Question

The Ksp of Mg(OH)2 is 9.0 x 10‐12. If a solution is 0.045M with respect to Mg2+ ion. Is the ionization of Mg(OH)2 at equilibrium if the pH is adjusted to 9.1505? Show all work including the balanced equation.

Answer 1

Ang? At equillibricen Mg(OH)ees) = Mg?? + 2 OH Ksp = [Mget] [oH] ² eq (1) Griven. Ksp = 9.0x1012 concentraction of Matt - [m ] = 0.045M To be in a equillibrium [ot] must satisfy the above - @4 (1) given, pH = 9.1505 pOH = 14-9.1505 = 4.8495 { poH = 14 - PH} pOH = -logio [OH) 4.8495= -logio Com] [on'] = 1044.8495) = # FAA 1.414164 x105m [on'] = (1-414164x105)2 = 2X100 M LMS= Ksp = g.ox 1012 RHS = [Mg2t] [or] ² = 0.045 x 8 x 10-10 = 0.09 x 1080 = 9x162 LHS = RMS

conclusion res the ionization of Mg(OH)2 is at equilibrium when the pH is adjusted to 9.1505.