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Chapter 17: Problem 52

Explain, with balanced ionic equations, why (a) \(\mathrm{CuI}_{2}\) dissolves in ammonia solution, (b) AgBr dissolves in NaCN solution, (c) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) dissolves in \(\mathrm{KCl}\) solution.

Short Answer

Expert verified
CuI2, AgBr and Hg2Cl2 dissolve in NH3, NaCN and KCl solutions due to the formation of complex ions. The balanced ionic equations are: For CuI2 in NH3 : CuI2 + 4NH3 -> [Cu(NH3)4]2+ + 2I-. For AgBr in NaCN : AgBr + 2CN- -> [Ag(CN)2]- + Br-. For Hg2Cl2 in KCl : Hg2Cl2 + 2Cl- -> [HgCl2]2+ + 2Cl-.

Step by step solution

01

Dissolution of CuI2 in Ammonia solution

When \(\mathrm{CuI}_{2}\) dissolves in ammonia solution, the equation is expressed as follows: \(\mathrm{CuI}_{2} + 4NH_{3}\) \( \rightarrow \) \([Cu(NH_{3})_{4}]^{2+} + 2I^{-}\) Here, the \(\mathrm{CuI}_{2}\) ionizes in the ammonia solution to form a complex ion and iodine ion.
02

Dissolution of AgBr in NaCN solution

When AgBr dissolves in NaCN solution, the ionic equation is written as follows: \(\mathrm{AgBr} +2CN^{-}\) \( \rightarrow \) \([Ag(CN)_{2}]^{-} + Br^{-}\) Here, the \(\mathrm{AgBr}\) ionizes to form a complex ion and bromine ion.
03

Dissolution of Hg2Cl2 in KCl solution

When \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) dissolves in \(\mathrm{KCl}\) solution, the equation is expressed as follows: \(\mathrm{Hg}_{2} \mathrm{Cl}_{2} + 2Cl^{-}\) \( \rightarrow \) \([HgCl_{2}]^{2+} + 2Cl^{-}\) Here, the \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) ionizes in the \(\mathrm{KCl}\) solution to form a complex ion and chloride ion.

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Most popular questions from this chapter

The \(\mathrm{p} K_{\mathrm{a}} \mathrm{s}\) of two monoprotic acids \(\mathrm{HA}\) and \(\mathrm{HB}\) are 5.9 and \(8.1,\) respectively. Which of the two is the stronger acid?

Specify which of these systems can be classified as a buffer system: (a) \(\mathrm{KCl} / \mathrm{HCl}\), (b) \(\mathrm{NH}_{3} / \mathrm{NH}_{4} \mathrm{NO}_{3}\) (c) \(\mathrm{Na}_{2} \mathrm{HPO}_{4} / \mathrm{NaH}_{2} \mathrm{PO}_{4}\)

Amino acids are the building blocks of proteins. These compounds contain at least one amino group and one carboxyl group. Consider glycine, whose structure is shown in Figure 11.18 . Depending on the \(\mathrm{pH}\) of the solution, glycine can exist in one of three possible forms: Fully protonated: \(\mathrm{NH}_{3}-\mathrm{CH}_{2}-\mathrm{COOH}\) Dipolar ion: \(\mathrm{NH}_{3}-\mathrm{CH}_{2}-\mathrm{COO}^{-}\) Fully ionized: \(\mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{COO}^{-}\) Predict the predominant form of glycine at \(\mathrm{pH} 1.0,\) \(7.0,\) and \(12.0 .\) The \(\mathrm{p} K_{\mathrm{a}}\) of the carboxyl group is 2.3 and that of the ammonium group is 9.6.

The solubility product of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is \(1.2 \times 10^{-11}\) What minimum \(\mathrm{OH}^{-}\) concentration must be attained (for example, by adding \(\mathrm{NaOH}\) ) to make the \(\mathrm{Mg}^{2+}\) concentration in a solution of \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) less than \(1.0 \times 10^{-10} M ?\)

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?
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