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Chapter 16: Problem 3

Write the equilibrium equations on which the following \(K_{s p}\) expressions are based. (a) \(\left[\mathrm{Hg}_{2}{ }^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2}\) (b) \(\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{CrO}_{4}^{2-}\right]\) (c) \(\left[\mathrm{Mn}^{4+}\right]\left[\mathrm{O}^{2-}\right]^{2}\) (d) \(\left[\mathrm{Al}^{3+}\right]^{2}\left[\mathrm{~S}^{2-}\right]^{3}\)

Short Answer

Expert verified
Q: Write the equilibrium equations for each given Ksp expression: a) [Hg2^2+][Cl^-]^2 b) [Pb^2+][CrO4^2-] c) [Mn^4+][O^2-]^2 d) [Al^3+]^2[S^2-]^3 A: a) Hg2Cl2(s) ⇌ Hg2^2+(aq) + 2Cl^-(aq) b) PbCrO4(s) ⇌ Pb^2+(aq) + CrO4^2-(aq) c) MnO2(s) ⇌ Mn^4+(aq) + 2O^2-(aq) d) Al2S3(s) ⇌ 2Al^3+(aq) + 3S^2-(aq)

Step by step solution

01

(a) Writing the equilibrium equation for \(\left[\mathrm{Hg}_{2}{ }^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2}\)

Begin by identifying the ions in the expression. We have Hg\(_{2}^{2+}\) and Cl\(^-\). The exponents in the expression are their respective stoichiometric coefficients from the balanced chemical equation. In this case, the equation will be: \begin{equation} \mathrm{Hg}_{2}\mathrm{Cl}_{2}(s) \rightleftharpoons \mathrm{Hg}_{2}^{2+}(aq) + 2\mathrm{Cl}^{-}(aq) \end{equation}
02

(b) Writing the equilibrium equation for \(\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{CrO}_{4}^{2-}\right]\)

Identify the ions: Pb\(^{2+}\) and CrO\(_4^{2-}\). Write the balanced chemical equation based on the exponents in the expression: \begin{equation} \mathrm{PbCrO}_{4}(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + \mathrm{CrO}_{4}^{2-}(aq) \end{equation}
03

(c) Writing the equilibrium equation for \(\left[\mathrm{Mn}^{4+}\right]\left[\mathrm{O}^{2-}\right]^{2}\)

Identify the ions: Mn\(^{4+}\) and O\(^{2-}\). Write the balanced chemical equation based on the exponents in the expression: \begin{equation} \mathrm{MnO}_{2}(s) \rightleftharpoons \mathrm{Mn}^{4+}(aq) + 2\mathrm{O}^{2-}(aq) \end{equation}
04

(d) Writing the equilibrium equation for \(\left[\mathrm{Al}^{3+}\right]^{2}\left[\mathrm{~S}^{2-}\right]^{3}\)

Identify the ions: Al\(^{3+}\) and S\(^{2-}\). Write the balanced chemical equation based on the exponents in the expression: \begin{equation} \mathrm{Al}_{2}\mathrm{S}_{3}(s) \rightleftharpoons 2\mathrm{Al}^{3+}(aq) + 3\mathrm{S}^{2-}(aq) \end{equation}

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Most popular questions from this chapter

Before lead in paint was discontinued, lead chromate was a common pigment in yellow paint. A \(1.0\) -L solution is prepared by mixing \(0.50 \mathrm{mg}\) of lead nitrate with \(0.020 \mathrm{mg}\) of potassium chromate. Will a precipitate form? What should \(\left[\mathrm{Pb}^{2+}\right]\) be to just start precipitation?

Write a net ionic equation for the reaction with ammonia by which (a) silver chloride dissolves. (b) aluminum ion forms a precipitate. (c) copper(II) forms a complex ion.

Write a net ionic equation for the reaction with \(\mathrm{OH}^{-}\) by which (a) \(\mathrm{Ni}^{2+}\) forms a precipitate. (b) \(\mathrm{Sn}^{4+}\) forms a complex ion. (c) \(\mathrm{Al}(\mathrm{OH})_{3}\) dissolves.

The box below represents one liter of a saturated solution of the species \(\square\), where squares represent the cation and circles represent the anion. Water molecules, though present, are not shown. Complete the next three figures below by filling one-liter boxes to the right of the arrow, showing the state of the ions after water is added to form saturated solutions. The species represented to the left of the arrow is the solid form of the ions represented above. Do not show the water molecules.

Given \(K_{s p}\) and the equilibrium concentration of one ion, calculate the equilibrium concentration of the other ion. (a) cadmium(II) hydroxide: \(K_{\text {sp }}=2.5 \times 10^{-14} ;\left[\mathrm{Cd}^{2+}\right]=1.5 \times 10^{-6} M\) (b) copper(II) arsenate \(\left(\mathrm{Cu}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\right): K_{\mathrm{sp}}=7.6 \times 10^{-36} ;\left[\mathrm{AsO}_{4}^{3-}\right]=\) \(2.4 \times 10^{-4} M\) (c) zinc oxalate: \(K_{s p}=2.7 \times 10^{-8} ;\left[\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right]=8.8 \times 10^{-3} M\)
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