Question

In the lime soda process at one time used in large scale municipal water softening, calcium hydroxide prepared from lime and sodium carbonate are added to precipitate \(\mathrm{Ca}^{2+}\) as \(\mathrm{CaCO}_{3}(s)\) and \(\mathrm{Mg}^{2+}\) as \(\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{~s})\) : $$ \begin{gathered} \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}{ }^{2-}(a q) \longrightarrow \mathrm{CaCO}_{3}(s) \\ \mathrm{Mg}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{MgOH}_{2}(a q) \end{gathered} $$ How many moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) should be added to soften \(1200 \mathrm{~L}\) of water in which $$ \begin{aligned} {\left[\mathrm{Ca}^{2+}\right] } &=5.0 \times 10^{-4} \mathrm{M} \text { and } \\\ {\left[\mathrm{Mg}^{2+}\right] } &=7.0 \times 10^{-4} \mathrm{M} \end{aligned} $$

Short answer

To soften 1200 L of water with given concentrations of calcium and magnesium ions, we need to add 0.6 moles of \(\ce{Ca(OH)2}\) and 0.84 moles of \(\ce{Na2CO3}\).

Step-by-step solution

Calculate moles of calcium and magnesium ions in the water

Given the concentrations of \(\ce{Ca^2+}\) and \(\ce{Mg^2+}\), we can calculate the moles present in 1200 L of water as follows: Number of moles of \(\ce{Ca^2+}\) \(= \) concentration \(×\) volume = \((5.0 × 10^{-4} \mathrm{M}) (1200 \mathrm{L})\) Number of moles of \(\ce{Mg^2+}\) \(= \) concentration \(×\) volume = \((7.0 × 10^{-4} \mathrm{M}) (1200 \mathrm{L})\)

Calculate moles of calcium hydroxide and sodium carbonate required

Now, we will use stoichiometry to calculate the moles of the added compounds: \[\ce{ Ca^2+(aq) + CO3^{2-}(aq) → CaCO3(s)} \] \[\ce{ Mg^2+(aq) + 2OH^-(aq) → Mg(OH)2(s)} \] From the balanced equations, we can see that one mole of calcium ion reacts with one mole of carbonate ion and one mole of magnesium ion reacts with two moles of hydroxide ions. So, moles of \(\ce{Ca(OH)2}\) needed = moles of \(\ce{Ca^2+}\) in water = \((5.0 × 10^{-4} \mathrm{M})(1200 \mathrm{L})\) And, moles of \(\ce{Na2CO3}\) needed = moles of \(\ce{Mg^2+}\) in water = \((7.0 × 10^{-4} \mathrm{M})(1200 \mathrm{L})\)

Calculate the final values

Now, we can calculate the moles of \(\ce{Ca(OH)2}\) and \(\ce{Na2CO3}\) required: Moles of \(\ce{Ca(OH)2}\) = \((5.0 × 10^{-4} \mathrm{M})(1200 \mathrm{L}) = 0.6 \, \mathrm{moles}\) Moles of \(\ce{Na2CO3}\) = \((7.0 × 10^{-4} \mathrm{M})(1200 \mathrm{L}) = 0.84 \, \mathrm{moles}\) Therefore, 0.6 moles of \(\ce{Ca(OH)2}\) and 0.84 moles of \(\ce{Na2CO3}\) must be added to soften the 1200 L of water.

Key concepts

The Lime Soda Process

The lime soda process is a method of water softening that involves adding calcium hydroxide, which is derived from lime, and sodium carbonate to water. The purpose of these chemicals is to remove calcium (\rcalcium ions and magnesium (\rmagnesium ions by precipitating them out of the solution in the form of insoluble compounds: calcium carbonate (\(\rCaCO3\)) and magnesium hydroxide (\(\rMg(OH)2\)).

When handling an exercise that applies the lime soda process, it's essential to understand the reactions involved. For calcium ions in the water, the reaction can be represented as:\[\ce{Ca^{2+}(aq) + CO3^{2-}(aq) \rightarrow CaCO3(s)}\]Similarly, for magnesium ions, the corresponding reaction is:\[\ce{Mg^{2+}(aq) + 2OH^{-}(aq) \rightarrow Mg(OH)2(s)}\]In practical applications, this process not only improves water quality by softening it but also aids in preventing scale formation in pipelines and appliances. The effectiveness of this process directly depends on the proper understanding and application of stoichiometry and chemical equilibrium principles.

Understanding Stoichiometry

Stoichiometry is foundational in executing the lime soda process correctly. It refers to the quantitative relationship between reactants and products in a chemical reaction. It's what allows us to calculate the exact amounts of calcium hydroxide and sodium carbonate needed to react with the calcium and magnesium ions in water.

For instance, in the original exercise, stoichiometry is used to determine the number of moles of \(\ce{Ca(OH)2}\) and \(\ce{Na2CO3}\) necessary to react with the given concentrations of \(\ce{Ca^2+}\) and \(\ce{Mg^2+}\) in the water sample. The balanced chemical reactions dictate that each mole of \(\ce{Ca^2+}\) requires one mole of carbonate ion from sodium carbonate, while each mole of \(\ce{Mg^2+}\) needs two moles of hydroxide ions from calcium hydroxide.

To enhance students' understanding, it's advantageous to provide a clear connection between the mole concept in stoichiometry and the actual quantities of substances used in a reaction. Visual aids or interactive exercises can also be beneficial in illustrating the stoichiometric relationships and improving comprehension.

Chemical Equilibrium in Water Treatment

Chemical equilibrium plays a crucial role in the lime soda process of water softening. It describes the state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products over time.

In the context of the exercise, once calcium hydroxide and sodium carbonate are added to the water, they react with the calcium and magnesium ions until the insoluble carbonates and hydroxides precipitate out. Ideally, the reactions proceed to completion. However, in reality, some reactions may reach a state of equilibrium before all reactants have fully converted into products. This can affect the efficiency of the water softening process.

Understanding equilibrium principles can help to adjust the amounts of reactants added, ensuring that the reactions proceed as far as possible towards completion. Introducing concepts such as Le Chatelier's principle might deepen the understanding of how changing conditions can shift the equilibrium, thus impacting the water treatment process. It's important for students to realize that chemical equilibrium considerations are integral in optimizing industrial processes like water softening.