In the lime soda process once 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 \(\mathrm{Mg}(\mathrm{OH})_{2}(s);\) $$ \begin{aligned} \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q) & \longrightarrow \mathrm{CaCO}_{3}(s) \\ \mathrm{Mg}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) & \longrightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s) \end{aligned} $$ How many moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and $\mathrm{Na}_{2} \mathrm{CO}_{3}\( should be added to soften (remove the \)\mathrm{Ca}^{2+}$ and \(\mathrm{Mg}^{2+}\) ) 1000 L of water in which $$ \begin{array}{l} {\left[\mathrm{Ca}^{2+}\right]=3.5 \times 10^{-4} \mathrm{M}} \\ {\left[\mathrm{Mg}^{2+}\right]=7.5 \times 10^{-4} \mathrm{M}} \end{array} $$

First, we need to find the moles of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) ions in the 1000 L of water. We do this using the given concentrations and the water volume: Moles of \(\mathrm{Ca}^{2+}\) = Concentration of \(\mathrm{Ca}^{2+}\times\) Volume of water Moles of \(\mathrm{Mg}^{2+}\) = Concentration of \(\mathrm{Mg}^{2+}\times\) Volume of water The volume of water is given in liters, but we need to convert it to meters cubed (m³) for our calculation. To convert liters to cubic meters, we multiply by 0.001: Volume of water = 1000 L \(\times\) 0.001 = 1 m³ Now, we can calculate the moles of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) ions: Moles of \(\mathrm{Ca}^{2+}\) = \((3.5 \times 10^{-4}\ \mathrm{M}) \times (1\ \mathrm{m}^{3}) = 3.5 \times 10^{-4}\ \mathrm{moles}\) Moles of \(\mathrm{Mg}^{2+}\) = \((7.5 \times 10^{-4}\ \mathrm{M}) \times (1\ \mathrm{m}^{3}) = 7.5 \times 10^{-4}\ \mathrm{moles}\)

Now that we have the moles of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) ions in the water, we can use the stoichiometry of the provided reactions to determine the moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) required. From the first reaction, one mole of \(\mathrm{Ca}^{2+}\) reacts with one mole of \(\mathrm{CO}_{3}^{2-}\) to form one mole of \(\mathrm{CaCO}_{3}(s)\). Since \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) provides one mole of \(\mathrm{CO}_{3}^{2-}\) per mole of itself, the moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) required will be equal to the moles of \(\mathrm{Ca}^{2+}\) ions: Moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) = Moles of \(\mathrm{Ca}^{2+}\) = \(3.5 \times 10^{-4}\ \mathrm{moles}\) From the second reaction, one mole of \(\mathrm{Mg}^{2+}\) reacts with two moles of \(\mathrm{OH}^{-}\) to form one mole of \(\mathrm{Mg}(\mathrm{OH})_{2}(s)\). Since \(\mathrm{Ca}(\mathrm{OH})_{2}\) provides two moles of \(\mathrm{OH}^{-}\) per mole of itself, the moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) required will be half of the moles of \(\mathrm{Mg}^{2+}\) ions: Moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) = \(\frac{1}{2}\times\) Moles of \(\mathrm{Mg}^{2+}\) = \((\frac{1}{2}) (7.5 \times 10^{-4}\ \mathrm{moles}) = 3.75 \times 10^{-4}\ \mathrm{moles}\) So, to soften the 1000 L of water, \(3.75 \times 10^{-4}\ \mathrm{moles}\) of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(3.5 \times 10^{-4}\ \mathrm{moles}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) should be added.