MgCl2 + 2NaOH ==> Mg(OH)2 + 2NaCl
millimols MgCl2 = 10.0 x 0.18 = 1.8
mmols NaOH = 20.1 x 0.56 = about 11.3 but you should do these more accurately.
The remaining steps are approximations too.
1.8 mmols MgCl2 will use 3.6 mmols NaOH and form 1.8 mols Mg(OH)2.
The solubility of Mg(OH)2 is determined by the excess of the common ion which in this case is NaOH.
NaOH not used in the reaction is 11.3-3.6 = about 7.7
(OH^-) = mmols/mL = 7.7/(10+20.1) = about 0.255M
Ksp = (Mg^2+)(OH^-)^2
Plug in Ksp and OH and solve for Mg concn.
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.
1 answer
1 answer