Question 19.11: Using Molecular Scenes to Predict Whether a Precipitate Will......
Using Molecular Scenes to Predict Whether a Precipitate Will Form
Problem These four scenes (Fig 19.11) represent solutions of silver (gray) and carbonate (black and red) ions above solid silver carbonate. (The solid, other ions, and water are not shown.)
(a) Which scene best represents the solution in equilibrium with the solid?
(b) In which, if any, other scene(s) will additional solid silver carbonate form?
(c) Explain how, if at all, addition of a small volume of concentrated strong acid affects the [Ag^+] in scene 4 and the mass of solid present.
Plan (a) The solution of silver and carbonate ions in equilibrium with the solid (Ag_2CO_3) should have the same relative numbers of cations and anions as in the formula. We examine the scenes to see which has a ratio of 2 Ag^+ to 1 CO_3^{2−}. (b) A solid forms if the value of Q_{sp} exceeds that of K_{sp}. We write the dissolution equation and Q_{sp} expression. Then we count ions to calculate Q_{sp} for each scene and see which Q_{sp} value, if any, exceeds the value for the scene identified in part (a). (c) The CO_3^{2−} ion reacts with added H_3O^+, so adding strong acid will shift the equilibrium to the right. We write the equations and determine how a shift to the right affects [Ag^+] and the mass of solid Ag_2CO_3.
Solution (a) Scene 3 is the only one with an Ag^+/CO_3^{2−} ratio of 2/1, as in the solid’s formula.
(b) Calculating the ion products:
Scene 1: Q_{sp} = (2)^2(4) = 16 Scene 2: Q_{sp} = (3)^2(3) = 27
Scene 3: Q_{sp} = (4)^2(2) = 32 Scene 4: Q_{sp} = (3)^2(4) = 36
Therefore, for scene 3, K_{sp} = 32; the Q_{sp} value for scene 4 is the only other one that equals or exceeds 32, so a precipitate of Ag_2CO_3 will form there.
(c) Writing the equations:
(1) Ag_2CO_3(s) \xrightleftharpoons[]{} 2Ag^+(aq) + CO_3^{2−}(aq)
(2) CO_3^{2−}(aq) + 2H_3O^+(aq) ⟶ [H_2CO_3(aq)] + 2H_2O(l) ⟶ 3 H_2O(l) + CO_2(g)
The CO_2 leaves as a gas, so adding H_3O^+ shifts the equilibrium position of reaction 2 to the right. This change lowers the [CO_3^{2−}] in reaction 1, thereby causing more CO_3^{2−} to form. As a result, more solid dissolves, which means that the [Ag^+] increases and the mass of Ag_2CO_3 decreases.
Check (a) In scene 1, the formula has two CO_3^{2−} per formula unit, not two Ag^+.
(b) Even though scene 4 has fewer Ag^+ ions than scene 3, its Q_{sp} value is higher and exceeds the K_{sp}.