Problem 40 - Chemistry: The Central Science

Question

How many milliliters of \(0.105 \mathrm{M} \mathrm{HCl}\) are needed to titrate each of the following solutions to the equivalence point: (a) \(45.0 \mathrm{~mL}\) of \(0.0950 \mathrm{M} \mathrm{NaOH}\), (b) \(22.5 \mathrm{~mL}\) of \(0.118 \mathrm{M} \mathrm{NH}_{3}\) (c) \(125.0 \mathrm{~mL}\) of a solution that contains \(1.35 \mathrm{~g}\) of \(\mathrm{NaOH}\) per liter?

Step-by-Step Solution

Answer

The volumes of \(0.105 \mathrm{M} \mathrm{HCl}\) required to titrate each solution to the equivalence point are: (a) \(40.714 \mathrm{~mL}\) for \(45.0 \mathrm{~mL}\) of \(0.0950 \mathrm{M} \mathrm{NaOH}\), (b) \(25.286 \mathrm{~mL}\) for \(22.5 \mathrm{~mL}\) of \(0.118 \mathrm{M} \mathrm{NH}_{3}\), and (c) \(40.179 \mathrm{~mL}\) for \(125.0 \mathrm{~mL}\) of a solution that contains \(1.35 \mathrm{~g}\) of \(\mathrm{NaOH}\) per liter.

Step 1: (Step 1: Write the balanced chemical equations for each titration)

For each titration, we will have different balanced chemical equations, which are: (a) For \(\mathrm{HCl}\) and \(\mathrm{NaOH}\), the balanced chemical equation is: \[ \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H}_{2}\mathrm{O} \] The stoichiometry between \(\mathrm{HCl}\) and \(\mathrm{NaOH}\) is 1:1. (b) For \(\mathrm{HCl}\) and \(\mathrm{NH}_{3}\), the balanced chemical equation is: \[ \mathrm{HCl} + \mathrm{NH}_{3} \rightarrow \mathrm{NH}_{4}\mathrm{Cl} \] The stoichiometry between \(\mathrm{HCl}\) and \(\mathrm{NH}_{3}\) is 1:1. (c) For \(\mathrm{HCl}\) and \(\mathrm{NaOH}\), the balanced chemical equation is: \[ \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H}_{2}\mathrm{O} \] The stoichiometry between \(\mathrm{HCl}\)and \(\mathrm{NaOH}\) is 1:1.

Step 2: (Step 2: Calculate the moles of the base in each solution)

Now, we will calculate the moles of the base in each solution for a, b, and c. (a) Moles of \(\mathrm{NaOH}\) in \(45.0 \mathrm{~mL}\) of \(0.0950 \mathrm{M} \mathrm{NaOH}\): Moles = Molarity × Volume (in Liters) Moles of \(\mathrm{NaOH}\) = \(0.0950 \cdot 0.045 = 0.004275 \mathrm{~mol}\) (b) Moles of \(\mathrm{NH}_{3}\) in \(22.5 \mathrm{~mL}\) of \(0.118 \mathrm{M} \mathrm{NH}_{3}\): Moles of \(\mathrm{NH}_{3}\) = \(0.118 \cdot 0.0225 = 0.002655 \mathrm{~mol}\) (c) Moles of \(\mathrm{NaOH}\) in \(125.0 \mathrm{~mL}\) of solution containing \(1.35 \mathrm{~g}\) of \(\mathrm{NaOH}\) per liter: First, we need to find the molarity of \(\mathrm{NaOH}\) in this solution: Molarity = Mass of \(\mathrm{NaOH}\)/[(Molecular weight of \(\mathrm{NaOH}\)) × (Volume in liters)] Molarity = \(1.35 / (40.0 \cdot 1) = 0.03375 \mathrm{M}\) Now, we will calculate the moles of \(\mathrm{NaOH}\): Moles of \(\mathrm{NaOH}\) = \(0.03375 \cdot 0.125 = 0.00421875 \mathrm{~mol}\)

Step 3: (Step 3: Calculate the volume of \(0.105 \mathrm{M} \mathrm{HCl}\) required to neutralize each solution)

