Problem 39 - Chemistry: The Central Science

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 con- tains \(1.85 \mathrm{~g}\) of \(\mathrm{HCl}\) per liter?

Step-by-Step Solution

Answer

The required volume of \(0.0850\mathrm{M}\mathrm{NaOH}\) to titrate each solution to the equivalence point are: (a) \(42.4\mathrm{~mL}\) for the \(40.0\mathrm{~mL}\) of \(0.0900\mathrm{M}\mathrm{HNO}_{3}\) solution. (b) \(35.0\mathrm{~mL}\) for the \(35.0\mathrm{~mL}\) of \(0.0850\mathrm{M}\mathrm{CH}_{3}\mathrm{COOH}\) solution. (c) \(29.9\mathrm{~mL}\) for the \(50.0\mathrm{~mL}\) of a solution that contains \(1.85\mathrm{~g}\) of \(\mathrm{HCl}\) per liter.

Step 1: (a) Calculate moles of HNO3

First, we need to calculate the moles of \(\mathrm{HNO}_{3}\), using the formula: moles = concentration \(\times\) volume. In our case: moles of \(\mathrm{HNO}_{3} = 0.0900\mathrm{M} \times 40.0\mathrm{~mL} \times \frac{1\mathrm{~L}}{1000\mathrm{~mL}} = 3.60 \times 10^{-3} \mathrm{~mol}\)

Step 2: (a) Calculate volume of NaOH

Now we need to find the volume of \(\mathrm{NaOH}\). We use the formula moles of acid = moles of base or \(C_{acid} \times V_{acid} = C_{base} \times V_{base}\): \(3.60 \times 10^{-3} \mathrm{~mol} = 0.0850\mathrm{M} \times V_{NaOH}\) \(V_{NaOH} = \frac{3.60 \times 10^{-3} \mathrm{~mol}}{0.0850\mathrm{M}} \approx 42.4\mathrm{~mL}\)

Step 3: (b) Calculate moles of CH3COOH

Next, we need to calculate the moles of \(\mathrm{CH}_{3}\mathrm{COOH}\), using the formula: moles = concentration \(\times\) volume: moles of \(\mathrm{CH}_{3}\mathrm{COOH} = 0.0850\mathrm{M} \times 35.0\mathrm{~mL} \times \frac{1\mathrm{~L}}{1000\mathrm{~mL}} = 2.975 \times 10^{-3} \mathrm{~mol}\)

Step 4: (b) Calculate volume of NaOH

Now we need to find the volume of \(\mathrm{NaOH}\) for the second solution: \(2.975 \times 10^{-3} \mathrm{~mol} = 0.0850\mathrm{M} \times V_{NaOH}\) \(V_{NaOH} = \frac{2.975 \times 10^{-3} \mathrm{~mol}}{0.0850\mathrm{M}} \approx 35.0\mathrm{~mL}\)

Step 5: (c) Calculate moles of HCl

First, we need to convert the mass of \(\mathrm{HCl}\) to moles. The molar mass of \(\mathrm{HCl}\) is approximately \(36.46\mathrm{~g/mol}\): moles of \(\mathrm{HCl} = \frac{1.85\mathrm{~g/L} \times 50.0\mathrm{~mL} \times \frac{1\mathrm{~L}}{1000\mathrm{~mL}}}{36.46\mathrm{~g/mol}} \approx 2.54 \times 10^{-3} \mathrm{~mol}\)

Step 6: (c) Calculate volume of NaOH

Now we need to find the volume of \(\mathrm{NaOH}\) required to titrate the third solution: \(2.54 \times 10^{-3} \mathrm{~mol} = 0.0850\mathrm{M} \times V_{NaOH}\) \(V_{NaOH} = \frac{2.54 \times 10^{-3} \mathrm{~mol}}{0.0850\mathrm{M}} \approx 29.9\mathrm{~mL}\) In conclusion, the required volume of \(0.0850\mathrm{M}\mathrm{NaOH}\) to titrate each solution to the equivalence point are: (a) \(42.4\mathrm{~mL}\) for the \(40.0\mathrm{~mL}\) of \(0.0900\mathrm{M}\mathrm{HNO}_{3}\) solution. (b) \(35.0\mathrm{~mL}\) for the \(35.0\mathrm{~mL}\) of \(0.0850\mathrm{M}\mathrm{CH}_{3}\mathrm{COOH}\) solution. (c) \(29.9\mathrm{~mL}\) for the \(50.0\mathrm{~mL}\) of a solution that contains \(1.85\mathrm{~g}\) of \(\mathrm{HCl}\) per liter.

Key Concepts

Equivalence Point

In the field of chemistry, specifically during a titration process, the concept of an equivalence point is a fundamental one. It refers to the juncture in the titration where the amount of titrant added is enough to exactly neutralize the analyte solution. This moment is crucial because it indicates that the moles of acid and base in the solution are equal to each other.

When conducting a titration, such as in the exercise given where sodium hydroxide (NaOH) is used to titrate different acids, achieving the equivalence point is the goal. It's important to note, however, that the equivalence point is different from the endpoint. The endpoint is the indicator's color change, which should be chosen to match as closely as possible to the pH at the equivalence point, but in practice, it might not be perfectly aligned.

Finding the equivalence point plays an essential role in determining the concentration of an unknown solution using a standard solution, as seen in the example provided, where it helps to find the required volume of NaOH to neutralize different acid solutions.

Molarity

The concept of molarity is a standard unit of concentration in chemistry. It is expressed as moles of solute per liter of solution, denoted by the unit M (molar). This measurement is pivotal when describing the concentration of solutions in chemical reactions, including titrations.

In the exercise, molarity comes into play when determining the concentration of the NaOH solution (0.0850 M) used for titration and the concentration of the acid solutions. To calculate the amount of NaOH needed to reach the equivalence point, we used the molarity of the acid to find the number of moles of acid present, which must equal the number of moles of base added. Understanding molarity allows students to grasp changes in solution concentrations and the calculations involved in determining how much of one substance is required to react with a given amount of another substance.

Stoichiometry

Stoichiometry is a sector of chemistry that involves the quantitative relationships between the reactants and products in chemical reactions. It draws upon the principles of the conservation of mass and the laws of chemical proportion to predict the quantities of substances consumed and produced in chemical reactions.

The exercise showcases stoichiometry in practice. It requires a balance between the moles of NaOH used and the moles of the acid in solution, based on their respective balanced chemical equations. This balance reflects stoichiometry's core principle, which is that substances react in precise ratios determined by their coefficients in the balanced equation. In simple acid-base reactions, such as those in the given problems, these ratios are often one-to-one, but in more complex reactions, stoichiometry becomes indispensable to comprehend these ratios and to calculate the amounts of each reactant required.

Acid-Base Reactions

Acid-base reactions are a class of chemical reactions that involve the transfer of protons between a donor (acid) and an acceptor (base). They are one of the most fundamental types of reactions in chemistry. Acids are substances that can donate a proton (H+ ion), and bases are substances that can accept a proton.

Throughout the exercise, we are dealing with acid-base reactions when titrating a strong base (NaOH) with different acids (HNO3 and CH3COOH) to find the equivalence point. Acid-base titrations are a practical application of the theoretical concepts of acids and bases and often use a pH indicator or pH meter to determine when the reaction has reached the equivalence point. Understanding how acids and bases interact in different concentrations, as well as the properties that categorize them, is vital for interpreting the outcomes and calculations in such titrations.