Ionic Equilibrium - use of solids as indicators

Indicators are used for you to help us know that a particular reactant is completely used up for a reaction. Through acid-base titration, we have seen that indicators inform us that a reactant is completed used through  a change in colour of the solution.  These indicators first create a colour in the original solution and when the end-point is reached, they will show a change in colour.

However, a colour change  in the solution is not the only way to indicate that the reaction is completed. Another way is to form a coloured solid while doing a titration. An example would be to determine the concentration of Cl^- ions present in a solution by titrating the solution against a standard solution of AgNO₃.

(A) Scenario: How it is used.

When you add Ag⁺ ions to a solution containing Cl^- ions, a white precipitate is seen. This happens because the K_sp value of AgCl is very small and hence the product of the concentrations of Ag⁺ ions and Cl^- ions (also known as ionic product) can easily be larger than the K_sp. This results in precipitation of the AgCl solid.  Chromate ions (CrO₄^2-) are added so that when a red precipitate of Ag₂CrO₄ is formed, you would that all the AgCl has been precipitated out.  This enables us to determine how much Ag⁺ ions was added into the reaction mixture and as a result allow us to know the amount of Cl^- ions present initially.

(B) Explanation:  How this works.

The numerical values of the K_sp of AgCl and Ag₂CrO₄ are 1.77 x 10^-10 and 1.12 x 10^-12. Although the K_sp value of Ag₂CrO₄ is smaller, only a small concentration of CrO₄^2- ions is added while a larger concentration of Cl^- ions is present. Hence, it is easier for the ionic product of AgCl to exceeds its K_sp than that of Ag₂CrO₄.

Furthermore, when the red precipitate of Ag₂CrO₄ is formed, we can be quite certain that all of the Cl^- ions are precipitated out. This is because of the following:

• Although K_sp of the Ag₂CrO₄ is of a smaller value, the K_sp expression of silver chromate is essentially a concentration³ term. This is in comparison to silver chloride whose K_sp expression is a concentration². Hence, the solubility of silver chromate is greater than that of AgCl. This implies that the former is more soluble than the latter and it is quite fair to say it is more difficult to precipitate out too.

• In addition, for the same small concentration of Ag⁺ ions present (since both solids are insoluble you do not expect a large concentration of Ag⁺ ions), a larger concentration of CrO₄^2- ions is present in the solution than Cl^- ions. This would mean that a greater percentage of the Cl^- ions is precipitated out.

(C) The steps taken: The mathematics behind solving for the percentage of Cl^- ion precipitated.

When Ag₂CrO₄ is first formed, we have obtained a saturated solution of Ag₂CrO₄. Hence, by solving the K_sp of Ag₂CrO₄, we can obtain the concentration of Ag⁺ ions left in the solution. As both the equilibria which show the dissociation of AgCl and Ag₂CrO₄ exist in the same system, the concentration of Ag⁺ ions is the same for both equilibrium. It is not possible to obtain two distinct concentration of Ag⁺ ions (one for each insoluble solid doing slight dissociation) when both solids exist in the same solution. Therefore, the concentration of Ag⁺ ions is common for both ionic solids.

Hence, with this piece of information, when we can solve for the concentration of Cl^- ions in the solution using the K_sp expression of AgCl. This turns out to be a very small number which suggest almost all of the Cl^- ions has been precipitated out of the solution, therefore giving us an endpoint.

(D) Conclusion: Final thoughts.

Curiously, is it possible to have absolutely no Cl^- ions present in this context? The answer is no . As the solid AgCl remains in the reaction mixture, some would definitely dissolve giving rise to a concentration of Cl^- ions in the solution.

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^{Article written by Kwok YL 2011.}

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