Problem 31 - Chemical Principles: The Quest for Insight

Question

Calculate the \(\mathrm{pH}\) at each stage in the titration for the addition of \(0.150 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) to \(25.0 \mathrm{~mL}\). of \(0.110 \mathrm{M} \mathrm{Na} \mathrm{OH}(\mathrm{aq})(\mathrm{a})\) 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

The pH is: (a) >7 because the solution is basic with NaOH in excess; (b) approximately 13.3 after adding 5.0 mL of HCl; (c) approximately 12.3 after adding another 5.0 mL of HCl; (d) 7 at the stoichiometric point; (e) approximately 2 after adding 5.0 mL of HCl past stoichiometric point; (f) approximately 1 after adding 10 mL HCl past stoichiometric point.

Step 1: Initial pH Calculation

First sum the moles of NaOH and HCl present initially. Moles of NaOH = 0.025 L * 0.110 M = 0.00275 mol. Since no HCl has been added yet, the solution is basic and the pH can be calculated using the concentration of NaOH. pH = 14 - pOH = 14 - (-log(0.110 M)).

Step 2: pH After Adding 5.0 mL of HCl

Calculate the moles of HCl added: 0.005 L * 0.150 M = 0.00075 mol. Subtract the moles of HCl from moles of NaOH to find the moles of OH- remaining. New concentration of OH- = (0.00275 mol - 0.00075 mol) / 0.030 L. Then, calculate the pOH and pH from the hydroxide ion concentration.

Step 3: pH After Adding Another 5.0 mL of HCl

Repeat the previous step with an additional 5.0 mL HCl, so a total of 10.0 mL HCl has been added. Moles of HCl added = 0.010 L * 0.150 M = 0.0015 mol. Subtract from initial moles of NaOH, find new concentration of OH-, and calculate the pOH and pH accordingly.

Step 4: pH at the Stoichiometric Point

Determine the total volume of HCl needed to reach the stoichiometric point, which is when moles HCl = moles NaOH. Calculate the volume of HCl required to neutralize all the NaOH based on their initial concentrations and volumes. At the stoichiometric point, the solution is neutral, so pH = 7.

Step 5: pH After Adding 5.0 mL of HCl Beyond the Stoichiometric Point

After adding 5 mL beyond the stoichiometric point, calculate the moles of excess HCl, which is the total moles of HCl added minus the moles of NaOH initially present. Find the new concentration of HCl in the total volume, and then use it to calculate the pH directly.

Step 6: pH After Adding 10 mL More HCl Beyond the Stoichiometric Point

Calculate the additional moles of HCl added (total of 10 mL beyond the stoichiometric point). Determine the concentration of HCl in the total volume, and calculate the pH using the concentration of HCl.

Key Concepts

Stoichiometric Point

The stoichiometric point in a titration is a moment of perfect balance - it's where the amount of acid exactly equals the amount of base. Imagine a seesaw that is absolutely level; that's like reaching the stoichiometric point during a titration. At this point, all of the base (like NaOH) has reacted with the acid (like HCl), leaving you with a neutral solution, which typically has a pH of 7. This is essential to understand because it marks the endpoint of the reaction in acid-base titrations. The stoichiometric point is exceptionally important in various chemical industries and laboratory work, where it ensures that reactions have been completed perfectly.

Formulas and calculations aside, the idea of reaching this equilibrium is all about mixing the two solutions in just the right proportions. This concept is similar to finding just the right level of sugar in your coffee, not too sweet, not too bitter—it has to be just right.

Acid-Base Neutralization

Have you ever wondered what happens when an antacid (a base) meets stomach acid? It's all a game of acid-base neutralization! This is the chemical reaction where an acid and a base, like HCl and NaOH, mingle and neutralize each other. The result of this peace-making process is usually water and a salt. Just think of it like two opposing teams shaking hands and walking off the field together—no more conflict, just neutral ground.

Understanding these neutralization reactions is vital not only for cooking up the perfect chemistry experiment but also for appreciating the balancing act that goes on in nature and even within our own bodies. In the context of titrations, the volume and concentration of one solution needed to neutralize the other are what we are aiming to find.

pOH and pH Relationship

Diving into the pool of pH and pOH, it's crucial to know they're like two sides of the same coin. The pH measures how acidic or basic a solution is, while pOH gives us the scoop on the hydroxide ion concentration. In a nutshell, pH talks about protons (H+) and pOH chats about hydroxide ions (OH-). The relationship between them is straightforward: pH + pOH = 14 (at 25°C). So, if you know one, you can quickly figure out the other.

It's like knowing the time difference between two cities; if it's noon in City A and you know City B is 3 hours ahead, it must be 3 o'clock there! Likewise, if the pH is low, indicating acidity, the pOH must be high because there's a smaller amount of OH- floating around. In our daily lives, we use products labeled with pH values, such as shampoos and skincare items, which is a practical application of this chemistry concept.

Molarity and Solution Concentration

Molarity is a way of expressing the concentration of a solution, telling us how many moles of a solute are in a liter of solution. It's like a recipe that tells you how much sugar to dissolve in your tea to hit the sweet spot. In chemistry, it's crucial to get this 'recipe' right, because the concentration of your reactants directly affects the outcome of your experiments.

Molarity (M) calculations are straightforward: you divide the number of moles of your solute by the volume of the solution in liters. This is a fundamental concept in chemistry that enables scientists to predict how different substances will react together. Imagine baking a cake—if you don't get the proportions of ingredients right, you might end up with a cake that's too dense or too fluffy. Same goes for chemical solutions—molarity ensures the 'ingredients' of your reaction are perfectly proportioned.