Answer to board problem

45.0ml of 0.125M Magnesium chloride is added to 28.0 ml of 0.500M Sodium hydroxide.  Write the molecular, complete ionic and net ionic equations (noting spectator ions).  If a ppt is produced determine the theoretical yield.  Tell me what the pH of the initial  28.0 mls of base was before the magnesium chloride was added and tell me the final pH of the soln after the reactants were mixed.

Mg
2+aq + 2Cl-(aq) + 2Na+(aq) + 2OH-(aq) Mg(OH)2(s) + 2Na+(aq) + 2Cl-(aq)

pOH= -log0.5= 0.301 pH=14-0.301= 13.7


Spectators are Na
+ and Cl-

0.028L x 0.5M = 0.014mol NaOH

0.125M x 0.045L = .00563 mol MgCl
2

Need 2NaOH/1MgCl
2 or, 0.01126 mol…have 0.014 mol. So, NaOH is in excess.

MgCl
2 is limiting. So, make 0.00563 mol of Mg(OH)2
0.00563mol x 58.32g/mol= 0.328g Mg(OH)
2

Consume 2x(0.00563) mol NaOH= 0.01126 (.0113)

mol OH
- left = 0.014-0.01125=.00275 mol

divide by new volume (0.073)
0.00275mol/0.073L= 3.77e-2 M

[OH
-] = 3.77e-2 pOH= 1.42

pH 12.6