Titration Calculation: H3O+ After KOH Addition

Hey guys! Let's dive into a titration problem where we need to figure out the concentration of hydronium ions ($H_3O^+$) after adding a certain amount of potassium hydroxide (KOH) solution. This is a classic chemistry problem that combines stoichiometry and acid-base chemistry, so buckle up and let's get started!

Understanding the Titration Process

In essence, titration is a method used to determine the concentration of a solution (the analyte) by reacting it with a solution of known concentration (the titrant). Think of it like a controlled reaction where we carefully add the titrant to the analyte until the reaction is complete. In our specific case, we are titrating a perchloric acid ($HClO_4$) solution, which is a strong acid, with a potassium hydroxide (KOH) solution, which is a strong base. The reaction between a strong acid and a strong base is a neutralization reaction, where the acid and base react to form water and a salt.

Let's break down the key concepts and steps involved in solving this problem. The core idea revolves around stoichiometry, which is the calculation of relative quantities of reactants and products in chemical reactions. We need to figure out how many moles of $HClO_4$ we initially have, how many moles of KOH we're adding, and then determine how these amounts affect the concentration of $H_3O^+$ in the final solution. Remember, strong acids like $HClO_4$ completely dissociate in water, meaning each molecule of $HClO_4$ donates one proton ($H^+$) to form a hydronium ion ($H_3O^+$). Similarly, strong bases like KOH completely dissociate in water to produce hydroxide ions ($OH^-$).

When we mix $HClO_4$ and KOH, the $H_3O^+$ ions from the acid react with the $OH^-$ ions from the base to form water ($H_2O$). This reaction is the heart of the neutralization process. If we add just the right amount of KOH, we'll completely neutralize the acid, meaning all the $H_3O^+$ ions will react with $OH^-$ ions. This point is called the equivalence point. However, in this problem, we're adding a specific amount of KOH (80.0 mL), and we need to determine if this is enough to completely neutralize the acid or if there will be excess acid or base left over. So, before we jump into calculations, let's make sure we have a solid grasp of the chemical principles at play. We're essentially dealing with a mole-to-mole reaction between $H_3O^+$ and $OH^-$, and understanding the stoichiometry is crucial for correctly determining the final $H_3O^+$ concentration.

Step-by-Step Calculation of H3O+ Concentration

Okay, guys, now let's get our hands dirty with the actual calculations! We'll break it down into manageable steps to make sure everything's crystal clear.

1. Calculate Initial Moles of $HClO_4$

First, we need to figure out how many moles of perchloric acid ($HClO_4$) we started with. We know the volume and molarity of the $HClO_4$ solution, so we can use the formula:

Moles = Molarity × Volume (in Liters)

We have 25.0 mL of 0.723 M $HClO_4$. Let's convert the volume to liters:

25. 0 mL = 25.0 mL / 1000 mL/L = 0.025 L

Now, we can calculate the moles of $HClO_4$:

Moles of $HClO_4$ = 0.723 mol/L × 0.025 L = 0.018075 mol

Since $HClO_4$ is a strong acid, it completely dissociates in water, meaning that the moles of $H_3O^+$ are equal to the moles of $HClO_4$:

Initial moles of $H_3O^+$ = 0.018075 mol

2. Calculate Moles of KOH Added

Next, we need to determine how many moles of potassium hydroxide (KOH) we added. We're given the volume and molarity of the KOH solution, so we can use the same formula as before:

Moles = Molarity × Volume (in Liters)

We added 80.0 mL of 0.27 M KOH. Let's convert the volume to liters:

80. 0 mL = 80.0 mL / 1000 mL/L = 0.080 L

Now, we can calculate the moles of KOH:

Moles of KOH = 0.27 mol/L × 0.080 L = 0.0216 mol

Since KOH is a strong base, it completely dissociates in water, meaning that the moles of $OH^-$ are equal to the moles of KOH:

Moles of $OH^-$ added = 0.0216 mol

3. Determine the Limiting Reactant and Moles of Excess Reactant

Now, this is a crucial step! We need to figure out which reactant, $H_3O^+$ or $OH^-$, is the limiting reactant. The limiting reactant is the one that gets used up first in the reaction, and it determines how much product can be formed. In our case, the reaction is:

$H_3O^+ (aq) + OH^- (aq) → 2 H_2O (l)$

This equation tells us that one mole of $H_3O^+$ reacts with one mole of $OH^-$. So, we can directly compare the moles of $H_3O^+$ and $OH^-$ to see which one is limiting.

We have 0.018075 mol of $H_3O^+$ and 0.0216 mol of $OH^-$. Since we have more moles of $OH^-$ than $H_3O^+$, the $H_3O^+$ is the limiting reactant. This means that all the $H_3O^+$ will react, and we'll have some $OH^-$ left over.

To find out how many moles of $OH^-$ are left over, we subtract the moles of $H_3O^+$ from the moles of $OH^-$:

Moles of excess $OH^-$ = Moles of $OH^-$ added - Moles of $H_3O^+$

Moles of excess $OH^-$ = 0.0216 mol - 0.018075 mol = 0.003525 mol

4. Calculate the Final Volume of the Solution

To calculate the final concentration, we need the final volume of the solution. This is simply the sum of the initial volumes of the $HClO_4$ and KOH solutions:

Final volume = Initial volume of $HClO_4$ + Volume of KOH added

Final volume = 25.0 mL + 80.0 mL = 105.0 mL

Let's convert this to liters:

Final volume = 105.0 mL / 1000 mL/L = 0.105 L

5. Calculate the Concentration of Excess $OH^-$

Since we have excess $OH^-$, the concentration of $H_3O^+$ will be determined by the autoionization of water. However, first, let's find the concentration of the excess $OH^-$.

Concentration of excess $OH^-$ = Moles of excess $OH^-$ / Final volume

Concentration of excess $OH^-$ = 0.003525 mol / 0.105 L = 0.03357 M

6. Calculate the $pOH$

Next, let's calculate the pOH of the solution:

pOH = -log[OH-] = -log(0.03357) = 1.474

7. Calculate the $pH$

Using the relationship between pH and pOH, we can calculate the pH of the solution:

pH + pOH = 14

pH = 14 - pOH = 14 - 1.474 = 12.526

8. Calculate the $H_3O^+$ Concentration

Now, we can find the $H_3O^+$ concentration using the following formula:

$[H_3O^+] = 10^{-pH}$

$[H_3O^+] = 10^{-12.526} = 2.978 × 10^{-13} M$

Final Answer and Key Takeaways

So, guys, after all the calculations, we've found that the $H_3O^+$ concentration after adding 80.0 mL of 0.27 M KOH to our 25.0 mL sample of 0.723 M $HClO_4$ is approximately $2.978 × 10^{-13}$ M. That's a pretty low concentration, indicating that the solution is quite basic due to the excess KOH.

The key takeaway here is the importance of stoichiometry in solving titration problems. We needed to carefully track the moles of each reactant to determine the limiting reactant and the amount of excess reactant. Understanding the reaction between strong acids and strong bases, and how they dissociate in water, is also crucial. Remember, strong acids completely donate their protons to form $H_3O^+$, and strong bases completely dissociate to form $OH^-$.

Also, don't forget the importance of calculating the final volume of the solution. This is necessary to accurately determine the final concentration of the ions. And finally, remember the relationship between pH, pOH, and the concentrations of $H_3O^+$ and $OH^-$. These are essential tools in your chemistry toolkit!

If you have any questions about this problem or titration calculations in general, feel free to ask. Keep practicing, and you'll become a titration master in no time!