Question

A sample of hard water contains \(0.005\) mole of \(\mathrm{CaCl}_{2}\) per litre. What is the minimum concentration of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) which must be exceeded for removing \(\mathrm{Ca}^{2+}\) ions from this water sample? The solubility product of \(\mathrm{CaSO}_{4}\) is \(2.4 \times 10^{-5}\). (a) \(4.8 \times 10^{-3} \mathrm{M}\) (b) \(1.2 \times 10^{-3} \mathrm{M}\) (c) \(0.0144 \mathrm{M}\) (d) \(2.4 \times 10^{-3} \mathrm{M}\)

Short answer

The minimum concentration of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) which must be exceeded for removing \(\mathrm{Ca}^{2+}\) ions from this water sample is \(1.6 \times 10^{-3} \mathrm{M}\) (None of the options given match, please check the calculation and options again).

Step-by-step solution

Write down the reaction

Write the chemical reaction between \(\mathrm{CaCl}_{2}\) and \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) which would precipitate \(\mathrm{CaSO}_{4}\): \[\mathrm{Ca}^{2+} + \mathrm{SO}_{4}^{2-} \rightarrow \mathrm{CaSO}_{4}\(s\)\]

Determine the concentration of \(\mathrm{Ca}^{2+}\) ions

Since the sample contains 0.005 mole of \(\mathrm{CaCl}_{2}\) per litre, and \(\mathrm{CaCl}_{2}\) dissociates completely in water to give \(\mathrm{Ca}^{2+}\) and \(\mathrm{Cl}^{-}\) ions, the concentration of \(\mathrm{Ca}^{2+}\) ions is also 0.005 M.

Use the solubility product to find the concentration of \(\mathrm{SO}_{4}^{2-}\) ions

Using the solubility product of \(\mathrm{CaSO}_{4}\), we have: \[K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}]\] Substitute \(K_{sp} = 2.4 \times 10^{-5}\) and \(\mathrm{Ca}^{2+} = 0.005\ M\): \[[\mathrm{SO}_{4}^{2-}] = \frac{2.4 \times 10^{-5}}{0.005}\]

Calculate the concentration of \(\mathrm{SO}_{4}^{2-}\) ions required

Calculate the concentration of \(\mathrm{SO}_{4}^{2-}\): \[[\mathrm{SO}_{4}^{2-}] = \frac{2.4 \times 10^{-5}}{0.005} = 4.8 \times 10^{-3}\mathrm{M}\]

Find the minimum concentration of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) to remove \(\mathrm{Ca}^{2+}\)

To find the minimum concentration of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) that must be exceeded, we need to account for the stoichiometry of the reaction. Since \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) provides 3 moles of \(\mathrm{SO}_{4}^{2-}\) for every mole of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), the concentration of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) will be one third of the concentration of \(\mathrm{SO}_{4}^{2-}\): \[[\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}] = \frac{4.8 \times 10^{-3}\mathrm{M}}{3}\]

Key concepts

Solubility Product Constant

Understanding the solubility product constant (K_sp) is key to solving many chemical equilibrium problems, particularly those involving sparingly soluble salts. The K_sp is a type of equilibrium constant that measures the extent to which a compound can dissolve in water. In other words, it represents the product of the concentrations of the ions in a saturated solution, each raised to the power of their respective coefficients in the balanced equation.

For instance, for calcium sulfate (\textrm{CaSO}_{4}), the dissolution can be represented as follows: \[\textrm{CaSO}_4(s) \rightleftharpoons \textrm{Ca}^{2+}(aq) + \textrm{SO}_4^{2-}(aq)\] When the salt is in equilibrium with its dissolved ions, the K_sp expression is given by: \[K_{sp} = [\textrm{Ca}^{2+}][\textrm{SO}_4^{2-}]\] In this exercise, knowing the K_sp of calcium sulfate allows us to calculate the concentration of sulfate ions needed to begin precipitation, which is an essential step in the water treatment process described in the problem.

Stoichiometry Calculations

Stoichiometry calculations are crucial for quantitatively describing the relationships between reactants and products in a chemical reaction. These calculations are based on the principle of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction.

To solve stoichiometry problems, we use the coefficients from the balanced chemical equation to determine the ratios in which reactants combine and products form. In our exercise, for the reaction of aluminum sulfate with calcium ions, we know that every one mole of \(\mathrm{Al}_2(\mathrm{SO}_4)_3\) releases three moles of \(\mathrm{SO}_4^{2-}\). This 1:3 stoichiometric ratio permits us to calculate the concentration of aluminum sulfate necessary to precipitate calcium sulfate from hard water. To enhance comprehension here, picture a see-saw where on one side you have 1 mole of aluminum sulfate causing 3 moles of sulfate ions to rise on the other—this visual can help solidify the concept of stoichiometry balances.

Concentration Calculations

Concentration calculations are a fundamental aspect of many chemical equilibrium problems, especially when dealing with solutions. Concentration typically refers to the amount of a substance in a specified volume of solvent and is often expressed as molarity (M), which is moles per liter.

During the exercise, we calculated the concentration of the calcium ions to be 0.005 M, given the quantity provided in the water sample. We then used this figure, in conjunction with the solubility product constant, to find the concentration of sulfate ions required to initiate precipitation. These concentration values are not static; they tell a dynamic story of how different substances interact in a solution. By understanding the math behind these concentrations, students can predict whether a salt will precipitate under certain conditions and can design protocols for inducing or preventing such reactions.

Precipitation Reactions

Precipitation reactions involve the formation of an insoluble solid from the reaction of two soluble substances in solution. These reactions are fundamental to various processes in both industrial applications and nature, such as water purification and mineral formation.

In the context of this exercise, we're concerned with the precipitation of calcium sulfate from a hard water sample to soften it. By calculating the minimum concentration of aluminum sulfate needed to create the insoluble calcium sulfate, we directly apply the principles of precipitation reactions to solve a practical problem. It's like a dance where the ions in the water meet, pair up with the right partner, and settle down to form a solid. The solubility product constant helps us predict when the music stops and the solid precipitate forms, thereby providing valuable insight into the chemical choreography within a solution.