As the stoichiometry between \(\mathrm{HCl}\) and the base in each case is 1:1, we will now use the moles of the base calculated in step 2 to determine the volume of \(0.105 \mathrm{M} \mathrm{HCl}\) needed. (a) Volume of \(0.105 \mathrm{M} \mathrm{HCl}\) for \(0.004275 \mathrm{~mol}\) \(\mathrm{NaOH}\): Volume (in Liters) = Moles/Molarity Volume = \(0.004275 / 0.105 = 0.04071429 \mathrm{~L}\) or \(40.714 \mathrm{~mL}\) (b) Volume of \(0.105 \mathrm{M} \mathrm{HCl}\) for \(0.002655 \mathrm{~mol}\) \(\mathrm{NH}_{3}\): Volume = \(0.002655 / 0.105 = 0.02528571 \mathrm{~L}\) or \(25.286 \mathrm{~mL}\) (c) Volume of \(0.105 \mathrm{M} \mathrm{HCl}\) for \(0.00421875 \mathrm{~mol}\) \(\mathrm{NaOH}\): Volume = \(0.00421875 / 0.105 = 0.04017857 \mathrm{~L}\) or \(40.179 \mathrm{~mL}\) Therefore, the required volumes of \(0.105 \mathrm{M} \mathrm{HCl}\) to titrate each solution to the equivalence point are: (a) \(40.714 \mathrm{~mL}\), (b) \(25.286 \mathrm{~mL}\), and (c) \(40.179 \mathrm{~mL}\).

Key Concepts

Stoichiometry

Stoichiometry lies at the heart of titration calculations and determines how we calculate the quantities of reactants needed in chemical reactions. In the context of titrations, stoichiometry helps us understand the relationship between the amount of acid and base needed for a complete reaction.

For example, when titrating hydrochloric acid (HCl) with sodium hydroxide (NaOH), we rely on the balanced chemical equation \( \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H}_2\mathrm{O} \). This equation shows a 1:1 stoichiometry, meaning one mole of HCl reacts with one mole of NaOH.

We use stoichiometry to calculate how many moles of HCl are required based on the moles of NaOH or any base present. This ensures that the reaction reaches the equivalence point, where equal amounts of acid and base have reacted.

• Identify the balanced equation.
• Understand molar relationships between reactants.
• Use mole ratios to find the required reactant quantities.

Molarity

Molarity is a measurement of concentration in chemistry, crucial for titration problems. It is defined as the number of moles of solute (the substance being dissolved) per liter of solution. The formula for molarity \( M \) is \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \).

In titrations, knowing the molarity of a solution allows you to find out how many moles of a reactant you have based on its volume. For instance, if you have a 0.105 M solution of HCl, it means there are 0.105 moles of HCl in every liter of solution.

Calculating molarity also lets you account for how much of a titrant is needed to react completely with a given volume of another solution. During titrations, the molarity informs us how concentrated the solution is, thus helping in precise stoichiometric calculations.

• Use the formula \( M = \frac{\text{moles}}{\text{liters}} \).
• Essential for identifying the required volume of titrant.
• Facilitates determining concentration changes during reactions.

Acid-Base Reactions

Acid-base reactions are unique because they involve the transfer of protons (H⁺) from the acid to the base. The equivalence point in titrations is reached when the number of moles of the acid equals the number of moles of the base in the mixture, resulting in a neutral solution.

Different acids and bases react in stoichiometric ratios, which can vary based on the strength and number of ionizable hydrogens of the acid and the base's ability to accept protons. In the titration exercises solved, we considered common reactions such as with \( \mathrm{HCl} \) and \( \mathrm{NaOH} \) or \( \mathrm{NH}_3\), both having a 1:1 stoichiometry.

• Involves proton donation and acceptance.
• The equivalence point marks stoichiometric completion.
• Important to understand reaction characteristics like whether it's exothermic or releases gas.
• Stoichiometry directly links to determining reactants' required amounts.

Grasping these reactions' nature often requires observing the neutralization process, often tested with indicators like phenolphthalein or through pH meters.