when an acid and a base react to form water and a salt. This type of reaction is central to the concept of titration. In a neutralization reaction, the hydronium ions from the acid combine with the hydroxide ions from the base to produce water. The stoichiometry of the reaction will determine the equivalence point or stoichiometric point, where the molar amount of acid equals that of the base.

The titration exercise involves a neutralization reaction between HCl, a strong acid, and KOH, a strong base. At the stoichiometric point (Step 2d), the solution becomes neutral with a pH of 7 as the amounts of HCl and KOH are equivalent. After this point, any additional HCl will result in an acidic solution since there is an excess of hydronium ions \(H^+\). Understanding neutralization reactions is key to accurately calculating pH during titration and recognizing the changes from basic to neutral and then to acidic as the titration proceeds.