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 \(0.0900 \mathrm{M} \mathrm{HNO}_{3}\), (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 HCl per liter?

Short answer

The required volumes of 0.0850 M NaOH to titrate each solution to the equivalence point are: (a) 42.4 mL for 40.0 mL of 0.0900 M HNO3, (b) 35.0 mL for 35.0 mL of 0.0850 M CH3COOH, (c) 29.8 mL for 50.0 mL of a solution that contains 1.85 g of HCl per liter.

Step-by-step solution

(a) Moles of HNO3

To calculate the moles of HNO3, we will use the given volume (40 mL) and concentration (0.0900 M): Moles of HNO3 = Volume × Concentration Moles of HNO3 = \(40.0 \times 10^{-3} L\) × 0.0900 M = 0.0036 mol

(b) Moles of CH3COOH

To calculate the moles of CH3COOH, we will use the given volume (35 mL) and concentration (0.0850 M): Moles of CH3COOH = Volume × Concentration Moles of CH3COOH = \(35.0 \times 10^{-3} L\) × 0.0850 M = 0.002975 mol

(c) Moles of HCl

First, calculate the mass of HCl in the 50.0 mL solution: Mass of HCl = \(1.85 \frac{g}{L}\) × \(50.0 \times 10^{-3} L\) = 0.0925 g Next, convert the mass of HCl to moles by dividing by its molar mass (36.461 g/mol): Moles of HCl = \(\frac{0.0925}{36.461}\) = 0.002536 mol ##Step 2: Determine the volume of NaOH required## Since the equivalence point is when the moles of acid equal the moles of the base, we can use the given concentration of NaOH to find the required volume for each scenario.

(a) Volume of NaOH for HNO3 titration

Using the moles of HNO3 calculated previously, we can determine the required volume of NaOH: Volume = \(\frac{Moles}{Concentration}\) Volume = \(\frac{0.0036}{0.0850}\) L = 0.0424 L = 42.4 mL

(b) Volume of NaOH for CH3COOH titration

Using the moles of CH3COOH calculated previously, we can determine the required volume of NaOH: Volume = \(\frac{Moles}{Concentration}\) Volume = \(\frac{0.002975}{0.0850}\) L = 0.0350 L = 35.0 mL

(c) Volume of NaOH for HCl titration

Using the moles of HCl calculated previously, we can determine the required volume of NaOH: Volume = \(\frac{Moles}{Concentration}\) Volume = \(\frac{0.002536}{0.0850}\) L = 0.0298 L = 29.8 mL So the required volumes of 0.0850 M NaOH to titrate each solution to the equivalence point are: (a) 42.4 mL for 40.0 mL of 0.0900 M HNO3, (b) 35.0 mL for 35.0 mL of 0.0850 M CH3COOH, (c) 29.8 mL for 50.0 mL of a solution that contains 1.85 g of HCl per liter.

Key concepts

Equivalence Point

In a titration, the equivalence point is a critical concept. It is the stage where the amount of titrant added equals the amount of substance in the analyte that it is reacting with. Basically, the moles of the acid will match the moles of the base during an acid-base reaction.

When the equivalence point is reached, the reaction between the acid and base is complete. It's crucial to accurately determine this point to calculate the amount of unknown concentration in solutions. Notably, the equivalence point is different from the endpoint, which is when a color change is observed due to an indicator reaching its endpoint.

The equivalence point is typically determined using pH meters or titration indicators. These tools help ensure precision, ensuring the titration results accurately reflect the solution's chemistry.

Molarity

Molarity, a central concept in chemistry, refers to the concentration of a solution expressed as the number of moles of solute per liter of solution. It is a commonly used unit for indicating the concentration of chemical solutions. The formula to calculate molarity is:

• Molarity () = \(\frac{\text{moles of solute}}{\text{liters of solution}}\)

Molarity provides vital information about the strength of a solution. For example, when preparing solutions for titration, it helps chemists understand how much of a reagent is needed to reach the equivalence point.

In titration problems, molarity helps in calculating the moles of a particular substance present in a given volume. With this, chemists can perform stoichiometric calculations and ensure reactions are proceeding with the correct amounts of reactants. Knowing the molarity allows precise control over reactions, ensuring accuracy in experimental results.

Acid-Base Reaction

Acid-base reactions are fundamental chemical processes commonly explored in titrations. Such reactions involve an acid reacting with a base to produce water and sometimes a salt. Understanding the nature of acids and bases is crucial for predicting the outcome of these reactions.

Acids donate protons (H⁺ ions) whereas bases accept protons. This transfer of protons from the acid to the base is what characterizes acid-base reactions. An example in the provided exercise could involve hydrochloric acid (HCl) donating protons to sodium hydroxide (NaOH), a base.

Knowing the molarities of the reactants allows you to calculate the volume of base needed to neutralize the given amount of acid, reaching the equivalence point. During such reactions, observing pH changes can help identify when equivalent amounts of acid and base have reacted completely, supporting precise titration.

Stoichiometry

Stoichiometry is the section of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It's a tool that allows chemists to predict how much of each reactant is needed or will be produced in a reaction.

In titrations, stoichiometry plays a significant role. Using the balanced equation of the chemical reaction, stoichiometry helps calculate the exact amounts of reactants required to reach the equivalence point. By understanding stoichiometry, chemists ensure that the proportion of reactants leads to complete reactions without excess or deficit.

For example, in the titration of NaOH and HCl, stoichiometry shows that one mole of NaOH will neutralize one mole of HCl. This 1:1 molar ratio ensures the calculation of how much titrant is needed for a given amount of analyte. Accurate stoichiometric calculations are essential for preparing solutions and conducting experiments that yield valid results.