How to Neutralize HCl with NaOH: A Step-by-Step Calculation

Understanding the principles of acid-base neutralization involves a clear understanding of chemical stoichiometry and the reaction ratio between acids and bases. This article will guide you through the process of calculating the volume of 1 M NaOH required to neutralize a given volume of a 0.500 M HCl solution. We'll go through every step with clarity, ensuring you grasp the underlying concepts.

Understanding the Reaction

The balanced chemical equation for the neutralization reaction between HCl and NaOH is as follows:

HCl NaOH → NaCl H2O

From the equation, it is evident that HCl and NaOH react in a 1:1 molar ratio. This means that one mole of HCl reacts with one mole of NaOH to form sodium chloride and water.

Let's begin the calculation process step-by-step.

Step 1: Calculate the Moles of HCl

First, we need to determine the moles of HCl present in the given 25.0 mL of a 0.500 M HCl solution. The formula for calculating moles is:

moles Molarity × Volume (in liters)

In this case:

Molarity of HCl 0.500 M Volume of HCl 25.0 mL 0.0250 L (since 1 mL 0.001 L)

Substituting these values into the formula:

moles of HCl 0.500 M × 0.0250 L 0.0125 moles

Step 2: Determine the Moles of NaOH Required

Since the reaction is a 1:1 molar ratio, the number of moles of NaOH required to neutralize HCl is the same as the number of moles of HCl present. Therefore, we need 0.0125 moles of NaOH.

Step 3: Calculate the Volume of NaOH Required

Now, we need to calculate the volume of 1 M NaOH solution required to provide 0.0125 moles. The formula to calculate volume is:

Volume (in liters) moles / Molarity

In this case:

Molarity of NaOH 1 M Moles of NaOH 0.0125 moles

Substituting these values into the formula:

Volume of NaOH (in liters) 0.0125 moles / 1 M 0.0125 L

Convert this volume to milliliters:

Volume of NaOH (in mL) 0.0125 L × 1000 mL/L 12.5 mL

Conclusion

To neutralize 25.0 mL of a 0.500 M HCl solution, you need 12.5 mL of 1 M NaOH.

Alternative Methods and Formulas

There are other mathematical approaches to solving this problem, such as using the formula:

V1M1 V2M2

Where:

V1 is the volume of the first solution (HCl) in mL M1 is the molarity of the first solution (HCl) V2 is the volume of the second solution (NaOH) in mL M2 is the molarity of the second solution (NaOH)

Using the values from the question:

V1 25 mL M1 0.5 M M2 1.0 M

The calculation is as follows:

V2 (V1 × M1) / M2 (25 mL × 0.5 M) / 1.0 M 12.5 mL

Similarly, using the concept of moles:

moles HCl 0.0125 moles (from step 1)

moles NaOH moles HCl (from 1:1 molar ratio)

Using the formula V moles / (Molarity NaOH):

V2 0.0125 moles / 1 M 12.5 mL

This confirms the previous calculation and provides a clear understanding of the stoichiometric relationships involved in acid-base neutralization reactions.