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Chapter 16: Problem 77

The molecular solid \(\mathrm{I}_{2}(\mathrm{s})\) is only slightly soluble in water but will dissolve to a much greater extent in an aqueous solution of \(\mathrm{KI}\), because the \(\mathrm{I}_{3}^{-}\) anion forms. Write an equation for the formation of the \(I_{3}^{-}\) anion, and indicate the Lewis acid and Lewis base.

Short Answer

Expert verified
The equation for formation of the I₃⁻ ion is \( I_{2(s)} + I_{-(aq)} \rightarrow I_{3^{-}(aq)} \). In this reaction, I- acts as a Lewis base while I₂ acts as a Lewis acid.

Step by step solution

01

Formulate Equation

You start with the compounds provided: I₂ and KI. When I₂ dissolves in KI, it reacts with I⁻ (from KI) to form the I₃⁻ ion. The reaction can be written as follows: \[ I_{2(s)} + I_{-(aq)} \rightarrow I_{3^{-}(aq)} \]
02

Identify Lewis Acids and Bases

Next we identify the Lewis acid and Lewis base in the reaction. The Lewis base is a species that has an electron pair available for bonding, while the Lewis acid is a species that can accept this pair. In this reaction, I- (from KI) acts as a Lewis base as it donates an electron pair. I₂ acts as a Lewis acid, as it accepts the electron pair.

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Most popular questions from this chapter

Calculate \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\mathrm{pH}\) in saturated \(\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq})\) which contains \(3.9 \mathrm{g} \mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) per \(100 \mathrm{mL}\) of solution.

Caproic acid, \(\mathrm{HC}_{6} \mathrm{H}_{11} \mathrm{O}_{2},\) found in small amounts in coconut and palm oils, is used in making artificial flavors. A saturated aqueous solution of the acid contains \(11 \mathrm{g} / \mathrm{L}\) and has \(\mathrm{pH}=2.94 .\) Calculate \(K_{\mathrm{a}}\) for the acid. $$\mathrm{HC}_{6} \mathrm{H}_{11} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{O}_{2}^{-} \quad K_{\mathrm{a}}=?$$

The conjugate acid of \(\mathrm{HPO}_{4}^{2-}\) is (a) \(\mathrm{PO}_{4}^{3-}\) (b) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-} ;(\mathrm{c}) \mathrm{H}_{3} \mathrm{PO}_{4} ;(\mathrm{d}) \mathrm{H}_{3} \mathrm{O}^{+} ;\) (e) none of these.

For the ionization of phenylacetic acid, $$\begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \\\ K_{\mathrm{a}}=4.9 \times 10^{-5} \end{array}$$ (a) What is \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2}^{-}\right]\) in \(0.186 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} ?\) (b) What is the \(\mathrm{pH}\) of \(0.121 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} ?\)

According to the Bronsted-Lowry theory, label each of the following as an acid or a base. (a) \(\mathrm{HNO}_{2}\) (b) \(\mathrm{OCl}^{-} ;(\mathrm{c}) \mathrm{NH}_{2}^{-} ;\) (d) \(\mathrm{NH}_{4}^{+} ;\) (e) \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\)
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