Question

Will a precipitate form when 100.0 ml of 7.8 x 10⁻⁴ m mg(NO₃)₂ is added to 100.0 ml of 1.6 x 10⁻⁴ m NaOH?

The ion product for Mg(OH)₂ is Since Q is than Ksp, Mg(OH)₂ precipitate from the solution

Answer 1

A precipitate of Mg(OH)2 will form in the mixture since the reaction quotient (Q) is greater than the solubility product constant (Ksp) after calculating the ion concentrations post-mixing.

Step-by-step explanation:

To determine if a precipitate will form when solutions of Mg(NO3)2 and NaOH are mixed, the ion product (Q) of Mg(OH)2 needs to be compared with its solubility product constant (Ksp). Given that the concentrations of Mg2+ and OH- are 7.8 x 10-4 M and 1.6 x 10-4 M when the solutions are mixed, the final concentration of each ion in the mixture can be calculated due to the dilution which occurs when the solutions are mixed.

The reaction for the dissolution of Mg(OH)2 is: Mg(OH)2 (s) ⇒ Mg2+ (aq) + 2OH- (aq). The Ksp is given as 8.9 x 10-12. The Q of the reaction equals [Mg2+][OH-]2, and if Q > Ksp, then precipitation will occur.

First, we determine the new concentrations after mixing the solutions:

• The combined volume of the solutions is 200.0 mL (100.0 mL + 100.0 mL).
• The concentration of Mg2+ after dilution is 7.8 x 10-4 M / 2 = 3.9 x 10-4 M.
• The concentration of OH- after dilution is 1.6 x 10-4 M / 2 = 0.8 x 10-4 M.

Calculating the reaction quotient, Q:

• Q = [Mg2+][OH-]2 = (3.9 x 10-4)(0.8 x 10-4)2
• Q = 3.9 x 10-4 x 0.64 x 10-8
• Q = 2.496 x 10-11

Since Q (2.496 x 10-11) is greater than Ksp (8.9 x 10-12), a precipitate of Mg(OH)2 will form in the mixture.