# How many milliliters mL of a 0.610 M NaOH solution are needed to neutralize 20.0 mL of a 0.245 M H2SO4 solution?

Determining the volume of sodium hydroxide (NaOH) solution needed to neutralize a given volume of sulfuric acid (H2SO4) solution is a common type of stoichiometry problem. Stoichiometry involves using the quantitative relationships between reactants and products in a chemical reaction to determine the amount of one substance needed to react with a given amount of another substance.

## Step-By-Step Solution

Here is a step-by-step approach to solving this problem:

1. Write the balanced chemical equation for the neutralization reaction between NaOH and H2SO4:
2. NaOH(aq) + H2SO4(aq) → NaHSO4(aq) + H2O(l)

3. Identify the given quantities:
• Volume of H2SO4 solution = 20.0 mL
• Molarity of H2SO4 solution = 0.245 M
• Molarity of NaOH solution = 0.610 M
4. Use the molarity equation to calculate the number of moles of H2SO4 in the given volume:
• Molarity = moles solute / L solution
• Rearrange: moles solute = Molarity x L solution
• Moles of H2SO4 = (0.245 M) x (0.0200 L) = 0.00490 moles
5. Use the stoichiometric ratio from the balanced equation to determine the number of moles of NaOH needed to completely react with the H2SO4:
• 1 mole NaOH reacts with 1 mole H2SO4
• So if you have 0.00490 moles H2SO4, you need 0.00490 moles NaOH to neutralize it.
6. Use the molarity equation again to convert from moles of NaOH to volume of NaOH solution:
• Moles NaOH = 0.00490 moles
• Molarity of NaOH = 0.610 M
• Rearrange: Volume NaOH solution = Moles NaOH / Molarity NaOH
• Volume NaOH solution = (0.00490 moles) / (0.610 M) = 0.00803 L = 8.03 mL

So in summary, 20.0 mL of 0.245 M H2SO4 solution reacts with 8.03 mL of 0.610 M NaOH solution.

## Explanation of Calculations

Let’s go through each calculation step in more detail:

### 1. Calculate moles of H2SO4

We are given:

• Volume of H2SO4 solution = 20.0 mL
• Molarity of H2SO4 solution = 0.245 M

To find the moles of H2SO4, we use the molarity equation:

Molarity = moles solute / L solution

Rearrange to solve for moles:

moles solute = Molarity x L solution

Plug in the known values:

moles H2SO4 = (0.245 M) x (0.0200 L) = 0.00490 moles

### 2. Use stoichiometric ratio to find moles NaOH

From the balanced equation:

NaOH(aq) + H2SO4(aq) → NaHSO4(aq) + H2O(l)

We see that 1 mole of NaOH reacts with 1 mole of H2SO4 in a 1:1 ratio.

Since we calculated 0.00490 moles H2SO4, we need 0.00490 moles NaOH to completely react with the H2SO4.

### 3. Convert moles NaOH to volume NaOH solution

Now we calculate the volume of 0.610 M NaOH solution needed to provide 0.00490 moles NaOH:

Given:

• Moles NaOH = 0.00490 moles
• Molarity of NaOH = 0.610 M

Use molarity equation:

Molarity = moles solute / L solution

Rearrange to solve for volume:

Volume solution = moles solute / Molarity

Plug in known values:

Volume NaOH solution = (0.00490 moles) / (0.610 M) = 0.00803 L = 8.03 mL

## Common Mistakes

Some common mistakes students make when solving this type of stoichiometry problem include:

• Forgetting to convert mL to L when using the molarity equation
• Not balancing the chemical equation correctly
• Using the wrong stoichiometric ratio between reactants
• Mixing up the molarities of the NaOH and H2SO4 solutions
• Converting incorrectly between moles and volume using molarity

To avoid these mistakes:

• Always check that the chemical equation is balanced first
• Note the molarities of each solution clearly
• Pay close attention to unit conversions between mL, L, and moles
• Double check that you used the correct stoichiometric ratio

## Conceptual Background

Let’s discuss some of the important chemistry concepts involved in solving this type of neutralization problem:

### Molarity

Molarity (M) is a concentration unit that expresses the number of moles of solute dissolved per liter of solution. It is used extensively in stoichiometry calculations. The molarity equation relates moles of solute, volume of solution, and molarity:

Molarity (M) = moles solute (mol) / volume solution (L)

### Balanced Chemical Equations

A balanced chemical equation shows the number and type of reactants and products involved in the reaction. All chemical equations must be balanced in terms of atoms or charges. The coefficients in a balanced equation provide the stoichiometric mole ratios between reactants and products.

