posted by Gregory .
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
(Mg^2+) = ?
(OH^-) = 12.47/(11.7+ 24.9) = about 0.341M
.......Mg(OH)2 ==> Mg^2+ + 2OH^-
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.