Question

(a) The solubility product, K_sp, of Sn(OH)₄(s) is 1.0 x 10^-57. What is its solubility (in g/L) in an aqueous solution of NaOH with a pH of 12.80 ? The temperature is 25^oC.

(b) For the reaction 2 A(g) ⇋ 3 B(g) + 2 C(g), we start off with just pure A(g) (there is no B(g) or C(g)). When we reach equilibrium, the partial pressure of C(g) is 4.19 atm. The equilibrium constant for this reaction is 8.39 . What is the partial pressure of A(g) at equilibrium? What was the initial pressure of A(g)?

Answer 1

a) The initial concentration of OH- is calculated:

[OH-] = 10 ^ - (14 - pH) = 10 ^ - (14 - 12.8) = 0.063 M

We have the expression of Ksp:

Ksp = [Sn + 4] * [OH -] ^ 4

1E-57 = X * (0.063) ^ 4

X = 6.35E-53 M is cleared.

It is calculated in g / L:

g / L = 6.35E-53 mol / L * (185.81 g / mol) = 1.18 g / L

b) The progress of the reaction is calculated:

P C = 2 * X = 4.19

X = 2.10 atm is cleared

We have that the expression of the constant is:

Kp = P B ^ 3 * P C ^ 2 / P A ^ 2

8.39 = (3 * 2.10) ^ 3 * (4.19) ^ 2 / P A ^ 2

Clears:

P A eq = 22.87 atm

The initial pressure is calculated:

Initial P A = P A eq + 2 * X = 22.87 + 2 * 2.10 = 27.07 atm

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