Chapter 4: Problem 64
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 \(0.475 M \mathrm{NaOH}\)
Short Answer
Expert verified
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 by step solution
01
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
02
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
03
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
04
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 \(0.475 M \, NaOH\)
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
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 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.
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.
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:
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.