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Chapter 16: Problem 133

Fluoride ion is added to municipal water supplies to prevent tooth decay. The concentration of \(\mathrm{Ca}^{2+}\) in water equilibrated with the atmosphere and limestone \(\left(\mathrm{CaCO}_{3}\right)\) is \(5.0 \times 10^{-4} \mathrm{M}\). If fluoride ion were added to the water until a concentration of \(1.0 \mathrm{ppm}\) was reached, would a precipitate of \(\mathrm{CaF}_{2}\left(K_{\mathrm{sp}}=3.5 \times 10^{-11}\right)\) form?

Short Answer

Expert verified
No precipitate will form because \(Q < K_{sp}\).

Step by step solution

01

Convert ppm to Molarity

First, convert the fluoride concentration from parts per million (ppm) to molarity. Knowing that 1 ppm is equivalent to 1 mg/L, and the molar mass of fluoride (F) is approximately 19 g/mol, we convert 1 ppm of fluoride:\[\text{Molarity of } \text{F}^- = \frac{1 \text{ mg/L}}{19000 \text{ g/mol}} = \frac{1 \times 10^{-3} \text{ g/L}}{19 \text{ g/mol}} \approx 5.26 \times 10^{-5} \text{ M}\]
02

Calculate the Ionic Product (Q)

Calculate the ionic product \( Q \) for the reaction using the concentration of \( \mathrm{Ca}^{2+} \) and the calculated F\(^-\):\[ Q = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 = (5.0 \times 10^{-4}) \times (5.26 \times 10^{-5})^2 \]Calculate the numerical value of \( Q \):\[ Q = (5.0 \times 10^{-4}) \times (2.77 \times 10^{-9}) = 1.39 \times 10^{-12} \]
03

Compare Q with Ksp

Compare the calculated value of \( Q \) with the solubility product constant \( K_{\mathrm{sp}} = 3.5 \times 10^{-11} \) of \( \mathrm{CaF}_2 \).Since \( Q = 1.39 \times 10^{-12} < K_{\mathrm{sp}} = 3.5 \times 10^{-11} \), the ionic product \( Q \) is less than \( K_{\mathrm{sp}} \), indicating that the solution is undersaturated and no precipitate will form.

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Key Concepts

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

Chemical Equilibria
In chemistry, equilibria involve balancing the forward and reverse reactions of compounds in a system. For sparingly soluble salts, like calcium fluoride (\(\text{CaF}_2\)), the equilibrium is between the solid salt and its ions in solution. This is described by the solubility product constant, \(K_{\text{sp}}\). It reflects the extent to which a compound can dissolve: the higher the \(K_{\text{sp}}\), the more soluble the compound.

For \(\text{CaF}_2\), the dissolution in water is represented by:
\[\text{CaF}_2(s) \leftrightarrow \text{Ca}^{2+}(aq) + 2\text{F}^-(aq)\].

The \(K_{\text{sp}}\) expression can be written as:
\[K_{\text{sp}} = [\text{Ca}^{2+}] [\text{F}^-]^2\].

This formula helps us determine whether a solution is saturated, unsaturated, or supersaturated by comparing it with the ionic product \(Q\). If \(Q < K_{\text{sp}}\), as seen in the exercise, no precipitate forms, indicating undersaturation.
Ion Concentration Conversion
Converting ion concentration units is crucial when dealing with chemical equilibria. In the exercise, we need to convert the concentration of fluoride ions from parts per million (ppm) to molarity. Understanding ppm:
  • 1 ppm refers to 1 mg of solute per liter of solution.
  • For fluoride, with a molar mass of 19 g/mol, we convert from ppm to molarity using this relationship:

\[\text{Molarity of \(\text{F}^-\)} = \frac{1 \text{ mg/L}}{19,000 \text{ mg/mol}} \approx 5.26 \times 10^{-5} \text{ M}\].

This conversion is vital because chemical computations often require molarity. When calculating the ionic product \(Q\), molarity allows us to work with concentrations consistently.
Precipitation Reactions
Precipitation reactions occur when the ionic product \(Q\) exceeds the solubility product \(K_{\text{sp}}\), leading to the formation of a solid from a solution. For \(\text{CaF}_2\), comparing \(Q\) and \(K_{\text{sp}}\) helps decide if a precipitate will form.

In this exercise, we calculated the ionic product as:
\[Q = [\text{Ca}^{2+}][\text{F}^-]^2 = 1.39 \times 10^{-12}\].

Since \(Q < K_{\text{sp}} = 3.5 \times 10^{-11}\), calcium fluoride is undersaturated, leading to no precipitation.
  • If \(Q > K_{\text{sp}}\), ions exceed solubility, precipitate forms.
  • \(Q = K_{\text{sp}}\), solution is saturated, no more solute dissolves.
  • \(Q < K_{\text{sp}}\), solution is unsaturated, more solute can dissolve.
Understanding these concepts helps predict outcomes of mixing solutions in laboratory or industrial settings.

