Problem 64 - Chemistry The Science in Context

Question

How many milliliters of \(0.250 M\) HNO \(_{3}\) are needed to neutralize the following solutions? a. \(25.0 \mathrm{mL}\) of \(0.395 M \mathrm{KOH}\) b. \(78.6 \mathrm{mL}\) of \(0.0100 M \mathrm{Al}(\mathrm{OH})_{3}\) c. \(65.9 \mathrm{mL}\) of \(0.475 M \mathrm{NaOH}\)

Step-by-Step Solution

Answer

a) 39.5 mL for 25.0 mL of 0.395 M KOH b) 9.43 mL for 78.6 mL of 0.0100 M Al(OH)₃ c) 125.24 mL for 65.9 mL of 0.475 M NaOH

Step 1: Write the balanced equations for each reaction with HNO\(_3\)

For each solution, we will first write out the balanced chemical equation, which will help us determine the stoichiometry (the ratio of reactants and products) for the reaction: a. KOH + HNO\(_3\) ⟶ KNO\(_3\) + H\(_2\)O b. Al(OH)\(_3\) + 3HNO\(_3\) ⟶ Al(NO\(_3\))\(_3\) + 3H\(_2\)O c. NaOH + HNO\(_3\) ⟶ NaNO\(_3\) + H\(_2\)O

Step 2: Identify moles of the given solutions

Next, we will find the moles of each solution using the given volumes and molar concentrations: a. Moles of KOH = volume (L) × molarity = (25.0 mL × \(\frac{1 L}{1000 mL}\)) × 0.395 mol/L = 0.00988 mol b. Moles of Al(OH)\(_3\) = (78.6 mL × \(\frac{1 L}{1000 mL}\)) × 0.0100 mol/L = 0.000786 mol c. Moles of NaOH = (65.9 mL × \(\frac{1 L}{1000 mL}\)) × 0.475 mol/L = 0.03131 mol

Step 3: Determine moles of HNO\(_3\) needed for neutralization

Now, from the balanced equations, calculate the moles of HNO\(_3\) required to neutralize each solution using stoichiometry: a. Moles of HNO\(_3\) = moles of KOH × \(\frac{1 \, mol \, HNO_3}{1 \, mol \, KOH}\) = 0.00988 mol b. Moles of HNO\(_3\) = moles of Al(OH)\(_3\) × \(\frac{3 \, mol \, HNO_3}{1 \, mol \, Al(OH)_3}\) = 0.002358 mol c. Moles of HNO\(_3\) = moles of NaOH × \(\frac{1 \, mol \, HNO_3}{1 \, mol \, NaOH}\) = 0.03131 mol

Step 4: Calculate the volume of HNO\(_3\) required

Finally, we can calculate the volume of \(0.250 M\) HNO\(_3\) needed to neutralize each solution by dividing moles of HNO\(_3\) required by its molarity: a. Volume of HNO\(_3\) = moles of HNO\(_3\) ÷ molarity = \(\frac{0.00988 \, mol}{0.250 \, M} = 0.0395 \, L\) or \(39.5 \, mL\) b. Volume of HNO\(_3\) = \(\frac{0.002358 \, mol}{0.250 \, M} = 0.00943 \, L\) or \(9.43 \, mL\) c. Volume of HNO\(_3\) = \(\frac{0.03131 \, mol}{0.250 \, M} = 0.12524 \, L\) or \(125.24 \, mL\) Therefore, the volume of \(0.250 M\) HNO\(_3\) required to neutralize each solution is: a. \(39.5 \, mL\) for \(25.0 \, mL\) of \(0.395 M \, KOH\) b. \(9.43 \, mL\) for \(78.6 \, mL\) of \(0.0100 M \, Al(OH)_3\) c. \(125.24 \, mL\) for \(65.9 \, mL\) of \(0.475 M \, NaOH\)

Key Concepts

Acid-Base Reactions

Neutralization reactions are a subtype of acid-base reactions. These occur when an acid and a base react to form water and a salt. The interaction often involves hydrogen ions (\(H^+\)) from the acid and hydroxide ions (\(OH^-\)) from the base, resulting in water (\(H_2O\)) as a byproduct. In our exercise, \(HNO_3\), a strong acid, reacts with bases like \(KOH\), \(Al(OH)_3\), and \(NaOH\), forming salts such as \(KNO_3\), \(Al(NO_3)_3\), and \(NaNO_3\) respectively.

Understanding these reactions involves several key components:

• The reaction of \(HNO_3\) with a base results in a neutral solution if stoichiometric amounts are used.
• Water formation shows the reactive interaction between \(H^+\) from acid and \(OH^-\) from base.
• Bonds are broken and formed in a manner that favours the production of a stable salt and water.

Mistakes sometimes arise when one doesn't correctly balance the equation or understand the stoichiometry of the involved components. Always write a balanced chemical equation first to avoid confusion.

Stoichiometry

Stoichiometry is the quantitative aspect of chemical reactions and equations. It involves calculating the amounts of reactants needed or products formed in a chemical reaction. In our given problem, stoichiometry helps us determine how much \(HNO_3\) is needed to neutralize different bases. This is done using the coefficients from the balanced chemical equation which gives the molar ratio of each reactant to product.

For this exercise:

• For \(KOH\) and \(NaOH\), the reactions are 1:1 based on their equations, meaning every mole of base neutralizes one mole of \(HNO_3\).
• For \(Al(OH)_3\), the reaction is 1:3, requiring three times the amount of \(HNO_3\) to completely react with "one mole of base" due to the trivalent nature of aluminum hydroxide.

Accurate stoichiometry ensures reactants are used efficiently and can help in determining the correct proportions in any chemical process.

Molarity

Molarity (M) is the concentration of a solution expressed as the number of moles of solute per liter of solution. It is a crucial concept for calculating the amounts of reactants and products in a solution. In any neutralization reaction like those in our exercise, knowing the molarity of \(HNO_3\) and the bases involved allows us to determine how much acid is needed for complete neutralization.

For example:- The molarity given for \(HNO_3\) is \(0.250\, M\).- By using this, we can determine the volume of \(HNO_3\) needed when scaling up from moles (calculated through stoichiometry) to volume, using: \[\text{Volume} (L) = \frac{\text{Moles of solute}}{\text{Molarity}}\]This method allows us to convert from the amount in moles to a measurable volume, facilitating real-world applications in laboratory settings or industrial processes.

Chemical Equations

Chemical equations convey essential information about a chemical reaction, including the reactants, products, and their stoichiometric relationships. Writing a balanced chemical equation is the foundation for calculating reactant or product quantities and is a necessary step in reaction analysis.

For the reactions given in the exercise:

• A balanced equation provides the exact number of moles of each reactant and product involved.
• Each equation must account for the conservation of mass, ensuring the same number of atoms for each element on both sides of the equation.

To solve our problem, we started with balanced chemical equations. These guided the stoichiometric calculations by showing the ratios needed, such as 1:1 for the reactions with \(KOH\) and \(NaOH\), or 1:3 for \(Al(OH)_3\). Properly balanced equations ensure the accuracy of further calculations concerning the reaction, leading to successful experimental outcomes.