Question

Calculate the original molarity of strontium hydroxide if the pH is 12.30

Define Le Chateliers principle

For the synthesis of ammonia (Haber process), the equilibrium constant at 500^oC is 6.0 x 10^-2. Predict the direction the system will shift to reach equilibrium in the following conditions.

[NH₃]_o = 1.0 x 10^-3 M

[N₂]_o = 1.0 x 10^-5 M

[H₂]_o = 2.0 x 10^-3 M

Answer 1

1) Sr(OH)2 -------> Sr⁺² + 2OH^-

pH = 12.30

pOH = 14 - 12.30

pOH = 1.70

[OH^-] = 10^-pOH = 10^-1.70

[OH^-] = 0.020 M

[Sr(OH)2] = [OH^-] / 2 = 0.20 /2 = 0.01 M

[Sr(OH)2] = 0.01 M

2)according to Lechatelier principle. if we disturb the equilbrium of any reaction then reaction proceeds towards to nulify that change to attain equilibrium again.

Kc = 6.0 x 10^-2

calculate the Qc from given data

Qc = [NH3]² / [N2] [H2]³

Qc = [1.0 X 10-3]² / [1.0 X 10^-5] [2.0 X 10^-3]³

Qc = 1.0 x 10^-6 / 8.0 x 0^-14

Qc = 1.25 x 10⁷

Qc > Kc

so reaction moves towards reactants to reach equilibrium