Problem 42 - Chemistry The Molecular Science

Question

Calculate the volume of 0.225 -M \(\mathrm{NaOH}\) required to titrate to the equivalence point for each of these samples. (a) \(20.0 \mathrm{~mL}\) of \(0.315-\mathrm{M} \mathrm{HBr}\) (b) \(30.0 \mathrm{~mL}\) of \(0.250-\mathrm{M} \mathrm{HClO}_{4}\) (c) \(6.00 \mathrm{~g}\) of concentrated acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH},\) which is \(99.7 \%\) pure

Step-by-Step Solution

Answer

(a) 28 mL, (b) 33.3 mL, (c) 442.6 mL.

Step 1: Understand the Reaction

For each titration problem, we need to recognize that HBr, HClO₄, and CH₃COOH are acids that react with NaOH, a base, in a neutralization reaction. The general reaction is: \[ \mathrm{Acid} + \mathrm{NaOH} \rightarrow \mathrm{Salt} + \mathrm{Water} \]. For each mole of HBr, HClO₄, and CH₃COOH, one mole of NaOH is required.

Step 2: Calculate Moles of HBr (a)

First, calculate the moles of HBr: \[ \text{moles of HBr} = 0.315 \text{ M} \times 20.0 \text{ mL} \]. Convert mL to L by dividing by 1000: \[ \text{moles of HBr} = 0.315 \times 0.020 \text{ L} = 0.0063 \text{ moles} \]

Step 3: Calculate Volume of NaOH Required (a)

Use the stoichiometry of the reaction (1:1 ratio with NaOH). \[ \text{Volume of NaOH} = \frac{\text{moles of HBr}}{\text{Molarity of NaOH}} = \frac{0.0063}{0.225} = 0.028 \text{ L} = 28 \text{ mL} \]

Step 4: Calculate Moles of HClO₄ (b)

Next, calculate the moles of HClO₄: \[ \text{moles of HClO₄} = 0.250 \text{ M} \times 30.0 \text{ mL} \]. Convert mL to L: \[ \text{moles of HClO₄} = 0.250 \times 0.030 \text{ L} = 0.0075 \text{ moles} \]

Step 5: Calculate Volume of NaOH Required (b)

Similarly, calculate the volume of NaOH needed: \[ \text{Volume of NaOH} = \frac{\text{moles of HClO₄}}{\text{Molarity of NaOH}} = \frac{0.0075}{0.225} = 0.0333 \text{ L} = 33.3 \text{ mL} \]

Step 6: Calculate Moles of CH₃COOH (c)

For CH₃COOH, calculate the number of moles: First, find the pure mass of acetic acid: \[ 6.00 \text{ g} \times 0.997 = 5.982 \text{ g} \]. Then use the molar mass of CH₃COOH (60.05 g/mol): \[ \text{moles of CH₃COOH} = \frac{5.982}{60.05} = 0.0996 \text{ moles} \]

Step 7: Calculate Volume of NaOH Required (c)

Finally, calculate the volume of NaOH needed: \[ \text{Volume of NaOH} = \frac{0.0996}{0.225} = 0.4426 \text{ L} = 442.6 \text{ mL} \]

Key Concepts

Molarity Calculations

Understanding molarity is crucial in solving titration problems. Molarity (M) is defined as the number of moles of solute per liter of solution. In simple terms, it tells you how concentrated a solution is. To calculate molarity, use the formula: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]In a titration problem, knowing the molarity of the solutions allows you to figure out how much of one solution is needed to react completely with a given volume of another solution. Always remember to convert volume from milliliters to liters by dividing by 1000, as molarity is expressed in terms of liters.

Stoichiometry

Stoichiometry is a cornerstone in chemistry that involves the calculation of reactants and products in chemical reactions. It is based on the balanced chemical equation and the concept of moles. When it comes to titration, stoichiometry helps us determine how much of each reactant is needed to reach the equivalence point, where the acid fully reacts with the base. For instance, if the reaction is between HBr and NaOH, both of which react in a 1:1 ratio, the moles of HBr will equal the moles of NaOH at the equivalence point. Thus, understanding stoichiometric relationships enables you to compute the required volume of titrant (e.g., NaOH) for neutralization.

Acid-Base Reactions

Acid-base reactions are a type of neutralization reaction where an acid and a base react to form water and a salt. This fundamental chemistry concept is key to solving titration problems. The main equation describing these reactions is:\[ \text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water} \]In the context of titration, you need to identify which compounds are the acid and the base. For example, in the exercise you are dealing with acids like HBr, HClO₄, and CH₃COOH reacting with NaOH, which is a strong base. The goal of a titration is to find the point where the number of moles of acid equals the number of moles of base, reaching what we call the equivalence point. At this point, the acidity and basicity are perfectly balanced, resulting in a neutral solution.