Question

Determine the value of the equilibrium constant (report your answer to three significant figures) for the following reaction if an equilibrium mixture contains 0.010 mol of solid PbBr2, and is 0.0100 M in Pb2+ ions and 0.0250 M in Br1- ions. Use the notation 4.31e-5 to indicate a number such as 4.31 x 10-5.

Answer 1

Answer:

6.25e-6 is the value of the equilibrium constant

Explanation:

we have this equation

$PbBr(s) ----- Pb^{2+}(aq) + 2Br(aq)$

When at a state of equilibrium,

we have the concentration of Pb^2+ to be 0.01

we have the concentration of Br^- to be 0.025

the equilibrium constant concentration of both pure solids and liquid s are said to be equal to 1

[PbBR2] = 1

such tht

Keq = [Pb^2+] x [Br-]^2

we already know the values of these from the above.

0.01x0.025^2

= 0.01 x 0.000625

= 0.00000625

= 6.25 x 10^-6

= 6.25e^-6