How many milliliters of 3.0 M NaOH are required to react with 4.0 mL of 16m HNO3?

To determine the amount of milliliters of 3.0 M NaOH required to react with 4.0 mL of 16m HNO3, we need to follow these steps:

Step 1: Write the balanced chemical equation

The balanced chemical equation for the reaction between NaOH and HNO3 is:

NaOH + HNO3 → NaNO3 + H2O

Step 2: Convert the volumes given to moles

To convert volumes to moles, we need to use the molarity (M) and volume given for each reactant. The molarity tells us the number of moles of solute per liter of solution.

For NaOH:

  • Molarity (M) = 3.0 M
  • Volume (V) = x mL (unknown)

For HNO3:

  • Molarity (M) = 16 M
  • Volume (V) = 4.0 mL

The equation relating molarity, volume and moles is:

Molarity (M) = Moles / Volume (L)

Rearranging to solve for moles:

Moles = Molarity (M) x Volume (L)

Using this equation:

Moles of NaOH:
Moles = Molarity x Volume
= 3.0 M x (x mL / 1000 mL/L)
= 0.003 x mL

Moles of HNO3:
Moles = Molarity x Volume
= 16 M x (4.0 mL / 1000 mL/L)
= 0.064 moles

Step 3: Determine mole ratio from balanced equation

From the balanced equation:

NaOH + HNO3 → NaNO3 + H2O

The mole ratio between NaOH and HNO3 is 1:1. This means that the number of moles of NaOH required equals the number of moles of HNO3 given.

Step 4: Use mole ratio to calculate moles of NaOH required

Since the mole ratio between NaOH and HNO3 is 1:1:

Moles of NaOH required = Moles of HNO3 given
= 0.064 moles

Step 5: Use moles of NaOH to calculate volume needed

To calculate the volume of NaOH required, we rearrange the moles equation:

Moles = Molarity x Volume

Volume = Moles / Molarity

Plugging in the values:

Volume of NaOH = Moles of NaOH required / Molarity of NaOH
= 0.064 moles / 3.0 M
= 0.021 L x (1000 mL/1 L)
= 21 mL

Conclusion

Based on the calculations, 21 mL of 3.0 M NaOH are required to completely react with 4.0 mL of 16 M HNO3.

To summarize:

  • The balanced equation showed a 1:1 mole ratio between NaOH and HNO3.
  • The moles of HNO3 were calculated from the given molarity and volume.
  • Using the 1:1 ratio, the moles of NaOH required equals the moles of HNO3.
  • The volume of NaOH was calculated from the required moles and known molarity.

This step-by-step approach allows us to systematically determine the amount of NaOH required for a complete reaction with the given amount of HNO3.

Background

This type of reaction involves an acid-base neutralization. HNO3 is a strong acid, completely dissociating into H+ and NO3- ions when dissolved in water. NaOH is a strong base that dissociates completely into Na+ and OH- ions.

When the H+ and OH- ions combine, they form water in an exothermic reaction. The Na+ and NO3- ions form sodium nitrate, which remains dissociated in solution.

Balancing the chemical equation for this neutralization reaction involves making sure equal amounts of H+ and OH- ions are present. This results in a 1:1 mole ratio between the acid (HNO3) and base (NaOH).

By converting the volumes given to moles using the molarity, then relating the moles through the balanced equation, we can determine the amount of one reactant needed to react completely with a given amount of the other reactant.

Acid-base neutralization reactions like this have many real-world applications. For example, NaOH can be used to neutralize acidic wastewaters and spills. HNO3 is commonly used as an industrial chemical, so spills and exposure are hazards. Neutralization with NaOH renders the acid harmless.

Practical Applications

Acid-base neutralization reactions have many practical uses and applications in both industry and laboratory settings.

Titration

The exact stoichiometric calculations used in this example are applied in acid-base titrations. Titration uses a standard solution of known concentration to determine the concentration of an unknown solution. The volumes required to reach the neutralization endpoint allow calculation of the unknown concentration.

pH Control

The pH of solutions can be adjusted by adding precise amounts of acids or bases calculated using neutralization reactions. This allows chemists to carefully control the pH of solutions needed for experiments and industrial processes.

Wastewater Treatment

Acids and bases are often used in industrial processes, resulting in wastewaters that require neutralization before discharge or reuse. Calculating additions of acids or bases allows effective neutralization of large wastewater volumes.

Product Formulation

Many consumer products require specific pH ranges to function properly and remain stable. Shampoos, cosmetics, and certain pharmaceuticals rely on carefully calculated acid-base chemistry during formulation.

Antacids

Acid-base neutralizations are used to counteract excess stomach acid and relieve symptoms of heartburn and indigestion. Antacids contain bases like magnesium hydroxide that neutralize stomach acid.

Solving Similar Problems

The approach used here can be applied to any acid-base neutralization reaction. The key steps are:

  1. Write a balanced chemical equation.
  2. Use molarities and volumes given to calculate moles of each reactant.
  3. Determine the mole ratio from the balanced equation.
  4. Use the mole ratio to calculate moles of one reactant needed to react with the given amount of the other.
  5. Convert moles of the needed reactant back to volume using molarity.

To practice these steps, try solving some additional examples:

Practice Example 1

How many milliliters of 12 M HCl are needed to completely react with 3.5 g of NaOH?

Practice Example 2

What volume of 5.6 M Na2CO3 solution reacts completely with 325 mL of 0.15 M HNO3?

Practice Example 3

If 25 mL of 0.1 M Ba(OH)2 are required to neutralize 35 mL of HClO4, what is the molarity of the HClO4?

Working through additional practice problems by applying the same approach helps solidify understanding of acid-base neutralization calculations.

Limitations

While the calculations shown provide an exact stoichiometric amount, there are some limitations to consider:

  • Volumes measurements may have small errors that propagate through the calculations.
  • If the reaction proceeds partially to equilibrium, calculations will no longer match experimental results.
  • Impurities in chemicals can alter the effective concentrations from stated values.
  • Secondary reactions may consume some of the reactants.

Due to these limitations, a small excess of titrant is often used to drive the reaction to completion. Calculations provide an estimate that must be refined through experiments.

Summary

Acid-base neutralization reactions involve combining an acid and a base to form water and a salt. Balancing the chemical equation allows stoichiometric calculations to determine volumes of reactants needed for complete reaction.

Key steps include:

  1. Writing a balanced equation
  2. Converting volumes and molarities to moles
  3. Determining the mole ratio from the equation
  4. Relating moles through the ratio
  5. Converting moles back to volume using molarity

This straightforward method allows calculation of acid and base amounts needed for neutralization reactions. The approach can be applied to a variety of acid-base systems with many practical uses.

However, the calculations involve some assumptions and limitations. Experimental refinements may be needed to account for errors and equilibrium effects. With careful work, stoichiometric calculations provide an excellent starting point for quantifying acid-base neutralization reactions.

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