# How many milliliters mL of 0.400 M NaOH are required to completely neutralize 20.0 mL of 0.200 M HCl?

In chemistry, a neutralization reaction occurs when an acid and a base react to produce a salt and water. The neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) can be represented by the following chemical equation:

HCl + NaOH → NaCl + H2O

In this reaction, the H+ ions from the HCl react with the OH- ions from the NaOH to form water. At the point where the amounts of H+ and OH- are stoichiometrically equivalent, the solution is neutral. This point is known as the equivalence point.

To determine the amount of NaOH required to completely neutralize a given amount of HCl, the molarity (M) and volume (mL) of both solutions must be known. The molarity of a solution is the number of moles of solute per liter of solution. For this problem:

• Molarity of HCl solution = 0.200 M
• Volume of HCl solution = 20.0 mL
• Molarity of NaOH solution = 0.400 M
• Volume of NaOH solution = x mL (unknown)

Using the molarity equation:

Molarity (M) = moles solute / liters solution

The number of moles of HCl and NaOH can be calculated:

Moles HCl = Molarity of HCl x Volume of HCl in liters
= 0.200 M x (20.0 mL / 1000 mL/L)
= 0.200 x 0.0200 L
= 0.00400 moles HCl

Moles NaOH = Molarity of NaOH x Volume of NaOH in liters
= 0.400 M x (x mL / 1000 mL/L)
= 0.400 x (x/1000) L

At the equivalence point:

Moles HCl = Moles NaOH

Setting the two mole values equal:

0.00400 moles HCl = 0.400 x (x/1000) moles NaOH

Solving this equation for the unknown volume of NaOH:

x / 1000 = 0.00400 / 0.400
x = (0.00400 / 0.400) x 1000

x = 10.0 mL NaOH

## Conclusion

To completely neutralize 20.0 mL of 0.200 M HCl, 10.0 mL of 0.400 M NaOH is required.

## Step-by-Step Workings

The detailed workings to arrive at the answer are as follows:

1. Write out the balanced chemical equation for the neutralization reaction:

HCl + NaOH → NaCl + H2O

2. Note the given information:
• Molarity of HCl solution = 0.200 M
• Volume of HCl solution = 20.0 mL
• Molarity of NaOH solution = 0.400 M
• Volume of NaOH solution = x mL (unknown)
3. Use the molarity equation to calculate moles of HCl:

Moles HCl = Molarity of HCl x Volume of HCl in liters

= 0.200 M x (20.0 mL / 1000 mL/L)

= 0.200 x 0.0200 L

= 0.00400 moles HCl

4. Use the molarity equation to calculate moles of NaOH:

Moles NaOH = Molarity of NaOH x Volume of NaOH in liters

= 0.400 M x (x mL / 1000 mL/L)

= 0.400 x (x/1000) L

5. At equivalence point, moles HCl = moles NaOH

0.00400 moles HCl = 0.400 x (x/1000) moles NaOH

6. Solve the equation for x:

x / 1000 = 0.00400 / 0.400

x = (0.00400 / 0.400) x 1000

x = 10.0 mL NaOH

7. Therefore, the volume of 0.400 M NaOH required to completely neutralize 20.0 mL of 0.200 M HCl is 10.0 mL

## Concepts

This stoichiometry problem applies the following important concepts:

• Molarity – The concentration of a solution expressed as moles solute per liter of solution.
• Balanced chemical equation – The number of reactant and product molecules must be equal on both sides of the reaction arrow.
• Stoichiometric moles – The mole ratio between reactants and products in a balanced equation.
• Equivalence point – When stoichiometrically equivalent amounts of acid and base have been mixed in a titration.

Understanding these concepts allows us to set up and solve the proper mathematical relationships to determine the volume of base needed to react completely with a given volume and molarity of acid.

