Question

You have found the following:

Pb⁺²(aq) + 4 I^-(aq) <=> [PbI₄]^-2(aq)

K = (3.119x10^4)

What is the value of K for the following reaction?

3 Pb⁺²(aq) + 12 I^-(aq) <=> 3 [PbI₄]^-2(aq)

Answer 1

The given equation is :

Pb²⁺(aq) + 4I^-(aq) <=> [PbI₄]^2-(aq) ....................(1)

For the above mentioned reaction, K = 3.119*10⁴

We know that when we multiply the coefficients of balanced reaction by a common factor, then K is raised to the power of that common factor.

Hence for the reaction : 3Pb²⁺(aq) + 12I^-(aq) <=> 3[PbI₄]^2-(aq) .................(2)

The above reaction is obtained simply by multiplying the coefficients of reaction(1) by 3.

Hence for reaction (2) , Equilibrium constant, K' = K³ = (3.119*10⁴ )³ = 3.03*10¹³