Problem 43 - Chemistry: The Central Science

Question

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 \(\left[\mathrm{Ca}^{2+}\right]=5.0 \times 10^{-4} \mathrm{M}\) and \(\left[\mathrm{HCO}_{3}^{-}\right]=7.0 \times 10^{-4} \mathrm{M} ?\)

Step-by-Step Solution

Answer

To soften the 1200 L of water, \(6.0 \times 10^{-1}\) moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(8.4 \times 10^{-1}\) moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) should be added.

Step 1: Calculate the number of moles of \(\mathrm{Ca}^{2+}\) and \(\mathrm{HCO}_{3}^{-}\)

We are given the volume and ion concentrations of water. We can calculate the number of moles using the formula: Number of moles = Concentration × Volume For \(\mathrm{Ca}^{2+}\) ions: Number of moles = \([Ca^{2+}]\) * Volume of water Number of moles = \(5.0 \times 10^{-4} M\) * \(1200 L\) For \(\mathrm{HCO}_{3}^{-}\) ions: Number of moles = \([HCO_{3}^{-}]\) * Volume of water Number of moles = \(7.0 \times 10^{-4} M\) * \(1200 L\)

Step 2: Calculate the moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) needed

To remove the \(\mathrm{Ca}^{2+}\) ions from water, \(\mathrm{Ca}(\mathrm{OH})_{2}\) precipitates the calcium ions as \(\mathrm{Ca(OH)}_{2}\). The balanced chemical reaction for this process is: \(\mathrm{Ca}^{2+} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Ca(OH)}_{2}\) From the balanced equation, we can see that for each mole of \(\mathrm{Ca}^{2+}\) ion, one mole of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is needed. Therefore, the number of moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) required is equal to the number of moles of \(\mathrm{Ca}^{2+}\) ions. From step 1: Moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) = Number of moles of \(\mathrm{Ca}^{2+}\) Moles of \(\mathrm{Ca}(\mathrm{OH})_{2} = 5.0 \times 10^{-4} M \times 1200 L\)

Step 3: Calculate the moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) needed

To remove the \(\mathrm{HCO}_{3}^{-}\) ions from water, \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) precipitates the ions as \(\mathrm{CaCO}_{3}\) after reacting with \(\mathrm{Ca}^{2+}\). The balanced chemical reaction for this process is: \(\mathrm{Ca}^{2+} + \mathrm{HCO}_{3}^{-} + \mathrm{CO}_{3}^{2-} \rightarrow \mathrm{CaCO}_{3} + \mathrm{H}_{2}\mathrm{O}\) From the balanced equation, we can see that for each mole of \(\mathrm{HCO}_{3}^{-}\) ion, one mole of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) is needed. Therefore, the number of moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) required is equal to the number of moles of \(\mathrm{HCO}_{3}^{-}\) ions. From step 1: Moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3} = 7.0 \times 10^{-4} M \times 1200 L\)

Key Concepts

Moles Calculation

Understanding moles calculation helps us determine the amount of a substance in a given volume. In chemistry, the concept of moles allows chemists to quantify the number of molecules or atoms in a sample. This is particularly important when preparing solutions or reacting chemicals in precise amounts.

To calculate moles from concentration and volume, we use the formula:

• Number of moles = Concentration (in Molarity, M) × Volume (in Liters, L)

For example, if we need to calculate the moles of Calcium ions \(Ca^{2+}\) in 1200 L of water with a concentration of \(5.0 \times 10^{-4}\) M:
Number of moles = \(5.0 \times 10^{-4} \, \text{M} \, \times \, 1200 \, \text{L} = 0.6 \, \text{moles}\).

Repeating the same for bicarbonate ions \(HCO_3^{-}\) with a concentration of \(7.0 \times 10^{-4}\) M results in \(0.84 \, \text{moles}\). Recognizing how to do these calculations is fundamental to reacting the necessary quantities.

Chemical Reactions

Chemical reactions involve the transformation of one or more substances into new compounds. In our water softening context, reactions help precipitate unwanted ions, cleaning the water. Often, these reactions are represented through balanced equations, which indicate the stoichiometry of reactants and products.

For example, the removal of Calcium ions \(Ca^{2+}\) from water is achieved by reacting with Calcium hydroxide \((Ca(OH)_2)\) to form a precipitate. The equation for this reaction is:

• \(Ca^{2+} + 2OH^{-} \rightarrow Ca(OH)_2 \)\

This shows a 1:1 ratio between \(Ca^{2+}\) ions and \(Ca(OH)_2\) required. Understanding these balanced reactions ensures we add just the right amount of chemicals needed in the water softening process.

Precipitation Reactions

Precipitation reactions occur when two soluble ions in a solution combine to form an insoluble compound. This reaction is widely used in water treatment to remove unwanted dissolved ions.

In the scenario of softening water, when Sodium carbonate \(Na_2CO_3\) is added, it reacts with bicarbonate ions \(HCO_3^{-}\) and Calcium ions \(Ca^{2+}\) to form solid Calcium carbonate \((CaCO_3)\). The reaction is:

• \(Ca^{2+} + HCO_3^{-} + CO_3^{2-} \rightarrow CaCO_3 + H_2O\)

This process effectively removes ions that contribute to water hardness by converting them into a solid form that can be filtered out. Recognizing the components and results of precipitation reactions is critical for creating effective water treatment solutions.