Chapter 4: Problem 80
What volume of a \(0.500 M \mathrm{HCl}\) solution is needed to neutralize each of the following: (a) \(10.0 \mathrm{~mL}\) of a \(0.300 \mathrm{M} \mathrm{NaOH}\) solution (b) \(10.0 \mathrm{~mL}\) of a \(0.200 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution
Short Answer
Expert verified
(a) 6.00 mL of 0.500 M HCl,\n(b) 8.00 mL of 0.500 M HCl
Step by step solution
01
Determine Molar Ratios
Use the chemical equation to determine the reaction ratio. In part (a) \(\mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H}_2\mathrm{O}\), and in part (b) \(\mathrm{2HCl} + \mathrm{Ba(OH)}_2 \rightarrow \mathrm{BaCl}_2 + 2\mathrm{H}_2\mathrm{O}\), these reactions show HCL and NaOH react in a 1:1 ratio and HCl reacts with Ba(OH)2 in a 2:1 ratio.
02
Calculate Moles of Base
In part (a), Use the formula: Molarity (M) = moles of solute/Liters of solution. Rearranging for moles gives: moles = Molarity (M) * Volume (L). Substituting the given values gives: moles of NaOH = 0.300 M * 0.010 L = 0.003 moles. In part (b), the calculation is similar: moles of Ba(OH)2 = 0.200 M * 0.010 L = 0.002 moles.
03
Calculate Volume of Acid Needed
In part (a), since the reaction occurs in a 1:1 ratio, equal moles of HCl are needed, so volume (L) = moles / Molarity. Substituting the given values gives: volume = 0.003 moles / 0.500 M = 0.006 L = 6.00 mL. In part (b), notice that the reaction occurs in a 2:1 ratio so double the moles of HCl are needed: volume = 2*0.002 moles / 0.500 M = 0.008 L = 8.00 mL.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molarity Calculation
Understanding molarity is essential when studying chemistry, especially in neutralization reactions where the accurate measurement of reactant concentration is critical. Molarity, which is represented as 'M', refers to the concentration of a solution and is defined as the number of moles of a solute per liter of solution. To calculate molarity, we use the formula: \[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]
When you're faced with a problem involving molarity, it's often about finding the volume of one solution that will react completely with a given volume of another solution. As seen in the exercise, to neutralize a specific amount of base (e.g., NaOH), you'll need to calculate the number of moles first by rearranging the molarity equation:
\[ \text{moles} = M \times \text{Volume (L)} \]
Once you have the moles of the base, you can then determine the volume of the acid needed for neutralization. It's a straightforward switch from moles back to volume using the molarity of the acid solution.
When you're faced with a problem involving molarity, it's often about finding the volume of one solution that will react completely with a given volume of another solution. As seen in the exercise, to neutralize a specific amount of base (e.g., NaOH), you'll need to calculate the number of moles first by rearranging the molarity equation:
\[ \text{moles} = M \times \text{Volume (L)} \]
Once you have the moles of the base, you can then determine the volume of the acid needed for neutralization. It's a straightforward switch from moles back to volume using the molarity of the acid solution.
Stoichiometry
The stoichiometry of a reaction is the part that deals with the quantitative relationships between the substances as they participate in chemical reactions. It's like a recipe, detailing the amounts of each reactant you need and what quantities of results you can expect. When it comes to neutralization reactions, it's critical to understand that acids and bases react according to specific molar ratios derived from their balanced chemical equations.
For example, the textbook exercise demonstrates two reactions. The balanced equation for the reaction of HCl with NaOH is: \[ \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H}_2\mathrm{O} \]
Here, they react in a 1:1 ratio. In contrast, the reaction of HCl with Ba(OH)2 is depicted as: \[ \mathrm{2HCl} + \mathrm{Ba(OH)}_2 \rightarrow \mathrm{BaCl}_2 + 2\mathrm{H}_2\mathrm{O} \]
which corresponds to a 2:1 ratio. Knowing this, you can calculate exactly how much of each reactant you'll need. The stoichiometric coefficients (the numbers before each substance in a balanced equation) are the keys to unlocking these calculations. They tell you, for instance, that two moles of HCl will react with one mole of Ba(OH)2, which is essential information for executing the molarity calculations.
For example, the textbook exercise demonstrates two reactions. The balanced equation for the reaction of HCl with NaOH is: \[ \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H}_2\mathrm{O} \]
Here, they react in a 1:1 ratio. In contrast, the reaction of HCl with Ba(OH)2 is depicted as: \[ \mathrm{2HCl} + \mathrm{Ba(OH)}_2 \rightarrow \mathrm{BaCl}_2 + 2\mathrm{H}_2\mathrm{O} \]
which corresponds to a 2:1 ratio. Knowing this, you can calculate exactly how much of each reactant you'll need. The stoichiometric coefficients (the numbers before each substance in a balanced equation) are the keys to unlocking these calculations. They tell you, for instance, that two moles of HCl will react with one mole of Ba(OH)2, which is essential information for executing the molarity calculations.
Acid-Base Titration
Acid-base titration is a process used to determine the concentration of an unknown acid or base solution by neutralizing it with a known concentration of the opposite solution. The point at which neutralization occurs is known as the equivalence point. Here, the number of hydrogen ions from the acid equals the number of hydroxide ions from the base, resulting in water and a salt.
In the exercise, we're implicitly performing an acid-base titration calculation. We determine the volume of HCl that is needed to neutralize a given volume of NaOH or Ba(OH)2, which effectively mimics the process of titration but without a practical experiment. Completing these calculations correctly means you've effectively 'titrated' the base with your theoretical acid solution. This fundamental process is widely used in laboratories to analyze the purity of chemicals, in environmental testing, and even in medicine.
Visual Indicators and pH Meters
Chemists often use a visual indicator or a pH meter to identify the equivalence point during a titration. Indicators change color at a certain pH level, while pH meters provide a digital reading.In the exercise, we're implicitly performing an acid-base titration calculation. We determine the volume of HCl that is needed to neutralize a given volume of NaOH or Ba(OH)2, which effectively mimics the process of titration but without a practical experiment. Completing these calculations correctly means you've effectively 'titrated' the base with your theoretical acid solution. This fundamental process is widely used in laboratories to analyze the purity of chemicals, in environmental testing, and even in medicine.