HCl Volume For Neutralizing NaOH: A Chemistry Problem

Hey guys! Let's dive into a classic chemistry problem involving neutralization reactions. We've got a reaction between hydrochloric acid ($HCl$) and sodium hydroxide ($NaOH$), and we need to figure out how much $HCl$ we need to completely neutralize a given amount of $NaOH$. This is a super important concept in chemistry, so let's break it down step-by-step.

The Neutralization Equation

First, let's take a look at the balanced chemical equation:

$HCl + NaOH \rightarrow NaCl + H_2O$

This equation tells us a crucial piece of information: one mole of hydrochloric acid ($HCl$) reacts with one mole of sodium hydroxide ($NaOH$). This 1:1 molar ratio is the key to solving this problem. It's like saying for every one car you need one driver – the quantities are directly related. Now, when we talk about neutralization, we mean the reaction where an acid ($HCl$ in this case) and a base ($NaOH$) react to form salt ($NaCl$) and water ($H_2O$). The goal of neutralization is to have neither excess acid nor excess base in the final solution, making the solution neutral (pH around 7).

In this particular reaction, the hydrogen ions ($H^+$) from the $HCl$ react with the hydroxide ions ($OH^−$) from the $NaOH$ to form water ($H_2O$). The remaining ions, $Na^+$ and $Cl^−$, combine to form the salt, sodium chloride ($NaCl$), which is just your everyday table salt! This reaction is highly exothermic, meaning it releases heat. If you were to actually perform this experiment, you'd notice the solution getting warmer as the reaction proceeds. It’s like a tiny bonfire happening in your beaker, but don't worry, it's perfectly safe under controlled conditions.

Understanding the stoichiometry – the relationship between the amounts of reactants and products – is absolutely vital in chemistry. It's like following a recipe; you need the right proportions of ingredients to get the desired result. In our case, the 1:1 molar ratio between $HCl$ and $NaOH$ is our recipe. If we have more $HCl$ than $NaOH$, we'll have some leftover acid, and the solution won't be completely neutralized. Similarly, if we have more $NaOH$ than $HCl$, we'll have leftover base. The key is to get the perfect balance.

Problem Breakdown: What We Know

Okay, let's recap what the problem gives us. We know:

• Molarity of $HCl$ solution: 6.0 M (This means there are 6.0 moles of $HCl$ in every liter of solution.)
• Volume of $NaOH$ solution: 125.0 mL (We'll need to convert this to liters later.)
• Molarity of $NaOH$ solution: 2.5 M (This means there are 2.5 moles of $NaOH$ in every liter of solution.)

What we need to find is the volume of the 6.0 M $HCl$ solution required to completely neutralize the 125.0 mL of 2.5 M $NaOH$ solution. Think of it like this: we have a certain amount of $NaOH$, and we need to figure out exactly how much $HCl$ we need to react with it perfectly, leaving no excess of either.

Now, let's talk about molarity. Molarity (M) is a super useful concept in chemistry because it tells us the concentration of a solution. It's defined as the number of moles of solute (the substance being dissolved, like $HCl$ or $NaOH$) per liter of solution. So, a 6.0 M $HCl$ solution is more concentrated than a 2.5 M $NaOH$ solution. This means that a smaller volume of the 6.0 M $HCl$ solution will contain the same number of moles of $HCl$ as a larger volume of the 2.5 M $NaOH$ solution contains of $NaOH$.

The volume of $NaOH$ is given in milliliters (mL), but we'll need to convert it to liters (L) to make our calculations consistent. Remember that 1 liter is equal to 1000 milliliters. So, to convert 125.0 mL to liters, we simply divide by 1000: 125.0 mL / 1000 mL/L = 0.1250 L. This conversion is crucial because molarity is defined in terms of liters, and we need to work with the same units to get the correct answer. It’s like speaking the same language – if we mix units, our calculations will get lost in translation!

Calculating Moles of NaOH

The next step is to calculate the number of moles of $NaOH$ present in the 125.0 mL (or 0.1250 L) of 2.5 M solution. This is where the molarity formula comes in handy: Molarity (M) = moles of solute / liters of solution. We can rearrange this formula to solve for moles:

Moles of solute = Molarity (M) × Liters of solution

Plugging in the values we know:

Moles of $NaOH$ = 2.5 M × 0.1250 L = 0.3125 moles

So, we have 0.3125 moles of $NaOH$ in our solution. This is a crucial piece of information because, as we discussed earlier, the balanced equation tells us that we need the same number of moles of $HCl$ to completely neutralize it. It’s like knowing you need exactly 0.3125 slices of bread to make sandwiches – any more or less, and you won't have the perfect sandwich-to-filling ratio!

Understanding this calculation is fundamental to stoichiometry. It allows us to bridge the gap between the concentration of a solution (molarity) and the actual amount of substance present (moles). Moles are like the chemist's counting unit; they allow us to keep track of the number of atoms or molecules involved in a reaction. Without knowing the number of moles, we'd be flying blind, unable to accurately predict how much of one substance will react with another. It’s like trying to bake a cake without measuring the ingredients – you might end up with a tasty surprise, but it probably won't be what you were aiming for!

Determining the Required Volume of HCl

Now that we know we need 0.3125 moles of $HCl$ to neutralize the $NaOH$, we can calculate the volume of the 6.0 M $HCl$ solution required. We'll use the same molarity formula, but this time we'll rearrange it to solve for volume:

Liters of solution = Moles of solute / Molarity (M)

Plugging in the values:

Liters of $HCl$ solution = 0.3125 moles / 6.0 M = 0.05208 L

We now have the volume in liters, but the answer choices are in milliliters, so we need to convert back to milliliters by multiplying by 1000:

Volume of $HCl$ solution = 0.05208 L × 1000 mL/L = 52.08 mL

Rounding to the appropriate number of significant figures (based on the داده given in the problem), we get approximately 52 mL.

So, the correct answer is C. 52 mL. We've successfully calculated the volume of $HCl$ needed to neutralize the $NaOH$ solution! This calculation highlights the power of molarity and stoichiometry in solving quantitative chemistry problems. It's like using a map to navigate a city; by understanding the relationships between moles, molarity, and volume, we can accurately predict the outcomes of chemical reactions.

The ability to convert between moles, volume, and molarity is a cornerstone of chemical calculations. It allows us to work with solutions in a quantitative way, meaning we can precisely measure and control the amounts of reactants we use. This is essential in many areas of chemistry, from research and development to industrial processes. Imagine trying to manufacture a drug without knowing the exact amount of each ingredient needed – it would be a recipe for disaster! So, mastering these calculations is not just about getting the right answer on a test; it's about developing a fundamental skill that will serve you well in any chemistry-related field.

Answer

Therefore, the correct answer is C. 52 ml.