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Chapter 16: Problem 9

Sodium chloride is listed in the solubility rules as a soluble compound. Therefore, the \(K_{\mathrm{sp}}\) value for \(\mathrm{NaCl}\) is infinite. Is this statement true or false? Explain.

Short Answer

Expert verified
The statement is false. Although sodium chloride (NaCl) is soluble in water according to solubility rules, this does not mean its \(K_{\mathrm{sp}}\) value is infinite. No compound has infinite solubility in water, and solubility rules only help to determine if a compound is soluble or not, without providing a specific \(K_{\mathrm{sp}}\) value.

Step by step solution

01

Understanding Solubility Rules

Solubility rules are general guidelines to determine which ionic compounds are soluble in water. According to the solubility rules, all salts containing alkali metal ions, like sodium (Na\(^+\)), and nitrates, are soluble in water. This means that sodium chloride (NaCl) is indeed soluble in water.
02

Understanding the Solubility Product Constant, \(K_{\mathrm{sp}}\)

The solubility product constant (\(K_{\mathrm{sp}}\)) is an equilibrium constant that quantifies the solubility of a slightly soluble compound. The larger the \(K_{\mathrm{sp}}\) value, the more soluble the compound is in water. For a salt reaction of the form \(\mathrm{AB} \leftrightarrow \mathrm{A}^+ + \mathrm{B}^-\), the \(K_{\mathrm{sp}}\) can be written as: \[K_{\mathrm{sp}} = [\mathrm{A}^+][\mathrm{B}^-]\]
03

Relating Solubility Rules and \(K_{\mathrm{sp}}\) values

While it is true that a higher \(K_{\mathrm{sp}}\) value indicates greater solubility, the statement that the \(K_{\mathrm{sp}}\) value for NaCl is "infinite" is incorrect. In reality, no compound has infinite solubility, as there would be a limit to how much of a compound can dissolve in a given volume of water. The solubility rules merely help us determine if a compound is soluble or not, but they do not provide us with the specific value of \(K_{\mathrm{sp}}\). In fact, sodium chloride has a finite solubility in water, and its \(K_{\mathrm{sp}}\) value is not infinite.
04

Conclusion

The statement "Sodium chloride is listed in the solubility rules as a soluble compound. Therefore, the \(K_{\mathrm{sp}}\) value for \(\mathrm{NaCl}\) is infinite." is false. While sodium chloride is indeed soluble, its \(K_{\mathrm{sp}}\) value is not infinite, as no compound has infinite solubility in water.

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Most popular questions from this chapter

The \(K_{\mathrm{sp}}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) is $2 \times 10^{-32} .\( At what pH will a \)0.2-M\( \)\mathrm{Al}^{3+}$ solution begin to show precipitation of \(\mathrm{Al}(\mathrm{OH})_{3} ?\)

Calculate the concentration of \(\mathrm{Pb}^{2+}\) in each of the following. a. a saturated solution of $\mathrm{Pb}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=1.2 \times 10^{-15}$ b. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) buffered at \(\mathrm{pH}=13.00\) c. Ethylenediaminetetraacetate \(\left(\mathrm{EDTA}^{4-}\right)\) is used as a complexing agent in chemical analysis and has the following structure: Solutions of \(\mathrm{ED} \mathrm{TA}^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The reaction of \(\mathrm{EDTA}^{4-}\) with \(\mathrm{Pb}^{2+} \mathrm{is}\) $$\mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-}(a q) \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) \quad K=1.1 \times 10^{18}$$ Consider a solution with 0.010 mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to 1.0 \(\mathrm{L}\) of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing 0.050 \(\mathrm{M}\) Na_thion. Does \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitate from this solution?

A solution contains $1.0 \times 10^{-6} M \mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}\( and \)5.0 \times 10^{-7} M$ \(\mathrm{K}_{3} \mathrm{PO}_{4} .\) Will \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) precipitate? $\left[K_{\mathrm{sp}} \text { for } \mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}=1.0 \times 10^{-31} . ] \right.$

In the presence of \(\mathrm{CN}^{-}, \mathrm{Fe}^{3+}\) forms the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-} .\) The equilibrium concentrations of \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) are $8.5 \times 10^{-40} \mathrm{M}\( and \)1.5 \times 10^{-3} M,\( respectively, in a \)0.11-M$ KCN solution. Calculate the value for the overall formation constant of \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) . $$\mathrm{Fe}^{3+}(a q)+6 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \qquad K_{\text { overall }}=?$$

A solution is prepared by mixing 100.0 \(\mathrm{mL}\) of \(1.0 \times 10^{-4} M\) \(\mathrm{Be}\left(\mathrm{NO}_{3}\right)_{2}\) and 100.0 \(\mathrm{mL}\) of $8.0 M\( \)\mathrm{NaF}.$ $$\mathrm{Be}^{2+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}^{+}(a q) \quad K_{1}=7.9 \times 10^{4}$$ $$\operatorname{Be} \mathrm{F}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \operatorname{Be} \mathrm{F}_{2}(a q) \quad K_{2}=5.8 \times 10^{3}$$ $$\operatorname{BeF}_{2}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \operatorname{Be} \mathrm{F}_{3}^{-}(a q) \quad K_{3}=6.1 \times 10^{2}$$ $$\operatorname{Be} \mathrm{F}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}_{4}^{2-}(a q) \qquad K_{4}=2.7 \times 10^{1}$$ Calculate the equilibrium concentrations of $\mathrm{F}^{-}, \mathrm{Be}^{2+}, \mathrm{BeF}^{+},\( \)\mathrm{BeF}_{2}, \mathrm{BeF}_{3}^{-},$ and \(\mathrm{BeF}_{4}^{2-}\) in this solution.
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