Question

Which of the following salts will be substantially more soluble in acidic solution than in pure water: (a) \(\mathrm{ZnCO}_{3},(\mathbf{b}) \mathrm{ZnS}\), (c) \(\mathrm{BiI}_{3}\) (d) \(\mathrm{AgCN},(\mathrm{e}) \mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2} ?\)

Short answer

The salts with increased solubility in acidic solution are (a) \(\mathrm{ZnCO}_{3}\), (d) \(\mathrm{AgCN}\), and (e) \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}\), as their anionic components react with acidic protons, shifting the equilibrium towards solubility.

Step-by-step solution

Write solubility equilibria for each salt

First, we'll write the solubility equilibrium expressions for each of the given salts: (a) \(\mathrm{ZnCO}_3 \rightleftharpoons \mathrm{Zn}^{2+} + \mathrm{CO}_3^{2-}\) (b) \(\mathrm{ZnS} \rightleftharpoons \mathrm{Zn}^{2+} + \mathrm{S}^{2-}\) (c) \(\mathrm{BiI}_3 \rightleftharpoons \mathrm{Bi}^{3+} + 3\mathrm{I}^-\) (d) \(\mathrm{AgCN} \rightleftharpoons \mathrm{Ag}^+ + \mathrm{CN}^-\) (e) \(\mathrm{Ba}_3\left(\mathrm{PO}_4\right)_2 \rightleftharpoons 3\mathrm{Ba}^{2+} + 2\mathrm{PO}_4^{3-}\)

Check each equilibrium against the conditions of acidic solution

Now we need to analyze how each equilibrium would be affected by the presence of a strong acid, specifically looking for cases where the anion can react with the acid. The possible reactions are: (a) \(\mathrm{H^+} + \mathrm{CO}_3^{2-} \rightleftharpoons \mathrm{HCO}_3^-\) (b) No reaction, as sulfide (\(\mathrm{S}^{2-}\)) is not basic. (c) No reaction, as iodide (\(\mathrm{I}^-\)) is not basic. (d) \(\mathrm{H^+} + \mathrm{CN}^- \rightleftharpoons \mathrm{HCN}\) (e) \(\mathrm{H^+} + \mathrm{PO}_4^{3-} \rightleftharpoons \mathrm{HPO}_4^{2-}\)

Identify the salts with increased solubility in acidic solution

As seen in step 2, salts (a), (d), and (e) have anionic components that react with acidic protons. As a result, these salts will have increased solubility in an acidic solution as the acid consumes their respective anions, shifting the equilibrium towards solubility. Therefore, the correct choice is: (a) \(\mathrm{ZnCO}_{3}\) (d) \(\mathrm{AgCN}\) (e) \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

Key concepts

Salts Solubility in Acidic Solution

Understanding the solubility of salts in acidic solutions can be quite fascinating. Let's talk about why certain salts become more soluble when they encounter an acidic environment. It's a sort of chemical dance, where acids in the solution invite some salt anions for a tango, forming new products and, in turn, making room for more anions to leave the solid phase and join the solution.

Consider the scenario in our exercise, where the solubility of salts like \(\mathrm{ZnCO}_3\), \(\mathrm{AgCN}\), and \(\mathrm{Ba}_3(\mathrm{PO}_4)_2\), increases in an acidic environment. This happens because the anions in these salts (\(\mathrm{CO}_3^{2-}\), \(\mathrm{CN}^-\), and \(\mathrm{PO}_4^{3-}\)) can react with the excess \(\mathrm{H}^+\) ions from the acid to form new substances like \(\mathrm{HCO}_3^-\), \(\mathrm{HCN}\), and \(\mathrm{HPO}_4^{2-}\).

What we're seeing here is a 'reaction on the sidelines' that changes the main event — the solubility equilibrium. Since the anions are busy bonding with \(\mathrm{H}^+\), they are less available in the solution to reform the original solid salt, which tricks the equilibrium into dissolving more salt to balance things out. Simply put, the salt 'thinks' it needs to produce more dancers (ions) since the dance floor (the solution) got some extra space!

Chemical Equilibria

Chemical equilibria might remind you of a seesaw at a playground, constantly moving back and forth in an effort to stay balanced. At any moment, the amount of kids going up equals the number going down, achieving a state of dynamic balance. In chemistry, we describe this balance using the equilibrium constant (\(K_{sp}\)), which tells us how much of a substance will dissolve in solution at equilibrium.

Looking at our salts, \(\mathrm{ZnCO}_3\), \(\mathrm{ZnS}\), \(\mathrm{BiI}_3\), \(\mathrm{AgCN}\), and \(\mathrm{Ba}_3(\mathrm{PO}_4)_2\), we see different equilibrium expressions, each indicating the ratio of dissolved ions (the 'kids' going down the seesaw) to the undissolved solid (the 'kids' staying put on the ground). An acid effectively acts like a teacher entering the playground and calling some kids away – the seesaw dynamic changes because there's more room for motion. In our case, the 'kids' are the anions that react with the added acid, thus shifting the equilibrium and increasing solubility.

Common Ion Effect

The common ion effect is like having an uninvited guest turn up at a party and suddenly you have less space to dance! In chemical terms, this happens when you add an ion to a solution that is already present in the equilibrium system. The extra guest, or common ion, will push the equilibrium to favor the formation of the solid, otherwise known as 'shifting to the left', and the solubility of the salt decreases as a consequence.

In our textbook exercise, however, the conditions involve adding an acid, which provides \(\mathrm{H}^+\) ions that do not appear as a product in the dissolution of any of the listed salts, unless they can react with a salt's anions. Therefore, rather than demonstrating the common ion effect, the presence of a strong acid illustrates the opposite scenario - where the solubility of a salt is increased due to the consumption of its anions, as discussed in the previous sections. This 'anti-common ion effect', if you will, is crucial for understanding why certain salts are more soluble in acidic solutions.