Question

How many milliliters of \(0.0850 \mathrm{M} \mathrm{NaOH}\) are required to titrate each of the following solutions to the equivalence point: (a) \(40.0 \mathrm{~mL}\) of \(0.0900 \mathrm{MHNO}_{3}\), (b) \(35.0 \mathrm{~mL}\) of \(0.0850 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH}\), (c) \(50.0 \mathrm{~mL}\). of a solution that contains \(1.85 \mathrm{~g}\) of \(\mathrm{HCl}\) per liter?

Short answer

(a) It takes \(42.4 mL\) of \(0.0850 M\) NaOH to titrate \(40.0 mL\) of \(0.0900 M\) HNO3 to the equivalence point. (b) It takes \(35.0 mL\) of \(0.0850 M\) NaOH to titrate \(35.0 mL\) of \(0.0850 M\) CH3COOH to the equivalence point. (c) It takes \(29.8 mL\) of \(0.0850 M\) NaOH to titrate \(50.0 mL\) of a solution that contains \(1.85 g\) of HCl per liter to the equivalence point.

Step-by-step solution

Part (a) - Titration of 40.0 mL of 0.0900 M HNO3

First, find the number of moles of HNO3: Number of moles of HNO3 = M1V1 = (0.0900 mol/L)(0.0400 L) = 0.00360 mol Now, we need to find the volume (V2) of NaOH required to titrate this amount of HNO3: 0.00360 mol = (0.0850 mol/L)(V2) V2 = 0.00360 mol / 0.0850 mol/L = 0.0424 L V2 = 42.4 mL It takes \(42.4 mL\) of \(0.0850 M\) NaOH to titrate \(40.0 mL\) of \(0.0900 M\) HNO3 to the equivalence point.

Part (b) - Titration of 35.0 mL of 0.0850 M CH3COOH

First, find the number of moles of CH3COOH: Number of moles of CH3COOH = M1V1 = (0.0850 mol/L)(0.0350 L) = 0.002975 mol Now, we need to find the volume (V2) of NaOH required to titrate this amount of CH3COOH: 0.002975 mol = (0.0850 mol/L)(V2) V2 = 0.002975 mol / 0.0850 mol/L = 0.0350 L V2 = 35.0 mL It takes \(35.0 mL\) of \(0.0850 M\) NaOH to titrate \(35.0 mL\) of \(0.0850 M\) CH3COOH to the equivalence point.

Part (c) - Titration of 50.0 mL of a solution that contains 1.85 g HCl per liter

First, find the molarity of the HCl solution: Molarity = mass of solute / (molar mass of solute × volume of solvent) Molarity = (1.85 g/L) / (36.461 g/mol × 1.0 L) = 0.0507 mol/L Now, find the number of moles of HCl in \(50.0 mL\) of this solution: Number of moles of HCl = M1V1 = (0.0507 mol/L)(0.0500 L) = 0.002535 mol Next, we need to find the volume (V2) of NaOH required to titrate this amount of HCl: 0.002535 mol = (0.0850 mol/L)(V2) V2 = 0.002535 mol / 0.0850 mol/L = 0.0298 L V2 = 29.8 mL It takes \(29.8 mL\) of \(0.0850 M\) NaOH to titrate \(50.0 mL\) of a solution that contains \(1.85 g\) of HCl per liter to the equivalence point.

Key concepts

Equivalence Point

When you perform an acid-base titration, the equivalence point is a key concept that you should clearly understand. It is the moment in the titration process where the amount of acid is exactly equal to the amount of base added, resulting in a complete neutralization reaction. At this point, the moles of acid equal the moles of base according to their stoichiometric relationship in the balanced chemical equation. This does not necessarily mean the solution is neutral (pH 7), but rather that the reactants have reacted completely with each other.
The equivalence point can be determined using indicators, pH meters, or conductivity tests. Indicators change color at the equivalence point, allowing you to visually determine when the reaction is complete. Understanding this point is crucial for accurately calculating the concentrations and volumes needed in a titration process.

Molarity

Molarity is a way of expressing the concentration of a solution. It is defined as the number of moles of solute per liter of solution. The formula for molarity is given by \[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\]This concept is crucial in titration calculations as it helps you determine how much of a titrant is needed to reach the equivalence point.

• Moles of solute = molarity × volume of solution
• For example: If you have 40.0 mL of a 0.0900 M HNO3 solution, you calculate the moles by multiplying 0.0900 mol/L by 0.0400 L, resulting in 0.00360 moles of HNO3.

Molarity provides an essential link between solution concentration and the volume required to achieve a chemical reaction.

Stoichiometry

Stoichiometry deals with the quantitative relationships between reactants and products in a chemical reaction. In titration, it is essential for determining exactly how much titrant is needed to neutralize a given amount of analyte. The stoichiometry is derived from the balanced chemical equation, which states the mole ratios of reactants and products.

For example, in a reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl), the balanced equation is:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}\]This indicates a 1:1 mole ratio, meaning one mole of NaOH neutralizes one mole of HCl. Stoichiometry ensures that you use the correct proportions of reactants, leading to precise calculations of the volumes needed in the titration process.

Neutralization Reaction

A neutralization reaction is a chemical reaction in which an acid and a base react to form water and a salt. This type of reaction is at the heart of acid-base titrations, as the purpose is to determine the point at which the acid completely reacts with the base, known as the equivalence point.
These reactions are generally exothermic and can be represented as:
\[ \text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}\]In this process, the hydrogen ions from the acid react with the hydroxide ions from the base to form water, effectively "neutralizing" the effects of the acid and base. Observation methods such as pH change or color change with an indicator are typically used to signal the completion of this reaction. Understanding how the neutralization reaction occurs helps you predict and analyze the outcomes of titration experiments.

Chemical Calculations

Chemical calculations are vital in accurately determining the outcomes of titrations and other reactions. These calculations involve finding quantities like the number of moles, molarity, and the volume of solutions. In the context of titration, you often begin by determining the number of moles of the reactant using its volume and molarity.
Using the relationship:

• Moles = Molarity × Volume

You can calculate how much of a titrant you need to add to reach the equivalence point. Other calculations may involve converting units or using stoichiometry to find unknown concentrations or volumes based on known values. Mastery of chemical calculations ensures accurate and efficient results, making them an indispensable part of chemistry.