Water containing \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) ions is called hard water and is unsuitable for some household and industrial use because these ions react with soap to form insoluble salts, or curds. One way to remove the \(\mathrm{Ca}^{2+}\) ions from hard water is by adding washing soda \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot 10 \mathrm{H}_{2} \mathrm{O}\right)\). (a) The molar solubility of \(\mathrm{CaCO}_{3}\) is \(9.3 \times 10^{-5} \mathrm{M}\). What is its molar solubility in a \(0.050 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) solution? (b) Why are \(\mathrm{Mg}^{2+}\) ions not removed by this procedure? (c) The \(\mathrm{Mg}^{2+}\) ions are removed as \(\mathrm{Mg}(\mathrm{OH})_{2}\) by adding slaked lime \(\left[\mathrm{Ca}(\mathrm{OH})_{2}\right]\) to the water to produce a saturated solution. Calculate the \(\mathrm{pH}\) of a saturated \(\mathrm{Ca}(\mathrm{OH})_{2}\) solution. (d) What is the concentration of \(\mathrm{Mg}^{2+}\) ions at this \(\mathrm{pH} ?\) (e) In general, which ion \(\left(\mathrm{Ca}^{2+}\right.\) or \(\mathrm{Mg}^{2+}\) ) would you remove first? Why?

Step by step solution

01

Molar Solubility of \(\mathrm{CaCO_{3}}\) in Sodium Carbonate Solution (a)

First, write a balanced chemical equation for the dissolution of calcium carbonate in water: \[ \mathrm{CaCO_{3} (s)} \leftrightarrow \mathrm{Ca^{2+} (aq)} + \mathrm{CO_{3}^{2-} (aq)} \] Given that the molar solubility of \(\mathrm{CaCO_{3}}\) is \(9.3 \times 10^{-5} \mathrm{M}\), we can assume that sodium carbonate is added in enough quantity to keep carbonate ion concentration essentially constant at 0.050 M. Therefore, the molar solubility of \(\mathrm{CaCO_{3}}\) is the value at which this equilibrium is established i.e., \(9.3 \times 10^{-5} \mathrm{M}\).

02

Reaction of \(\mathrm{Mg^{2+}}\) Ions with Sodium Carbonate (b)

Magnesium ions react with sodium carbonate to form MgCO3, that unlike \(\mathrm{CaCO_{3}}\), is soluble in water. Therefore, \(\mathrm{Mg^{2+}}\) ions are not removed by this procedure.

03

Calculating the pH of a Saturated \(\mathrm{Ca(OH)_{2}}\) Solution (c)

The dissolution of \(\mathrm{Ca(OH)_{2}}\) in water can be represented as: \(\mathrm{Ca(OH)_{2} (s)} \leftrightarrow \mathrm{Ca^{2+} (aq)} + 2\,\mathrm{OH^{-} (aq)}\). The Ksp for this reaction is 5.02 x 10-6. The concentration of \(\mathrm{OH^{-}}\) ions can be calculated using the formula \(\sqrt{Ksp / 0.5}\) that results in \(0.010\,\mathrm{M}\). Using this value, the pOH can be calculated using the formula \(-log[OH^{-}]\). After calculating the pOH, the pH can be found by invoking the water autoionization constant Kw and the formula pH = 14 - pOH.

04

Calculating the Concentration of \(\mathrm{Mg^{2+}}\) Ions at the Calculated pH (d)

The ionization of \(\mathrm{Mg(OH)_{2}}\) in water can be represented as: \(\mathrm{Mg(OH)_{2} (s)} \leftrightarrow \mathrm{Mg^{2+} (aq)} + 2\,\mathrm{OH^{-} (aq)}\). The Ksp for this reaction is 5.61 x 10^-12. Using the OH- concentration from the charactiristics of a \(\mathrm{Ca(OH)_{2}}\) solution, we can determine the concentration of \(\mathrm{Mg^{2+}}\) ions using the formula \(Ksp/ [OH^-]^2\).

05

Choosing the Ion to Remove First and Reasoning (e)

The ion to be removed first would be \(\mathrm{Ca^{2+}}\) ions. Because, \(\mathrm{Ca^{2+}}\) ions precipitates out of solution more easily than \(\mathrm{Mg^{2+}}\) ions, reducing its solubility and making it easier to remove. Furthermore, washing soda, a commonly available and inexpensive household product, effectively removes \(\mathrm{Ca^{2+}}\) ions.

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