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Chapter 17: Problem 50

Give an example to illustrate the general effect of complex ion formation on solubility.

Short Answer

Expert verified
Complex ion formation enhances the solubility of an otherwise insoluble or sparingly soluble compound. An instance of this can be seen in the formation of a soluble complex ion by Aluminium hydroxide when reacted with hydroxide ions. This leads to more of the sparingly soluble Aluminium hydroxide dissolving to make up for the 'loss' of \(Al^{3+}\), thus significantly increasing its solubility.

Step by step solution

01

Concept Explanation

Complex ion formation occurs when a simple ion of a compound reacts with another molecule to form a complex ion. This is important in the context of solubility as complex ion formation increases the solubility of a compound by effectively reducing the concentration of its simple ions. In accordance with Le Chatelier's principle, this drives the dissolution of more of the compound.
02

Selecting an Example

Choose Aluminium hydroxide (\(Al(OH)_3\)) as an example. It has very low solubility in water due to its sparingly soluble nature.
03

Demonstrating Complex Ion Formation

By adding excess alkali to \(Al(OH)_3\) solution, it reacts with hydroxide ions (\(OH^-\)) to form a complex ion \(Al(OH)_4^-\). The equation is, \[Al(OH)_3 (s) + OH^- (aq) \rightarrow Al(OH)_4^- (aq)\] This added step shrinks the concentration of \(Al^{3+}\) ions, making it look as if they have nearly disappeared so, In response to the dwindling amount of \(Al^{3+}\), more \(Al(OH)_3\) will dissolve via Le Châtelier’s principle, directly upping the amount that dissolves.
04

Observing Increased Solubility

Following Le Chatelier's principle, due to the decrease in concentration of \(Al^{3+}\) ions, as they form complex ions, more \(Al(OH)_3\) dissolve to make up for the 'loss', leading to an increased solubility of Aluminium hydroxide. Hence, the formation of a soluble complex ion increases the solubility of an otherwise insoluble compound.

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Most popular questions from this chapter

Specify which of these systems can be classified as a buffer system: (a) \(\mathrm{KNO}_{2} / \mathrm{HNO}_{2}\) (b) \(\mathrm{KHSO}_{4} / \mathrm{H}_{2} \mathrm{SO}_{4}\) (c) HCOOK/HCOOH.

A sample of \(0.96 \mathrm{~L}\) of \(\mathrm{HCl}\) at \(372 \mathrm{mmHg}\) and \(22^{\circ} \mathrm{C}\) is bubbled into \(0.034 \mathrm{~L}\) of \(0.57 \mathrm{MH}_{3}\). What is the \(\mathrm{pH}\) of the resulting solution? Assume the volume of solution remains constant and that the \(\mathrm{HCl}\) is totally dissolved in the solution.

If \(\mathrm{NaOH}\) is added to \(0.010 \mathrm{MAl}^{3+}\), which will be the predominant species at equilibrium: \(\mathrm{Al}(\mathrm{OH})_{3}\) or \(\mathrm{Al}(\mathrm{OH})_{4}^{-} ?\) The \(\mathrm{pH}\) of the solution is \(14.00 .\left[K_{\mathrm{f}}\right.\) for \(\mathrm{Al}(\mathrm{OH})_{4}^{-}=2.0 \times 10^{33}\)

A sample of \(0.1276 \mathrm{~g}\) of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with \(0.0633 \mathrm{M} \mathrm{NaOH}\) solution. The volume of base required to reach the equivalence point was \(18.4 \mathrm{~mL}\). (a) Calculate the molar mass of the acid. (b) After \(10.0 \mathrm{~mL}\) of base had been added to the titration, the \(\mathrm{pH}\) was determined to be 5.87 . What is the \(K_{\mathrm{a}}\) of the unknown acid?

Define \(\mathrm{p} K_{\mathrm{a}}\) for a weak acid and explain the relationship between the value of the \(\mathrm{p} K_{\mathrm{a}}\) and the strength of the acid. Do the same for \(\mathrm{p} K_{\mathrm{b}}\) and a weak base.
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