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Hard water contains \(\mathrm{Ca}^{2+}, \mathrm{Mg}^{2+},\) and \(\mathrm{Fe}^{2+},\) which interfere with the action of soap and leave an insoluble coating on the insides of containers and pipes when heated. Water softeners replace these ions with \(\mathrm{Na}^{+} .\) Keep in mind that charge balance must be maintained. (a) If 1500 Lof hard water contains 0.020\(M \mathrm{Ca}^{2+}\) and \(0.0040 \mathrm{M} \mathrm{Mg}^{2+},\) how many moles of \(\mathrm{Na}^{+}\) are needed to replace these ions? ( b) If the sodium is added to the water softener in the form of \(\mathrm{NaCl}\) , how many grams of sodium chloride are needed?

Short Answer

Expert verified
In short, to soften 1500 L of hard water containing 0.020 M \(\mathrm{Ca}^{2+}\) and 0.0040 M \(\mathrm{Mg}^{2+}\), 72 moles of \(\mathrm{Na}^+\) are required. This can be provided by 4207.68 g of \(\mathrm{NaCl}\).

Step by step solution

01

Calculate moles of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) ions in the hard water

Given the concentration of \(\mathrm{Ca}^{2+}\) ions as 0.020 M and \(\mathrm{Mg}^{2+}\) ions as 0.0040 M in 1500 L of hard water. To find the moles of these ions, we can use the formula: Moles = Concentration × Volume Moles of \(\mathrm{Ca}^{2+}\) = \(0.020 \mathrm{M} × 1500 \mathrm{L} = 30 \mathrm{mol}\) Moles of \(\mathrm{Mg}^{2+}\) = \(0.0040 \mathrm{M} × 1500 \mathrm{L} = 6 \mathrm{mol}\)
02

Calculate moles of \(\mathrm{Na}^+\) needed to replace the ions

Now that we know the moles of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\), we can find the moles of \(\mathrm{Na}^+\) needed to replace them and maintain charge balance. For every \(\mathrm{Ca}^{2+}\) ion being replaced, 2 moles of \(\mathrm{Na}^+\) are required, and for every \(\mathrm{Mg}^{2+}\) ion, 2 moles of \(\mathrm{Na}^+\) ions are also needed. Moles of \(\mathrm{Na}^+\) needed for \(\mathrm{Ca}^{2+}\) = \(2 × 30 \mathrm{mol} = 60 \mathrm{mol}\) Moles of \(\mathrm{Na}^+\) needed for \(\mathrm{Mg}^{2+}\) = \(2 × 6 \mathrm{mol} = 12 \mathrm{mol}\) Total moles of \(\mathrm{Na}^+\) required = \(60 \mathrm{mol}+ 12 \mathrm{mol} = 72 \mathrm{mol}\)
03

Calculate the grams of \(\mathrm{NaCl}\) needed

Now, to calculate how many grams of \(\mathrm{NaCl}\) are needed to provide 72 moles of \(\mathrm{Na}^+\), we can use the molecular weight of \(\mathrm{NaCl}\). The molecular weight of Na is 22.99 g/mol, and Cl is 35.45 g/mol. Molecular weight of \(\mathrm{NaCl} = 22.99 + 35.45 = 58.44 \ \mathrm{g/mol}\) Now, to find the grams of \(\mathrm{NaCl}\) needed, we can multiply the moles of \(\mathrm{Na}^+\) by the molecular weight of \(\mathrm{NaCl}\): Grams of \(\mathrm{NaCl} = 72 \ \mathrm{mol} × 58.44 \ \mathrm{g/mol} = 4207.68 \ \mathrm{g}\) Thus, 4207.68 g of \(\mathrm{NaCl}\) are needed to soften 1500 L of hard water.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hard Water Ions

Hard water can be a nuisance in daily life due to the presence of certain minerals that it contains. Specifically, these minerals include ions like calcium (Ca2+), magnesium (Mg2+), and sometimes iron (Fe2+). These ions are responsible for creating challenges such as reducing the effectiveness of soap and detergents, and forming scale buildup on appliances and pipes. Scale is an insoluble coating that can decrease the lifespan of household mechanisms and affect the aesthetics of sinks and bathtubs. Understanding these ions' chemistry is fundamental in addressing the hardships caused by hard water.

