Problem 199 - Introductory Chemistry Atoms First

Question

A student combines \(60.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{NaOH}\) with \(60.0 \mathrm{~mL}\) of \(0.125 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\). What is the hydroxide ion molar concentration in the resulting solution?

Step-by-Step Solution

Answer

The hydroxide ion molar concentration in the resulting solution when the student combines 60.0 mL of 0.250 M NaOH with 60.0 mL of 0.125 M Ba(OH)₂ is \(0.250 M\).

Step 1: Calculate moles of hydroxide ions from NaOH solution

We first calculate the moles of OH⁻ ions coming from NaOH solution. Molarity = moles of solute / volume of solution moles of hydroxide ions = Molarity x Volume moles of hydroxide ions in NaOH solution = \(0.250 M \times 60.0 mL\) moles of hydroxide ions in NaOH solution = \(0.250 \times 0.060 L\) moles of hydroxide ions in NaOH solution = 0.015 mol

Step 2: Calculate moles of hydroxide ions from Ba(OH)₂ solution

Now, we will calculate the moles of hydroxide ions from Ba(OH)₂ solution. Keep in mind that one molecule of Ba(OH)₂ contains two hydroxide ions. moles of hydroxide ions in Ba(OH)₂ solution = Molarity x Volume moles of hydroxide ions in Ba(OH)₂ solution = \(0.125 M \times 60.0 mL \times 2\) (2 is multiplied because Ba(OH)₂ has two OH⁻ ions.) moles of hydroxide ions in Ba(OH)₂ solution = \(0.125 \times 0.060 L \times 2\) moles of hydroxide ions in Ba(OH)₂ solution = 0.015 mol

Step 3: Calculate total moles of hydroxide ions in the mixture

Now we need to find the total moles of hydroxide ions in the resulting solution. Total moles of hydroxide ions = moles of hydroxide ions in NaOH solution + moles of hydroxide ions in Ba(OH)₂ solution Total moles of hydroxide ions = 0.015 mol + 0.015 mol Total moles of hydroxide ions = 0.030 mol

Step 4: Calculate hydroxide ion molar concentration in the mixture

Now we will find the hydroxide ion molar concentration in the resulting solution. The total volume of the mixture = volume of NaOH solution + volume of Ba(OH)₂ solution Total volume of the mixture = 60.0 mL + 60.0 mL Total volume of the mixture = 120.0 mL = 0.120 L The hydroxide ion molar concentration in the resulting solution = total moles of hydroxide ions / total volume of the mixture The hydroxide ion molar concentration in the resulting solution = \( \frac{0.030 \text{mol}}{0.120 \text{L}} \) The hydroxide ion molar concentration in the resulting solution = 0.250 M The hydroxide ion molar concentration in the resulting solution is \(0.250 M\).

Key Concepts

Molarity Calculation

Molarity is a fundamental concept in chemistry used to express the concentration of a solution. It represents the number of moles of solute per liter of solution, and its units are moles per liter (mol/L or M). The formula to calculate molarity is simple:

\[\begin{equation}Molarity = \frac{moles \ of \ solute}{volume \ of \ solution \ (in \ liters)}\end{equation}\]

In the provided exercise, molarity is used to determine the amount of hydroxide ions in a solution. By multiplying the molarity of the NaOH and Ba(OH)₂solutions by their respective volumes (in liters), students can find the moles of OH⁻ provided by each solution. It's crucial to pay attention to the stoichiometry of the compounds involved. For example, since each Ba(OH)₂ molecule gives two hydroxide ions, the calculated moles must be doubled for this solute.

Understanding this principle ensures accurate calculation of solutes in various chemical solutions.

Moles of Solute

Moles of solute refer to the amount of a substance present in a solution. The mole is the standard unit of amount in chemistry and represents approximately 6.022 x 10²³ entities (Avogadro's number). It is useful for counting atoms, ions, or molecules in a sample.

To calculate the moles of solute, you can use the formula:

\[\begin{equation}Moles \ of \ solute = Molarity \times Volume \ (in \ liters)\end{equation}\]

This calculation is integral in the exercise as it helps quantify the hydroxide ion content from each substance. In step 1, the moles of hydroxide from NaOH are found, and in step 2, the moles from Ba(OH)₂ are calculated, keeping in mind that Ba(OH)₂ contributes twice the amount of OH⁻. Knowing the precise moles of solute is a core skill, especially for more complex reactions in academic and industrial laboratories.

Chemical Solution Concentration

Chemical solution concentration illustrates how much of a substance— the solute—is present within a solvent. Molarity is one method of describing this concentration, and it's crucial for preparing solutions in scientific studies.

Concentration can impact reaction rates, equilibrium states, and the outcomes of titrations. In our particular case, the hydroxide ion concentration is pivotal because it influences the pH and reactivity of the solution. Throughout chemistry, understanding and adjusting concentrations is key to controlling reactions and achieving desired results. Students need to be comfortable with calculating and interpreting molarity because it is extensively used in many areas of chemistry, ranging from basic experiments to pharmacology and environmental science.

Acid-base Titration

Acid-base titration is a laboratory method used to determine the concentration of an unknown acid or base solution by neutralizing it with a standard solution of known concentration. The point at which the reaction is complete is the equivalence point, typically indicated by a color change of an indicator or by a pH meter reading.

During a titration, the molarity of the titrant (the known solution) helps to find the molarity of the analyte (the unknown solution). The relationship between the quantities of acid and base can be established by a balanced chemical equation, and stoichiometry can then be applied to find the concentration of the unknown solution.

Understanding the molar concentration of hydroxide ions, as in our example, is important in titrations since it aids in predicting the amount of acid needed to reach the equivalence point. Acid-base titrations are not only fundamental in educational labs but also play a significant role in various industrial and medical applications.