Explain, by writing appropriate equations, how the following...

Explain, by writing appropriate equations, how the following insoluble compounds can be dissolved by the addition of a solution of nitric acid. (Carbonates dissolve in strong acids to form water and gaseous carbon dioxide.) What is the "driving force" for each reaction? (a) $\mathrm{Cu}(\mathrm{OH})_{2}$, (b) $\mathrm{Sn}(\mathrm{OH})_{4}$ (c) $\mathrm{ZnCO}_{3}$ (d) $(\mathrm{PbOH})_{2} \mathrm{CO}_{3}$

Explain, by writing appropriate equations, how the following insoluble compounds can be dissolved by the addition of a solution of nitric acid. (Carbonates dissolve in strong acids to form water and gaseous carbon dioxide.) What is the "driving force" for each reaction?
(a) $\mathrm{Cu}(\mathrm{OH})_{2}$,
(b) $\mathrm{Sn}(\mathrm{OH})_{4}$
(c) $\mathrm{ZnCO}_{3}$
(d) $(\mathrm{PbOH})_{2} \mathrm{CO}_{3}$

Key Concepts

Solubility Equilibria and Reaction Driving Forces

This key concept relates to the inherent equilibrium between a sparingly soluble solid and its ions in solution. When an insoluble metal compound reacts with an acid, the formation of soluble salts and the removal of reaction products (like water or CO2) drives the equilibrium toward further dissolution, effectively increasing the solubility of the originally insoluble compound.

Acid–Base Neutralization

This concept describes the reaction between acids and bases, where an acid reacts with a basic substance (such as a metal hydroxide) to produce water and a salt. In the case of metal hydroxides reacting with nitric acid, the acid neutralizes the hydroxide ion to form water, while the metal forms a soluble nitrate salt, leading to the dissolution of the insoluble compound.

Carbonate Decomposition and Gas Evolution

Many metal carbonates dissolve in acids due to the reaction where the carbonate ion reacts with protons to form water and carbon dioxide gas. The evolution of CO2 gas, which escapes from the solution, acts as a driving force that shifts the reaction equilibrium towards further dissolution of the carbonate, ensuring that the reaction goes to completion.

Transcript

00:01 In this question we're asked to look at how the addition of a strong acid, nitric acid, which is listed right here being a strong acid, it will completely dissociate into its respective ions, the hydrogen cation and the nitrate anion.

00:15 How adding it will help dissolve some of the solids that we are addressing here.

00:21 The first solid in part a is a copper 2 hydroxide.

00:25 Now because it's a somewhat soluble solid, when it's in the presence of water it will to some extent dissociate into its respective ions as well, the copper 2 plus ion and a couple of hydroxide ions.

00:40 Now in nitric acid the nitrate is kind of a spectator ion here, it's not really involved in the chemistry that we're about to see, but it's the hydrogen cation right here that's going to be the reactive part that helps to dissolve the copper 2 hydroxide and the other solids that we're going to see here.

00:58 So if we kind of just take this part of the reaction and say if i add two hydrogen cations that are supplied from the nitric acid here, that will react with the two hydroxides that are formed from the dissolution of copper 2 hydroxide to form two water molecules.

01:16 That's a chemical reaction, and when that happens that uses up some of the hydroxide ion.

01:24 It disappears essentially and becomes water when it reacts with the hydrogen cations from nitric acid.

01:31 Now we know from le châtelier's principle that any time we remove a product ion from a solution, the system must adjust itself to replace that lost product ion.

01:45 And so to replace the lost hydroxide ions that reacted here, this solid must dissolve more to supply more copper ions and replace the lost hydroxide ions.

01:58 So that increases the solubility as we react away hydroxide ions.

02:04 That's the driving force for the reactions that we're going to see.

02:08 The le châtelier's principle of shifting a reaction to replace lost product ions.

02:15 In part b we have something very similar here.

02:17 We have another hydroxide.

02:19 It's a tin 4 hydroxide in this case, but like the copper hydroxide, somewhat soluble so it will dissociate to some extent into its respective ions.

02:29 In this case the tin 4 plus ion and 4 hydroxides.

02:33 Just like the previous example, it's the reaction between the hydrogen ions that are supplied...

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