Problem 41 - Chemistry The Central Science

Question

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

Step-by-Step Solution

Answer

The required volumes of 0.0850 M NaOH to titrate each acidic solution to the equivalence point are: (a) 42.35 mL for 0.0900 M HNO3, (b) 35.00 mL for 0.0850 M CH3COOH, and (c) 29.86 mL for the HCl solution containing 1.85 g of HCl per liter.

Step 1: Calculate moles of acidic solutions

For each acidic solution, we need to find the moles of the solute. We will use the formula: Moles = Molarity × Volume The volume should be in liters for it to be consistent with the unit of molarity. (a) For 40.0 mL of 0.0900 M HNO3, calculate moles of HNO3: Moles of HNO3 = 0.0900 M × (40.0 mL × (1 L / 1000 mL)) Moles of HNO3 = 0.00360 mol (b) For 35.0 mL of 0.0850 M CH3COOH, calculate moles of CH3COOH: Moles of CH3COOH = 0.0850 M × (35.0 mL × (1 L / 1000 mL)) Moles of CH3COOH = 0.002975 mol (c) For 50.0 mL of HCl containing 1.85 g of HCl per liter, first calculate the molarity of HCl: Molarity of HCl = (1.85 g / (36.461 g/mol)) / 1 L Molarity of HCl = 0.05076 M Now, calculate moles of HCl: Moles of HCl = 0.05076 M × (50.0 mL × (1 L / 1000 mL)) Moles of HCl = 0.002538 mol

Step 2: Neutralization reaction with NaOH

Now, we will write down the balanced chemical equations for the titrations with NaOH: (a) HNO3 + NaOH → NaNO3 + H2O (b) CH3COOH + NaOH → CH3COONa + H2O (c) HCl + NaOH → NaCl + H2O In each case, the stoichiometric coefficients for both the acid and the base are equal to 1. This means that the number of moles of NaOH required to neutralize the acid completely is equal to the number of moles of the acid.

Step 3: Calculate the volume of NaOH required

Now, we will calculate the volume of 0.0850 M NaOH required to neutralize the acid by using the formula: Volume = Moles / Molarity (a) For 0.00360 mol of HNO3: Volume of NaOH = 0.00360 mol / 0.0850 M Volume of NaOH = 0.04235 L (42.35 mL) (b) For 0.002975 mol of CH3COOH: Volume of NaOH = 0.002975 mol / 0.0850 M Volume of NaOH = 0.03500 L (35.00 mL) (c) For 0.002538 mol of HCl: Volume of NaOH = 0.002538 mol / 0.0850 M Volume of NaOH = 0.02986 L (29.86 mL) So, the required volumes of 0.0850 M NaOH to titrate each acidic solution to the equivalence point are: (a) 42.35 mL for 0.0900 M HNO3 (b) 35.00 mL for 0.0850 M CH3COOH (c) 29.86 mL for the HCl solution containing 1.85 g of HCl per liter

Key Concepts

Molarity

Molarity is an essential concept in chemistry that helps us understand how concentrated a solution is. In simple terms, molarity is the number of moles of a solute present in one liter of solution. We often express it in units of moles per liter (M). The formula for calculating molarity is:

\[ \text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \]

In the context of acid-base titration problems, knowing the molarity of solutions is crucial. It allows us to determine how much of a titrant (the solution added, like NaOH) is needed to completely react with the acid in the solution being titrated. For instance, in our problem, we used the given molarity of the acid solutions to find out the amount of NaOH required to reach the equivalence point. This involves converting the volume of the solution to liters, multiplying by molarity, and finding the moles of acid in the solution.

• Always ensure the volume is in liters when using the molarity formula. For example, convert 40.0 mL to 0.0400 L.
• To find moles from molarity, multiply the molarity by the volume in liters.

Neutralization

Neutralization is the process in a chemical reaction where an acid and a base react to form water and a salt. This is a key step in titration, where the goal is to reach a point where the acid and base perfectly neutralize each other, known as the equivalence point. At this point, the moles of acid are equivalent to the moles of the base in the reaction.

In acid-base titrations like the one we are considering, three main reactions occur:

• \( \text{HNO}_3 + \text{NaOH} \rightarrow \text{NaNO}_3 + \text{H}_2\text{O} \)
• \( \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \)
• \( \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \)

Each reaction illustrates the one-to-one stoichiometric relationship between the acid and the base, meaning the moles of sodium hydroxide needed will be exactly the same as the moles of the acid present in each solution.

Chemical Equations

Chemical equations give us a balanced representation of a chemical reaction, showing the reactants on the left, the products on the right, and the stoichiometry of each component involved. In acid-base reactions, these equations help us visualize and calculate the proportion of each substance required or produced.

The balanced equations in our titration example are simple and straightforward, each having a 1:1 molar ratio between the acid and the base (NaOH):

• \( \text{HNO}_3 + \text{NaOH} \rightarrow \text{NaNO}_3 + \text{H}_2\text{O} \)
• \( \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \)
• \( \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \)

This balanced form is crucial because it allows the direct use of moles calculated from given molarity to find out how much of the other reactant is needed or produced. When a solution is titrated, we reach the equivalence point exactly when the solution has been neutralized, aided by our understanding of these chemical equations.

Remembering that these reactions are balanced helps in performing accurate calculations by directly equating moles of acids to moles of bases, leading to precise results in titration procedures.