If 11.6 mL of 0.17 M MgCl2 are mixed with 24.9 mL of 0.58 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
11.6 mL x 0.17M = 1.972 mmol MgCl2.
24.9 mL x 0.58M = 14.44 mmol NaOH.
....MgCl2 + 2NaOH ==> Mg(OH)2 + 2NaCl
I...1.972...14.44.......0........0
C...-1.972.-1.972.....1.972.......
E.....0.....12.47.....1.972
(Mg^2+) = ?
(OH^-) = 12.47/(11.7+ 24.9) = about 0.341M
.......Mg(OH)2 ==> Mg^2+ + 2OH^-
I......solid.......0.......0.341
C.......solid......x........2x
C......solid.......x......0.341+2x
Ksp = (Mg^2+)(*OH^-)^2
Substitute the E line into Ksp and solve for Mg.
Note that this is a limiting reagent problem (you know that because amounts are given for BOTH reactants) as well as a common ion (the OH^-) is common to the Ksp equilibrium of Mg(OH)2 and that decreases the solubility of Mg(OH)2. Check my work.
I keep getting 1.0e-10 for x. I even tried 1.03e-10. The answer is incorrect. I checked your work, and everything was calculated correctly, I'm not sure what went wrong.
I think I see the problem. For the first ICE box, shouldn't C for NaOH and NaCl be + 2*(1.972)?
Pardon me, I meant -2(1.972) for NaOH and + 2(1.972) for NaCl
This revision was correct. Thank you so much for helping, I actually understand how to solve the problem now.
I guess you didn't check closely enough. I made an error on the C line of the ICE chart.
If you use 1.972 mmol Mg^2+ it will take 2 times mol NaOH (that's 1.972 x 2) = 3.944 of NaOH which leaves just 14.44-3.944 = 10.496 and that convert to M = 10.496/(36.5) = about 0.287 or so and THAT is the common ion. That makes about 1.451E-10 for the final answer. Check it again and watch the significant figures. If the 0.17 and 0.58 are valid numbers (i.e., you did NOT omit a zero on each), and the Ksp is only two places, then you are allowed only two places and I would round the 1.451E-10 to 1.4E-10. Some would round to 1.5E-10.
I hope this clears it up.
To find the final concentration of Mg2+ in solution when equilibrium is established, we need to consider the reaction between MgCl2 and NaOH, which will result in the formation of Mg(OH)2.
The balanced chemical equation for the reaction is:
MgCl2 + 2NaOH -> Mg(OH)2 + 2NaCl
First, we need to calculate the moles of MgCl2 and NaOH.
Moles of MgCl2 = volume of MgCl2 solution (in L) × concentration of MgCl2 (in mol/L)
= 11.6 mL × (1 L / 1000 mL) × 0.17 mol/L
= 0.001972 mol
Moles of NaOH = volume of NaOH solution (in L) × concentration of NaOH (in mol/L)
= 24.9 mL × (1 L / 1000 mL) × 0.58 mol/L
= 0.014442 mol
Next, we need to determine which reactant will be the limiting reactant, i.e., which one will be completely consumed in the reaction. This can be done by comparing the moles of MgCl2 and NaOH.
From the balanced equation, we can see that 1 mole of MgCl2 reacts with 2 moles of NaOH to form 1 mole of Mg(OH)2. Therefore, the ratio of moles of MgCl2 to NaOH is 1:2.
Since the ratio is 1:2, we can calculate the moles of MgCl2 needed to react with the available moles of NaOH:
Moles of MgCl2 needed = 2 × moles of NaOH
= 2 × 0.014442 mol
= 0.028884 mol
Since the moles of MgCl2 available (0.001972 mol) are less than the moles needed (0.028884 mol), MgCl2 is the limiting reactant.
Now, we can calculate the moles of Mg(OH)2 formed using the stoichiometry of the balanced equation:
Moles of Mg(OH)2 formed = moles of limiting reactant (MgCl2)
= 0.001972 mol
Finally, we can calculate the concentration of Mg2+ in solution when equilibrium is established, considering that Mg(OH)2 is sparingly soluble and will reach equilibrium based on its solubility product constant (Ksp).
Ksp = [Mg2+][OH-]^2
Since the stoichiometric ratio between Mg2+ and OH- ions is 1:2, the concentration of Mg2+ will be two times the concentration of OH-.
[Mg2+] = 2 × [OH-]
The concentration of OH- can be determined from the balanced equation, as 1 mole of Mg(OH)2 is formed for every 2 moles of NaOH, and the volume is additive.
Volume of final solution = volume of MgCl2 solution + volume of NaOH solution
= 11.6 mL + 24.9 mL
= 36.5 mL
Concentration of OH- = moles of NaOH / volume of final solution (in L)
= 0.014442 mol / (36.5 mL × (1 L / 1000 mL))
= 0.395 mol/L
Finally, the concentration of Mg2+ can be calculated:
[Mg2+] = 2 × [OH-]
= 2 × 0.395 mol/L
= 0.79 mol/L
Therefore, the final concentration of Mg2+ in solution, when equilibrium is established, will be 0.79 mol/L.