Problem 69 - Chemistry: The Central Science

Question

(a) Which will have the highest concentration of potassium ion: \(0.20 \mathrm{M} \mathrm{KCl}, 0.15 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\), or \(0.080 \mathrm{M} \mathrm{K}_{3} \mathrm{PO}_{4} ?\) (b) Which will contain the greater number of moles of potassium ion: \(30.0 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr} \mathrm{O}_{4}\) or \(25.0 \mathrm{~mL}\) of \(0.080 \mathrm{M} \mathrm{K}_{3} \mathrm{PO}_{4} ?\)

Step-by-Step Solution

Answer

(a) The \(0.15\,M\,K_2CrO_4\) solution has the highest concentration of potassium ions with a concentration of \(0.30\,M\). (b) The \(30.0\,mL\) of \(0.15\,M\,K_2CrO_4\) will contain a greater number of moles of potassium ions with \(0.009\,mol\).

Step 1: Determine the concentration of potassium ions in each solution

To find the concentration of potassium ions in each solution, multiply the given concentrations by the number of potassium ions in each compound. For the \(0.20\,M\, KCl\) solution: In 1 mole of \(KCl\), we have 1 mole of \(K^+\) ions. Concentration of potassium ions in \(0.20\,M\,KCl\) solution is 1 × \(0.20\,M = 0.20\,M\). For the \(0.15\,M\,K_2CrO_4\) solution: In 1 mole of \(K_2CrO_4\), we have 2 moles of \(K^+\) ions. Concentration of potassium ions in \(0.15\,M\,K_2CrO_4\) solution is 2 × \(0.15\,M = 0.30\,M\). For the \(0.080\,M\,K_3PO_4\) solution: In 1 mole of \(K_3PO_4\), we have 3 moles of \(K^+\) ions. Concentration of potassium ions in \(0.080\,M\,K_3PO_4\) solution is 3 × \(0.080\,M = 0.24\,M\).

Step 2: Compare potassium ion concentrations

Now, let's compare the concentrations of potassium ions in the three solutions: \(0.20\,M (KCl) < 0.24\,M (K_3PO_4) < 0.30\,M (K_2CrO_4)\). (a) The \(0.15\,M\,K_2CrO_4\) solution has the highest concentration of potassium ions with a concentration of \(0.30\,M\).

Step 3: Calculate the number of moles of potassium ions in the given volumes

Now, let's find the number of the moles of potassium ions in specific volumes. For \(30.0\,mL\) of the \(0.15\,M\,K_2CrO_4\) solution: Number of moles = (Volume × Concentration) × Number of potassium ions in the compound. Number of moles = \((30.0\,mL × 0.15\,M) × 2\) Since \(1\,L = 1000\,mL\), then \(30.0\,mL = 0.030\,L\) Number of moles = \((0.030\,L × 0.15\,M) × 2 = 0.009\,mol\). For \(25.0\,mL\) of the \(0.080\,M\,K_3PO_4\) solution: Number of moles = (Volume × Concentration) × Number of potassium ions in the compound. Number of moles = \((25.0\,mL × 0.080\,M) × 3\) Since \(1\,L = 1000\,mL\), then \(25.0\,mL = 0.025\,L\) Number of moles = \((0.025\,L × 0.080\,M) × 3 = 0.006\,mol\).

Step 4: Compare the number of moles of potassium ions

Now, let's compare the number of moles of potassium ions in the two solutions: \(0.006\,mol (K_3PO_4) < 0.009\,mol (K_2CrO_4)\). (b) The \(30.0\,mL\) of \(0.15\,M\,K_2CrO_4\) will contain a greater number of moles of potassium ions with \(0.009\,mol\).

Key Concepts

Molarity Calculation

Molarity is a measure of concentration used in chemistry to indicate the number of moles of solute per liter of solution. It's represented by the capital letter 'M' and is sometimes referred to as molar concentration. When we talk about a 0.20 M solution of KCl, it means there is 0.20 moles of potassium chloride dissolved in one liter of solution.

To calculate molarity, you can use the equation: \( M = \frac{moles\text{ of solute}}{liters\text{ of solution}} \).

For ions in a solution, the molarity can be directly related to the moles of ions present in one liter. For example, in a 0.20 M KCl solution, since there is one potassium ion for every formula unit of KCl, the concentration of potassium ions is also 0.20 M. Understanding the molarity of different ion solutions is crucial when comparing their concentrations, as seen in the exercise provided.

Stoichiometry

Stoichiometry is the part of chemistry that deals with the relationships between the quantities of reactants and products in chemical reactions. In a stoichiometric calculation, we use the coefficients from the balanced chemical equation to quantify these relationships. Stoichiometry can also be applied when figuring out the moles of ions present in a solution based on the formula of the compound dissolved.

For instance, in the compound \( K_2CrO_4 \), there are two potassium ions (\( K^+ \)) for each formula unit. This stoichiometric relationship allows us to determine that a 0.15 M solution of \( K_2CrO_4 \) actually contains a potassium ion molarity of 0.30 M, as each formula unit contributes two \( K^+ \) ions. This application of stoichiometry is critical for correctly assessing the concentration of ions in solution.

Moles of Ions in Solution

Moles represent a fundamental unit in chemistry denoting a specific quantity of particles, whether they're atoms, molecules or ions. When a compound dissociates in water, it produces a discrete number of ions. The total moles of ions in solution are directly related to the total moles of the compound dissolved, adjusted for the number of ions each compound yields upon dissolution.

For example, when \( K_3PO_4 \) is dissolved in water, each formula unit will produce three potassium ions. To find the total moles of potassium ions in a given volume of solution, you would perform the following calculation: (volume in liters) × (molarity) × (number of ions per formula unit). So, for a volume of 25.0 mL (or 0.025 L) of 0.080 M \( K_3PO_4 \) solution, the moles of potassium ions would be \( 0.025\text{ L} × 0.080\text{ M} × 3 = 0.006\text{ mol} \).The exercise provided illustrates how to use this information to compare the moles of potassium ions in different solutions, which is a practical skill in various chemical analysis and reaction prediction scenarios.