### Stoichiometric Ratios

Stoichiometric ratios refer to the quantitative relationships between amounts of reactants and products in a balanced chemical equation. These ratios allow us to convert between moles of different substances in the reaction. For example, in the equation:

NaOH(aq) + H2SO4(aq) → NaHSO4(aq) + H2O(l)

The stoichiometric ratio between NaOH and H2SO4 is 1:1. This tells us if we know the moles of H2SO4, we can find the moles of NaOH needed by using a factor of 1 mole NaOH for every 1 mole H2SO4.

### Limiting Reactant

The limiting reactant is the reactant that gets completely consumed in a reaction. Once the limiting reactant runs out, the reaction stops. Equivalently, it is the reactant present in the smallest stoichiometric amount. By calculating the moles of limiting reactant, we ensure enough of the excess reactant is present to completely react with it.

### Neutralization Reactions

A neutralization reaction occurs when an acid and a base react to form a salt and water. It typically involves combining the hydroxide ion (OH-) from a base with the hydrogen ion (H+) from an acid to generate H2O. Strong acid-strong base neutralizations go essentially to completion. Weak acid-weak base reactions achieve partial neutralization. The point at which moles of acid = moles of base added is the equivalence point.

## Practice Problems

Here are some additional practice problems you can work through to gain proficiency solving neutralization stoichiometry questions:

### Problem 1

How many mL of 2.50 M HCl solution are required to react completely with 325 mL of 1.55 M NaOH solution?

### Problem 2

If 35.5 mL of 0.775 M H2SO4 solution reacts with 22.8 mL of 0.358 M KOH solution, calculate the concentration of excess reactant remaining in solution after the neutralization reaction goes to completion.

### Problem 3

For the reaction: Fe(OH)3 + 3HCl → FeCl3 + 3H2O

If you initially had 55.5 g of Fe(OH)3 and 95.0 mL of 1.83 M HCl, determine which reactant is limiting and calculate the moles of H2O formed.

## Applications of Neutralization Stoichiometry

Understanding acid-base neutralization stoichiometry has many real-world uses and applications. Here are a few examples:

### Titration Analysis

Titration involves incrementally adding a known concentration and volume of acid or base solution to a sample of unknown concentration. Using indicator dyes, we can determine the equivalence point where neutralization is reached. From the stoichiometric calculations, the unknown concentration can be determined.

### Wastewater Treatment

Chemicals are often added to adjust the pH of wastewater to facilitate treatment processes. Knowing how to calculate the amount of acid or base required to reach the desired pH is essential using principles of acid-base stoichiometry.

### Antacid Tablets

Neutralizing excess stomach acid with antacid tablets relies on acid-base stoichiometry. Determining the amount of base needed to reach stomach pH neutrality requires knowledge of the stomach acid concentration and volume as well as the antacid strength.

### Making Soaps and Detergents

Saponification is the reaction of fats or oils with a strong base to produce soaps. Balancing the stoichiometry between the fat and the base is important to ensure all reactants are consumed and a fully neutralized soap is formed.

### Chemical Manufacturing

Many industrial chemical processes involve neutralization reactions, for example to remove impurities or produce salts. Careful stoichiometric calculations are done to optimize cost efficiency and material yields.

## Related Chemistry Concepts

While a full chemistry curriculum is beyond the scope here, it’s helpful to understand how acid-base stoichiometry fits into broader chemistry topics:

### Reaction Stoichiometry

The techniques used are applicable to other reaction types besides neutralization, including precipitation, redox, gas forming, and combustion reactions. The steps involved are: write and balance the equation, convert known quantity to moles, use stoichiometric mole ratio, then convert moles of unknown to desired units.

### Solution Chemistry

Understanding molarity and dilution calculations is essential. Other important topics are solubility rules, precipitation reactions, and colligative properties (boiling point elevation, freezing point depression, osmotic pressure). Acid-base chemistry requires knowledge of pH, equilibrium, buffers, indicators, and titration.

### Electrochemistry

The interaction of chemical species and electricity underlies all redox reactions. Important concepts are standard reduction potentials, galvanic/electrolytic cells, Nernst equation, Faraday’s laws, and electrolysis.

### Thermochemistry

Heat is exchanged in many chemical reactions and phase changes. Key principles are enthalpy, Hess’s law, calorimetry, heat capacity, and heating/cooling curves. Energetics of chemical reactions also tie into kinetics and equilibria.

### Quantum Chemistry

Understanding chemical bonding and reactivity requires quantum mechanics. This includes electronic structure theory, molecular orbital theory, periodic trends, and computational chemistry models.

## Conclusion

In summary, solving stoichiometry problems for neutralization reactions involves properly balancing the chemical equation, using molarity to interconvert solute concentrations between moles and volumes, applying the stoichiometric mole ratios, and converting the moles of the unknown reactant or product into the desired units. Mastering acid-base neutralization stoichiometry provides a foundation for performing titration calculations, determining appropriate pH adjustments, optimizing chemical processes, and comprehending more advanced chemistry principles.