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Chapter 18: Problem 43

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

Short Answer

Expert verified
To soften the \(1200\,\mathrm{L}\) of water with the given concentrations of ions, we need to add \(0.42\,\mathrm{mol}\) of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(0.6\,\mathrm{mol}\) of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\).

Step by step solution

01

Calculate the moles of ions present in water

First, we need to calculate the moles of \(\mathrm{Ca}^{2+}\) and \(\mathrm{HCO}_{3}^-\) ions present in \(1200 \mathrm{~L}\) of water: Moles of \(\mathrm{Ca}^{2+}\) ions = concentration \(\times\) volume = \((5.0 \times 10^{-4}\,\mathrm{M})(1200\, \mathrm{L})=0.6 \,\mathrm{mol}\) Moles of \(\mathrm{HCO}_{3}^-\) ions = concentration \(\times\) volume = \((7.0 \times 10^{-4}\,\mathrm{M})(1200\,\mathrm{L})=0.84 \,\mathrm{mol}\)
02

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

To find the moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) needed to neutralize the \(\mathrm{HCO}_{3}^-\) ions, we can use the stoichiometry of the chemical reaction: From the equation, \(\mathrm{Ca}(\mathrm{OH})_{2}+2 \mathrm{HCO}_{3}^{-} \longrightarrow \mathrm{CaCO}_{3}\downarrow+2 \mathrm{H}_{2}\mathrm{O}+\mathrm{CO}_{2}\), we understand that for every mole of \(\mathrm{Ca}(\mathrm{OH})_{2}\) added, 2 moles of \(\mathrm{HCO}_{3}^-\) are neutralized. Therefore, Moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) needed = \(\dfrac{\mathrm{Moles\, of\, HCO}_{3}^{-}}{2}= \dfrac{0.84\,\mathrm{mol}}{2} = 0.42\, \mathrm{mol}\)
03

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

To find the moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) needed to precipitate the \(\mathrm{Ca}^{2+}\) ions, we can use the stoichiometry of the chemical reaction: From the equation, \(\mathrm{Ca}^{2+}+\mathrm{CO}_{3}^{2-}\longrightarrow \mathrm{CaCO}_{3}\downarrow\), we understand that for every mole of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) added, 1 mole of \(\mathrm{Ca}^{2+}\) is precipitated. Therefore, Moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) needed = Moles of \(\mathrm{Ca}^{2+}\) ions = \(0.6\,\mathrm{mol}\)
04

Final answer

In conclusion, to soften the \(1200\,\mathrm{L}\) of water with the given concentrations of ions, we need to add: - \(0.42\,\mathrm{mol}\) of \(\mathrm{Ca}(\mathrm{OH})_{2}\) - \(0.6\,\mathrm{mol}\) of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\)

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