At equilibrium, a 4.50 L container has 2.6 g of carbon, CO2 at a partial pressure of 0.0020 atm, and a total pressure of 0.572 atm. Calculate KP for this reaction at 725oC.

The given question is incomplete.. The complete question is :

Consider the following reaction:

At equilibrium, a 4.50 L container has 2.6 g of carbon, CO2 at a partial pressure of 0.0020 atm, and a total pressure of 0.572 atm. Calculate KP for this reaction at 725oC.

162.45

Explanation:

Equilibrium pressure of = 0.0020 atm

Total pressure at equilibrium = 0.572 atm

Equilibrium pressure of = Total pressure at equilibrium - Equilibrium pressure of =  (0.572- 0.0020) = 0.570 atm

The given balanced equilibrium reaction is,

                         

At eqm               0        0.0020        0.570

The expression for equilibrium constant for this reaction will be,

Now put all the given values in this expression, we get :

By solving we get :

Thus for this reaction at is 162.45

The ph of a saturated solution of Ca(OH)2 is 12.35

CALCULATION:
For the reaction 
     Ca(OH)2 → Ca2+ + 2OH- 
we will use the Ksp expression to solve for the concentration [OH-] and then use the acid base concepts to get the pH:
     Ksp = [Ca2+][OH-]^2

The listed Ksp value is 5.5 x 10^-6. Substituting this to the Ksp expression, we have
     Ksp = 5.5 x 10^-6 = (s) (2s)^2 = 4s^3
     s3 = 5.5x10^-6 / 4

Taking the cube root, we now have
     s = cube root of (5.5x10^-6 / 4)s
        = 0.01112

We know that the value of [OH-] is actually equal to 2s:
     [OH-] = 2s = 2 * 0.01112 = 0.02224 M

We can now calculate for pOH:          
     pOH = - log [OH-]
              = -log(0.02224)
              = 1.65

Therefore, the pH is
     pH = 14 - pOH
           = 14 - 1.65
           = 12.35