Problem 8 - General Chemistry: Principles and Modern Applications

Question

Which solution has the greatest \(\left[\mathrm{SO}_{4}^{2-}\right]:\) (a) \(0.075 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4} ; \quad\) (b) \(\quad 0.22 \mathrm{M} \mathrm{MgSO}_{4} ; \quad\) (c) \(\quad 0.15 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) (d) \(0.080 \mathrm{M} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} ;\) (e) \(0.20 \mathrm{M} \mathrm{CuSO}_{4} ?\)

Step-by-Step Solution

Answer

The solution with the greatest \(\left[\mathrm{SO}_{4}^{2-}\right]\) is (d) \(0.080 \mathrm{M} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), with a concentration of 0.24 M.

Step 1: Break down each compound

Each compound breaks down differently to form sulphate ions. For (a) \(H_2SO_4\) breaks down to form 1 sulphate ion, (b) \(MgSO_4\) forms 1 sulphate ion, (c) \(Na_2SO_4\) results in 1 sulphate ion, (d) \(Al_2(SO_4)_3\) provides 3 sulphate ions, and (e) \(CuSO_4\) forms 1 sulphate ion.

Step 2: Calculate the molarity of sulphate ions

Molarity of sulphate ions for each compound will be: (a) 0.075 M - the same as the given solution, (b) 0.22 M - also the same, (c) 0.15 M - no change from the provided concentration, (d) 3*0.080 M = 0.24 M - due to the three available sulphate ions, and (e) 0.20 M - same as the solution's concentration.

Step 3: Compare molarity of sulphate ions

Comparing the molarity of sulphate ions we have calculated, the solution with the greatest concentration of sulphate ions is (d) \(0.080 \mathrm{M} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), with a sulphate ion concentration of 0.24 M.

Key Concepts

Molarity: Understanding the Basics

Molarity is a key concept in solution chemistry. It is used to describe the concentration of a solution. Molarity, often denoted as \( M \), is defined as the number of moles of solute (the substance dissolved) per liter of solution. To calculate molarity, use the formula:

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

Understanding molarity helps us determine the concentration of specific ions, like sulphate ions in the original exercise. This is vital for tasks like predicting the outcomes of chemical reactions and preparing solutions with precise concentrations. In the exercise, molarity is used to find out how many sulphate ions are present in various solutions.

Chemical Compounds and Their Breakdown

Chemical compounds consist of two or more elements chemically bonded together. When these compounds dissolve in water, they may dissociate into individual ions. For example, each compound in the original exercise dissociates, releasing ions like the sulphate ion \( (\mathrm{SO}_4^{2-}) \).

• \( H_2SO_4 \) releases 1 sulphate ion.
• \( MgSO_4 \) provides 1 sulphate ion.
• \( Na_2SO_4 \) yields 1 sulphate ion.
• \( Al_2(SO_4)_3 \) releases 3 sulphate ions.
• \( CuSO_4 \) gives 1 sulphate ion.

This dissociation is essential in understanding how each compound contributes to the total concentration of sulphate ions in a solution. Compounds with more ions per formula unit will generally contribute to a higher concentration of that ion in the solution.

Solution Chemistry: Finding Concentrations

Solution chemistry involves studying how substances dissolve and interact in a solution. When substances like sulphates are dissolved in water, they dissociate, increasing the solution’s ionic concentration. The original exercise's task was to identify which solution had the highest concentration of sulphate ions, derived from different chemical compounds.

• Calculate total molarity by multiplying the base molarity by the number of dissociated ions.
• Compare concentrations to find out which solution holds the greatest concentration.

The solution with \( Al_2(SO_4)_3 \) yielded the highest sulphate ion concentration due to its ability to provide three sulphate ions per formula unit. Understanding how to manipulate these concentrations enables us to work effectively in various scientific and industrial applications.