Asked by Marina
The solubility of insoluble substances changes depending on the nature of the solution. Below are two solutions in which copper (II) hydroxide is dissolved; in each case, the solubility is not the same as it is in pure water.
a. For each solution state whether the solubility is greater than or less than that of pure water and briefly account for the change in solubility.
i. 1-molar HCL (aq)
ii. 1-molar copper (II) nitrate (aq)
a. For each solution state whether the solubility is greater than or less than that of pure water and briefly account for the change in solubility.
i. 1-molar HCL (aq)
ii. 1-molar copper (II) nitrate (aq)
Answers
Answered by
DrBob222
Cu(OH)2 ==> Cu^2+ + 2OH^-
i. Adding HCl. The H^+ reacts with the OH^- to produce H2O (H^+ + OH^- => H2O), the OH^- is reduced and that shifts the equilibrium to the right (to increase the OH) which means the solubility is increased.
ii. Cu(NO3)2 ==> Cu^2+ + 2NO3^2-
Adding Cu^2+ to the Cu(OH)2 increases the Cu^2+ so the solution must try to reduce the Cu^2+ and it can do that by shifting to the left which decreases the solubility of the Cu(OH)2.
i. Adding HCl. The H^+ reacts with the OH^- to produce H2O (H^+ + OH^- => H2O), the OH^- is reduced and that shifts the equilibrium to the right (to increase the OH) which means the solubility is increased.
ii. Cu(NO3)2 ==> Cu^2+ + 2NO3^2-
Adding Cu^2+ to the Cu(OH)2 increases the Cu^2+ so the solution must try to reduce the Cu^2+ and it can do that by shifting to the left which decreases the solubility of the Cu(OH)2.
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