Problem 101 - Chemistry: An Atoms-Focused Approach

Question

How many milliliters of \(0.100 M\) NaOH are required to neutralize the following solutions? a. \(10.0 \mathrm{mL}\) of \(0.0500 M \mathrm{HCl}\) b. \(25.0 \mathrm{mL}\) of \(0.126 M \mathrm{HNO}_{3}\) c. \(50.0 \mathrm{mL}\) of \(0.215 M \mathrm{H}_{2} \mathrm{SO}_{4}\)

Step-by-Step Solution

Answer

Question: Calculate the required volume of \(0.100 M\) NaOH needed to neutralize each of the given solutions: a) \(10.0 mL\) of \(0.0500 M \mathrm{HCl}\), b) \(25.0 mL\) of \(0.126 M \mathrm{HNO}_{3}\), and c) \(50.0 mL\) of \(0.215 M \mathrm{H}_{2} \mathrm{SO}_{4}\). Answer: To neutralize each given solution, we require: a) 5.0 mL of \(0.100 M\) NaOH for \(10.0 mL\) of \(0.0500 M \mathrm{HCl}\), b) 31.5 mL of \(0.100 M\) NaOH for \(25.0 mL\) of \(0.126 M \mathrm{HNO}_{3}\), and c) 107.5 mL of \(0.100 M\) NaOH for \(50.0 mL\) of \(0.215 M \mathrm{H}_{2} \mathrm{SO}_{4}\).

Step 1. Calculate the moles of each Acid

For each given acid solution, we'll use the formula: moles = molarity × volume (in liters) a. For \(10.0 mL\) of \(0.0500 M \mathrm{HCl}\): moles(HCl) = \(0.0500 \frac{mol}{L} \times 0.010 L = 0.0005 mol\) b. For \(25.0 mL\) of \(0.126 M \mathrm{HNO}_{3}\): moles(HNO3) = \(0.126 \frac{mol}{L} \times 0.025 L = 0.00315 mol\) c. For \(50.0 mL\) of \(0.215 M \mathrm{H}_{2} \mathrm{SO}_{4}\): moles(H2SO4) = \(0.215 \frac{mol}{L} \times 0.050 L = 0.01075 mol\)

Step 2. Calculate the volume of NaOH

Now, we know the moles of each acid required to neutralize with NaOH. Using the molarity formula, we'll find the volume of NaOH required. volume = \(\frac{moles}{molarity}\) a. For \(HCl\): volume(NaOH) = \(\frac{0.0005 mol}{0.100 \frac{mol}{L}} = 0.005 L\) or 5.0 mL b. For \(HNO_{3}\): volume(NaOH) = \(\frac{0.00315 mol}{0.100 \frac{mol}{L}} = 0.0315 L\) or 31.5 mL c. For \(H_{2}SO_{4}\): volume(NaOH) = \(\frac{0.01075 mol}{0.100 \frac{mol}{L}} = 0.1075 L\) or 107.5 mL So, to neutralize each given solution, we require: a. 5.0 mL of \(0.100 M\) NaOH for \(10.0 mL\) of \(0.0500 M \mathrm{HCl}\) b. 31.5 mL of \(0.100 M\) NaOH for \(25.0 mL\) of \(0.126 M \mathrm{HNO}_{3}\) c. 107.5 mL of \(0.100 M\) NaOH for \(50.0 mL\) of \(0.215 M \mathrm{H}_{2} \mathrm{SO}_{4}\)

Key Concepts

Molarity

Molarity is a foundational concept in chemistry, and it helps us express the concentration of a solution. It tells us how many moles of solute are dissolved in one liter of solution. Typically, we denote molarity with the capital letter "M," for example, in the expression of a solution's concentration, such as 0.100 M NaOH. This indicates there are 0.100 moles of sodium hydroxide in every liter of the solution.

To calculate molarity, use this formula:

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

For practical applications, knowing how to use molarity is essential, especially in neutralization reactions, where precise amounts of each reactant are required. The example exercise uses molarity to determine how much sodium hydroxide (NaOH) solution is needed to completely neutralize various acid solutions.

By understanding molarity, you can predict and control how the solutions will react, ensuring the proper stoichiometric balance is maintained.

Acid-Base Titration

Acid-base titration is a process used to determine the concentration of an unknown acid or base solution. It involves the gradual addition of a titrant, which is a solution with a known concentration, to the unknown sample until the reaction reaches the neutralization point. This process is typically monitored using an indicator that changes color at the endpoint, signifying the neutralization.

The exercise in question involves using a known molarity of 0.100 M NaOH to neutralize different acidic solutions. The amount of NaOH needed will directly relate to the amount of acid present, showcasing a practical use of acid-base titration in laboratories.

• To find the endpoint in a titration, you observe a change in indicator color or measure the pH of the solution.
• This ensures the correct stoichiometric ratio is achieved, offering a precise method for determining unknown concentrations.

Understanding titration is crucial for quantitative chemical analysis and can be applied to various fields, including medical diagnostics, environmental testing, and industrial processes.

Stoichiometry

Stoichiometry is the aspect of chemistry that involves the calculation of reactants and products in chemical reactions. It is based on the principle that matter is conserved in chemical processes, allowing chemists to predict quantities needed or produced.

In the context of neutralization reactions, stoichiometry allows for the calculation of the precise amounts of acid and base that will completely react. This requires using balanced chemical equations to understand the relationship between reactants and products. For example, when NaOH neutralizes HCl:

• The balanced equation is: \( \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \)
• This equation shows a 1:1 molar ratio between NaOH and HCl.

Applying stoichiometry, we see that NaOH and HCl react perfectly when they are in equal molar quantities. For reactions involving polyprotic acids like \( \text{H}_2\text{SO}_4 \), stoichiometry indicates a different ratio, as illustrated by the exercise where twice the amount of NaOH is needed to fully neutralize sulfuric acid compared to hydrochloric acid.

Mastering stoichiometry is vital for successful chemical analysis and ensures accurate preparation of solutions in experimental settings.