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Chapter 16: Problem 69

When aqueous KI is added gradually to mercury (II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. (Hint: \(\mathrm{Hg}^{2+}\) reacts with \(\mathrm{I}^{-}\) to form \(\mathrm{HgI}_{4}^{2-}.)\)

Short Answer

Expert verified
In the first step, the formation of the orange precipitate occurs due to a double displacement reaction between mercury (II) nitrate (Hg(NO₃)₂) and potassium iodide (KI), resulting in insoluble mercury (II) iodide (HgI₂) and potassium nitrate (KNO₃): Hg(NO₃)₂ + 2 KI → HgI₂ + 2 KNO₃ In the second step, the dissolution of the precipitate occurs as more KI is added, causing the Hg²⁺ ions to react with I⁻ ions and form HgI₄²⁻ ions: HgI₂ + 2 I⁻ → HgI₄²⁻

Step by step solution

01

Write the equation for the formation of the orange precipitate

When aqueous potassium iodide (KI) is added to mercury (II) nitrate (Hg(NO₃)₂), a double displacement reaction occurs. This reaction forms mercury (II) iodide (HgI₂) and potassium nitrate (KNO₃). The balanced equation for this reaction is: Hg(NO₃)₂ + 2 KI → HgI₂ + 2 KNO₃ Since we observe an orange precipitate, we know that the mercury (II) iodide (HgI₂) is insoluble in water and precipitates out.
02

Write the equation for the dissolution of the orange precipitate

According to the hint, when more KI is added to the solution, the Hg²⁺ ions react with the I⁻ ions to form HgI₄²⁻ ions. This causes the previously formed orange precipitate of HgI₂ to dissolve. We'll write a balanced equation for this process: HgI₂ + 2 I⁻ → HgI₄²⁻ This equation shows that for every molecule of HgI₂, two iodide ions (I⁻) are needed to form HgI₄²⁻, allowing the precipitate to dissolve. Using these two balanced equations, we now understand both the formation and dissolution of the orange precipitate observed in this exercise.

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Most popular questions from this chapter

a. Using the \(K_{\mathrm{sp}}\) value for \(\mathrm{Cu}(\mathrm{OH})_{2}\left(1.6 \times 10^{-19}\right)\) and the overall formation constant for $\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\left(1.0 \times 10^{13}\right)$ calculate the value for the equilibrium constant for the following reaction: $$\mathrm{Cu}(\mathrm{OH})_{2}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)$$ b. Use the value of the equilibrium constant you calculated in part a to calculate the solubility (in mol/L) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0M\) \(\mathrm{NH}_{3} .\) In \(5.0M\) \(\mathrm{NH}_{3}\) the concentration of \(\mathrm{OH}^{-}\) is \(0.0095 M\) .

Write equations for the step wise formation of each of the following complex ions. a. \(N i(C N)_{4}^{2-}\) b. \(V\left(C_{2} O_{4}\right)_{3}^{3-}\)

A solution contains 0.25\(M \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.25\(M \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) Can the metal ions be separated by slowly adding \(\mathrm{Na}_{2} \mathrm{CO}_{3} ?\) Assume that for successful separation 99\(\%\) of the metal ion must be precipitated before the other metal ion begins to precipitate, and assume no volume change on addition of \(\mathrm{Na}_{2} \mathrm{CO}_{3}.\)

The copper(I) ion forms a chloride salt that has $K_{\mathrm{sp}}= 1.2 \times 10^{-6} .\( Copper (I) also forms a complex ion with \)\mathrm{Cl}^{-} :$ $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \qquad K=8.7 \times 10^{4}$$ a. Calculate the solubility of copper(I) chloride in pure water. (Ignore \(\mathrm{CuCl}_{2}^{-}\) formation for part a.) b. Calculate the solubility of copper(I) chloride in 0.10\(M\) \(\mathrm{NaCl}\).

When \(\mathrm{Na}_{3} \mathrm{PO}_{4}(a q)\) is added to a solution containing a metal ion and a precipitate forms, the precipitate generally could be one of two possibilities. What are the two possibilities?
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