## Acid-Base Titration Calculations

This type of acid-base neutralization calculation is commonly applied to acid-base titration analysis. In a titration, a base of known concentration is slowly added to an acid of known concentration until neutralization is reached. Indicators can show when the equivalence point is reached. The volume of base added to reach this point is used along with molarity to determine the moles and concentration of the unknown acid or base.

Some common titration calculations include:

• Determining the molarity of an unknown acid or base from the volume and molarity of the titrant
• Determining the volume of titrant required to reach the equivalence point in the titration if the molarities of acid and base are known
• Calculating the pH at different stages of a titration using the acid or base molarities and volumes added

Proper setup of the stoichiometric relationships and molarity calculations is crucial for accurate titration analyses.

## Real-World Applications

Acid-base neutralization reactions have many important real-world uses including:

• Antacid tablets – Contain bases like calcium carbonate or magnesium hydroxide which neutralize excess stomach acid.
• Controlling aquarium pH – Acids or bases added to adjust pH to levels safe for fish.
• Wastewater treatment – Acids or bases added to adjust pH levels before releasing water into the environment.
• Shampoo chemistry – Sodium lauryl sulfate reacts with hair proteins via an acid-base reaction.
• Food and beverages – Citric, malic, and other acids react with sodium bicarbonate in recipes to help batters rise.

Careful calculations are done to determine the exact amounts of acids and bases needed for these applications.

## Practice Examples

Here are some additional practice examples of acid-base neutralization calculations:

### Example 1

How many mL of 2.00 M HCl are needed to completely neutralize 0.350 L of 1.50 M NaOH?

### Solution

1. Write the balanced chemical equation:

HCl + NaOH → NaCl + H2O

2. Convert the NaOH volume to liters:

0.350 L NaOH x (1000 mL/1 L) = 350 mL NaOH

3. Calculate the moles of NaOH:

Moles NaOH = Molarity x Volume in Liters

= 1.50 M x 0.350 L = 0.525 moles NaOH

4. At equivalence point: Moles HCl = Moles NaOH

Moles HCl = 0.525 moles

5. Use moles HCl to calculate the volume of HCl needed:

Molarity x Volume = Moles

2.00 M x Volume HCl = 0.525 moles

Volume HCl = Moles HCl / Molarity HCl

= 0.525 moles / 2.00 M = 0.263 L = 263 mL HCl

Therefore, 263 mL of 2.00 M HCl will completely neutralize 0.350 L of 1.50 M NaOH.

### Example 2

A tank contains 3.60 L of 6.00 M NaOH solution. What volume of 12.0 M HCl solution is required to completely neutralize it?

### Solution

1. Write the balanced chemical equation:

HCl + NaOH → NaCl + H2O

2. Convert NaOH volume to liters:

Volume NaOH = 3.60 L

3. Calculate moles of NaOH:

Moles NaOH = Molarity x Volume

= 6.00 M x 3.60 L = 21.6 moles NaOH

4. Moles HCl = Moles NaOH at equivalence

Moles HCl = 21.6 moles

5. Calculate volume of HCl needed:

Molarity x Volume = Moles

12.0 M x Volume HCl = 21.6 moles

Volume HCl = Moles HCl / Molarity HCl

= 21.6 moles / 12.0 M = 1.80 L HCl

To completely neutralize 3.60 L of 6.00 M NaOH, 1.80 L of 12.0 M HCl is required.

## Common Mistakes

Some common mistakes when solving acid-base neutralization problems include:

• Forgetting to convert volumes to liters before using molarity
• Not setting up moles of acid = moles of base at the equivalence point
• Incorrectly calculating molarity x volume
• Messing up the unit conversions
• Not balancing the chemical equation correctly
• Plugging the numbers into the wrong places in the calculations

Being mindful of the units and carefully walking through each step can help avoid these errors.

## Conclusion

Acid-base neutralization reactions are important in chemistry and biology. Calculating the amounts required for complete neutralization requires a clear understanding of molarity, stoichiometric relationships, and setting up the proper mole ratio calculations. With practice, these types of problems can be systematically solved to determine the volume of acid or base needed for exact neutralization.