Ion Exchange in Water Softening

To combat the detrimental effects of hard water, a process known as ion exchange is employed in water softeners. During ion exchange, hard water ions, which carry a positive charge (cations), are exchanged with sodium (Na+) ions. Since calcium and magnesium ions are divalent (meaning they have a charge of +2), for every single ion of Ca2+ or Mg2+ that is removed, two sodium ions are needed to maintain electrical neutrality in the water. This exchange prevents the formation of scale and allows soaps and detergents to function more effectively by removing the source of the problem -- the hard water ions.

Mole Calculation

Mole calculation is a key element in chemistry that allows us to quantify substances in a very practical way. When dealing with hard water, determining the amount of a treatment substance, such as sodium, requires understanding of moles. A mole is a unit that represents approximately 6.022 x 1023 particles of a given substance, which is Avogadro's number. In water softening, we calculate the moles of hard water ions to figure out how many moles of the sodium ions we would need to replace them.

By utilizing stoichiometry, which is essentially 'chemical accounting', we ensure that the exchange between sodium ions and hard water ions is precise. This allows for effective softening without wastage of resources or over-saturation of sodium in the water, which could lead to other negative effects.

Stoichiometry in Water Treatment

Stoichiometry plays a pivotal role in ensuring chemical reactions in water treatment are accurately carried out. This field of chemistry involves calculating the relationships and proportions between reactants and products in a chemical reaction. When applying stoichiometry to water treatment, particularly in ion exchange, it ensures the correct amount of sodium chloride (NaCl) is used to replace the hard water ions, keeping the balance and avoiding an excess of sodium in the softened water. It's crucial for chemically balancing the treatment process and for maintaining an environmentally safe and cost-effective operation of water softening facilities.

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Most popular questions from this chapter

(a) How many milliliters of a stock solution of 6.0 \(\mathrm{MHNO}_{3}\) would you have to use to prepare 110 \(\mathrm{mL}\) of 0.500 \(\mathrm{M} \mathrm{HNO}_{3} ?\) (b) If you dilute 10.0 \(\mathrm{mL}\) of the stock solution to a final volume of \(0.250 \mathrm{L},\) what will be the concentration of the diluted solution?

Complete and balance the following molecular equations, and then write the net ionic equation for each: \begin{equation} \begin{array}{l}{\text { (a) } \operatorname{HBr}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow} \\ {\text { (b) } \mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{HClO}_{4}(a q) \longrightarrow} \\\ {\text { (c) } \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow}\end{array} \end{equation}

You know that an unlabeled bottle contains an aqueous solution of one of the following: \(A g N O_{3},\) CaCl \(_{2},\) or \(A l_{2}\left(S O_{4}\right)_{3}\) . A friend suggests that you test a portion of the solution with \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) and then with NaCl solutions. According to your friend's logic, which of these chemical reactions could occur, thus helping you identify the solution in the bottle? (a) Barium sulfate could precipitate. (b) Silver chloride could precipitate. (c) silver sulfate could precipitate. (d) More than one, but not all, of the reactions described in answers a-c could occur. (e) All three reactions described in answers a-c could occur.

Would you expect that an anion would be physically closer to the oxygen or to the hydrogens of water molecules that surround it in solution?

Federal regulations set an upper limit of 50 parts per million \((\mathrm{ppm})\) of \(\mathrm{NH}_{3}\) in the air in a work environment \([\mathrm{that}\) is, 50 molecules of \(\mathrm{NH}_{3}(g)\) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing \(1.00 \times 10^{2} \mathrm{mL}\) of 0.0105 \(\mathrm{M}\) HCl. The \(\mathrm{NH}_{3}\) reacts with HCl according to: $$\mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q)$$ After drawing air through the acid solution for 10.0 min at a rate of 10.0 \(\mathrm{L} / \mathrm{min}\) , the acid was titrated. The remaining acid needed 13.1 \(\mathrm{mL}\) of 0.0588 \(\mathrm{M} \mathrm{NaOH}\) to reach the equivalence point. (a) How many grams of \(\mathrm{NH}_{3}\) were drawn into the acid solution? (b) How many ppm of \(\mathrm{NH}_{3}\) were in the air? (Air has a density of 1.20 g/L and an average molar mass of 29.0 \(\mathrm{g} / \mathrm{mol}\) under the conditions of the experiment.) (c) Is this manufacturer in compliance with regulations?

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