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Most popular questions from this chapter

In qualitative analysis, \(\mathrm{Ca}^{2+}\) and \(\mathrm{Ba}^{2+}\) are separated from \(\mathrm{Na}^{+}, \mathrm{K}^{+}\), and \(\mathrm{Mg}^{2+}\) by adding aqueous \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) to a solution that also contains aqueous \(\mathrm{NH}_{3}\) (Figure 16.18). Assume that the concentrations after mixing are \(0.080 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) and \(0.16 \mathrm{M} \mathrm{NH}_{3}\). (a) List all the Br酶nsted-Lowry acids and bases present initially, and identify the principal reaction. (b) Calculate the \(\mathrm{pH}\) and the concentrations of all species present in the solution. (c) In order for the human eye to detect the appearance of a precipitate, a very large number of ions must come together to form solid particles. For this and other reasons, the ion product must often exceed \(K_{\mathrm{sp}}\) by a factor of about \(10^{3}\) before a precipitate can be detected in a typical qualitative analysis experiment. Taking this fact into account, show quantitatively that the \(\mathrm{CO}_{3}{ }^{2-}\) concentration is large enough to give observable precipitation of \(\mathrm{CaCO}_{3}\) and \(\mathrm{BaCO}_{3}\), but not \(\mathrm{MgCO}_{3}\). Assume that the metal-ion concentrations are \(0.010 \mathrm{M}\). (d) Show quantitatively which of the \(\mathrm{Mg}^{2+}, \mathrm{Ca}^{2+}\), and \(\mathrm{Ba}^{2+}\) ions, if any, should give an observable precipitate of the metal hydroxide. (e) Could the separation of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Ba}^{2+}\) from \(\mathrm{Mg}^{2+}\) be accomplished using \(0.80 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) in place of \(0.080 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} ?\) Show quantitatively why or why not.

One type of kidney stone is a precipitate of calcium oxalate \(\left(\mathrm{CaC}_{2} \mathrm{O}_{4}, K_{\mathrm{sp}}=2.3 \times 10^{-9}\right) .\) A urine sample has a \(\mathrm{Ca}^{2+}\) concentration of \(2.5 \times 10^{-3} \mathrm{M}\) and an oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\right.\), \(\left.K_{\mathrm{a} 1}=5.9 \times 10^{-2}, \quad K_{\mathrm{a} 2}=6.4 \times 10^{-5}\right)\) concentration of \(1.1 \times 10^{-4} \mathrm{M}\) (a) A typical \(\mathrm{pH}\) for urine is \(5.5 .\) Will a precipitate of calcium oxalate form under these conditions? (b) A vegetarian diet results in a higher \(\mathrm{pH}\) for urine, typically greater than 7 . Would kidney stones be more or less likely to form in urine with a higher \(\mathrm{pH}\) ?

Phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}, K_{\mathrm{a}}=1.3 \times 10^{-10}\right)\) is a weak acid used in mouthwashes, and pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}, K_{\mathrm{b}}=1.8 \times 10^{-9}\right)\) is a weak base used as a solvent. Calculate the value of \(K_{\mathrm{n}}\) for the neutralization of phenol by pyridine. Does the neutralization reaction proceed very far toward completion?

Consider the titration of \(100.0 \mathrm{~mL}\) of \(0.016 \mathrm{M} \mathrm{HOCl}\left(K_{\mathrm{a}}=3.5 \times 10^{-8}\right)\) with \(0.0400 \mathrm{M} \mathrm{NaOH} .\) How many milliliters of \(0.0400 \mathrm{M} \mathrm{NaOH}\) are required to reach the equivalence point? Calculate the \(\mathrm{pH}\) : (a) After the addition of \(10.0 \mathrm{~mL}\) of \(0.0400 \mathrm{M} \mathrm{NaOH}\) (b) Halfway to the equivalence point (c) At the equivalence point (d) The following acid-base indicators change color in the indicated \(\mathrm{pH}\) ranges: bromthymol blue \((6.0-7.6)\), thymolphthalein \((9.4-10.6)\), and alizarin yellow \((10.1-12.0)\). Which indicator is best for the titration? Which indicator is unacceptable?

In qualitative analysis, \(\mathrm{Ag}^{+}, \mathrm{Hg}_{2}^{2+}\), and \(\mathrm{Pb}^{2+}\) are separated from other cations by the addition of HCl. Calculate the concentration of \(\mathrm{Cl}^{-}\) required to just begin the precipitation of (a) \(\mathrm{AgCl}\), (b) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\), (c) \(\mathrm{PbCl}_{2}\) in a solution having metal-ion concentrations of \(0.030 \mathrm{M}\). What fraction of the \(\mathrm{Pb}^{2+}\) remains in solution when the \(\mathrm{Ag}^{+}\) just begins to precipitate?
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