If 10.0 mL of 0.18 M MgCl2 are mixed with 20.1 mL of 0.56 M NaOH, what will be the final concentration of Mg2+ in solution when equilibrium is established? Assume the volumes are additive. Ksp = 1.2 ✕ 10-11 for Mg(OH)2.
To find the final concentration of Mg2+ in solution, we need to determine if a precipitate of Mg(OH)2 will form and, if so, how much Mg2+ remains in solution.
First, let's write the balanced equation for the reaction between MgCl2 and NaOH:
MgCl2 + 2NaOH -> Mg(OH)2 + 2NaCl
Based on the stoichiometry of the balanced equation, we can determine that for every 1 mole of Mg(OH)2 that forms, 1 mole of Mg2+ is used:
1 mol Mg(OH)2 -> 1 mol Mg2+
We need to calculate the number of moles of Mg(OH)2 that can form based on the limiting reagent. To do this, let's compare the number of moles in the reactants.
Moles of MgCl2 = 0.010 L * 0.18 mol/L = 0.0018 mol MgCl2
Moles of NaOH = 0.0201 L * 0.56 mol/L = 0.011256 mol NaOH
From the balanced equation, we can see that 1 mole of MgCl2 reacts with 2 moles of NaOH to produce 1 mole of Mg(OH)2. Therefore, the moles of Mg(OH)2 that can form are limited by the amount of MgCl2, and the moles of Mg(OH)2 are equal to the moles of MgCl2.
Now, let's calculate the concentration of Mg2+ in solution after the reaction. We will assume that all the Mg(OH)2 that forms will dissociate into Mg2+ and OH- ions.
Concentration of Mg2+ in solution = Moles of Mg2+ / Total volume of solution
The moles of Mg2+ equal the moles of Mg(OH)2, which is equal to the moles of MgCl2. Therefore, Moles of Mg2+ = 0.0018 mol.
The total volume of the solution is the sum of the initial volumes of MgCl2 and NaOH. Therefore, Total volume = 10.0 mL + 20.1 mL = 30.1 mL = 0.0301 L.
Finally, we can calculate the concentration of Mg2+ in solution:
Concentration of Mg2+ = Moles of Mg2+ / Total volume of solution
= 0.0018 mol / 0.0301 L
≈ 0.0598 M
Therefore, the final concentration of Mg2+ in solution, when equilibrium is established, will be approximately 0.0598 M.