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Write a net ionic equation for the reaction with \(\mathrm{OH}^{-}\) by which (a) \(\mathrm{Sb}^{3+}\) forms a precipitate. (b) antimony(III) hydroxide dissolves when more \(\mathrm{OH}^{-}\) is added. (c) \(\mathrm{Sb}^{3+}\) forms a complex ion.

Short Answer

Expert verified
Question: Write the net ionic equation for the following reactions: a) Sb鲁鈦 forms a precipitate with OH鈦. b) Antimony(III) hydroxide dissolves when more OH鈦 is added. c) Sb鲁鈦 forms a complex ion with OH鈦. Answer: a) Sb鲁鈦(aq) + 3OH鈦(aq) 鈫 Sb(OH)鈧(s) b) Sb(OH)鈧(s) + OH鈦(aq) 鈫 Sb(OH)鈧勨伝(aq) c) Sb(OH)鈧(s) + OH鈦(aq) 鈫 Sb(OH)鈧勨伝(aq)

Step by step solution

01

Write the balanced molecular equation

Sb鲁鈦(aq) + 3OH鈦(aq) 鈫 Sb(OH)鈧(s)
02

Write the complete ionic equation

Sb鲁鈦(aq) + 3OH鈦(aq) 鈫 Sb(OH)鈧(s)
03

Write the net ionic equation

Sb鲁鈦(aq) + 3OH鈦(aq) 鈫 Sb(OH)鈧(s) As there are no spectator ions present in this reaction, the complete ionic equation is the same as the net ionic equation. #b) Dissolution of Sb(OH)鈧 with excess OH鈦#
04

Write the balanced molecular equation

Sb(OH)鈧(s) + OH鈦(aq) 鈫 Sb(OH)鈧勨伝(aq)
05

Write the complete ionic equation

Sb(OH)鈧(s) + OH鈦(aq) 鈫 Sb(OH)鈧勨伝(aq)
06

Write the net ionic equation

Sb(OH)鈧(s) + OH鈦(aq) 鈫 Sb(OH)鈧勨伝(aq) As the complete ionic equation already contains only net ionic species, it is also the net ionic equation. #c) Formation of a complex ion#
07

Write the balanced molecular equation

The balanced molecular equation is the same as the equation for dissolving antimony(III) hydroxide with excess OH鈦 since it forms the [Sb(OH)鈧刔鈦 complex ion. Sb(OH)鈧(s) + OH鈦(aq) 鈫 Sb(OH)鈧勨伝(aq)
08

Write the complete ionic equation

The complete ionic equation is the same as the equation for dissolving antimony(III) hydroxide with excess OH鈦. Sb(OH)鈧(s) + OH鈦(aq) 鈫 Sb(OH)鈧勨伝(aq)
09

Write the net ionic equation

The net ionic equation is the same as the equation for dissolving antimony(III) hydroxide with excess OH鈦. Sb(OH)鈧(s) + OH鈦(aq) 鈫 Sb(OH)鈧勨伝(aq)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reactions
Chemical reactions are integral to understanding how substances interact to form new products. In a reaction, the reactants undergo a transformation leading to the formation of products. This can be illustrated through an equation showing reactants on one side and products on the other. A key aspect of chemical reactions is the conservation of mass, meaning the number of atoms for each element must be the same on both sides of the equation.

For example, when \(\mathrm{Sb}^{3+}\) reacts with \(\mathrm{OH}^-\) ions, they combine to produce a precipitate, \(\mathrm{Sb(OH)_3}\), in a precipitation reaction. The equation \(\mathrm{Sb^{3+}(aq) + 3OH^- (aq) \rightarrow Sb(OH)_3(s)}\) is balanced and shows the stoichiometry of reactants and product, adhering to the law of conservation of mass.
Solubility Equilibrium
Solubility equilibrium is a fundamental concept when we talk about the dissolution or precipitation of substances. This dynamic equilibrium exists between the dissolved ions and the undissolved substance. If a substance's solubility limit is exceeded, excess solute will come out of the solution as a precipitate.

Understanding solubility is pivotal in predicting whether a precipitate will form in a reaction. For instance, when the \(\mathrm{Sb}^{3+}\) ions interact with a limited amount of \(\mathrm{OH}^-\) ions, a precipitate of \(\mathrm{Sb(OH)_3}\) forms because the product is not soluble in water. This is represented by the net ionic equation without any spectator ions, showing the direct transformation from reactants to the solid precipitate.
Complex Ion Formation
Complex ion formation involves the combination of simple ions to form a new, larger ion. This process is especially important in understanding chemical speciation in solutions. It often occurs when a metal ion binds to multiple ligands, which can be anions or neutral molecules, leading to the formation of a coordination complex.

For example, when more \(\mathrm{OH}^-\) is added to a solution-containing \(\mathrm{Sb(OH)_3}\), instead of forming a precipitate, it further reacts to form \(\mathrm{Sb(OH)_4^-}\), a complex ion. This process can be described as the dissolving of a precipitate to form a complex ion in solution, and the net ionic equation would show the conversion of the solid precipitate directly into the complex ion.

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Most popular questions from this chapter

Which of the following statements are true? (a) For an insoluble metallic salt, \(K_{\text {sp }}\) is always less than 1 . (b) More \(\mathrm{PbCl}_{2}\) can be dissolved at \(100^{\circ} \mathrm{C}\) than at \(25^{\circ} \mathrm{C}\). One can conclude that dissolving \(\mathrm{PbCl}_{2}\) is an exothermic process. (c) When strips of copper metal are added to a saturated solution of \(\mathrm{Cu}(\mathrm{OH})_{2}\), a precipitate of \(\mathrm{Cu}(\mathrm{OH})_{2}\) can be expected to form because of the common ion effect.

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Write a net ionic equation for the reaction with ammonia by which (a) \(\mathrm{Cu}(\mathrm{OH})_{2}\) dissolves. (b) \(\mathrm{Cd}^{2+}\) forms a complex ion. (c) \(\mathrm{Pb}^{2+}\) forms a precipitate.

A plot of the solubility of a certain compound \(\left(\mathrm{g} / 100 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\right)\) against temperature \(\left({ }^{\circ} \mathrm{C}\right)\) is a straight line with a positive slope. Is dissolving that compound an exothermic process?

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