chem please help meeeeeeeeeee?

Ksp = [Mg++]*[OH-]^2

let [OH-] =X then [Mg++] = 0.5X (1 to 2 mole ratio)

7.1E-12 = 0.5X * X^2 = 0.5X^3

X = 2.42E-4

p(OH) = 3.616

pH = 10.38

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Let X = moles per liter of Mg++ that goes into solution

Let 2X = moles per liter of OH- that goes into solution

The equilibrium reaction is Mg(OH)2(s) at equilib with Mg++(aq) plus 2OH-(aq)

Ksp = 7.1 x l0^-12 = (X)(2X)^2 or (Mg++)(OH-)^2

7.1 x l0^-12 = 4X^3

X^3 = 7.l x l0^-12 over 4 which equals. l.77 x l0^-12

and X = l.9 x l0^4 so 2X which is the concentration of OH- = 3.84 x l0^-4

:pOH = -log (OH-) or -log (3.84 x l0^-4) which = 3.42

and pH = l4-pOH or l4-3.42 which = 10.58

2. 2.61 Amps times 261 seconds = 681.2 coulombs.

One mole of electrons carries a charge of 96,485 coulombs.

So we have passed 681.2 over 96485 or .0071 moles of electrons through the solution which will plate out only half this many moles of Cu++ ions due to the fact that each ion has a charge of plus 2, so we are plating out .0071 moles over 2 which = .0035 moles Cu++ plated out or reduced to Cu metal.

The original number of moles of Cu++ in the solution = Molarity times volume in liters.

.742 Molar times .444 Liters = .329 moles original Cu++ in solution so Cu++ remaining would be .329 moles original minus .0035 moles plated out or ,326 moles Cu++ still in solution.

b. Moles Ag+ reduced = .l92 molar minus .121 molar or .07l moles Ag+ reduced which would also require .071 moles of electrons

Coulombs required = 96485 coulombs per mole e- times .071 moles e- = 6850.4 coulombs.

And if you are plating out at 4.72 Amps, this means that you are passing 4.72 coulombs of charge per second.l

So to get seconds, divide total coulombs passed, 6850.4 by 4.72 coulombs per second which =

1451.4 seconds.