Reaction 6-7 is allowed to come to equilibrium in a solution initially containing 0.0100 M BrO^-₃, 0.0100 M Cr³⁺, and 1.00 M H⁺.

To find the concentrations at equilibrium, we can construct a table showing initial and final concentrations (see below). We use the stoichiometry coefficients of the reaction to say that if x mol of Br^- are created, then we must also make x mol of Cr₂O₇^2- and 8x mol of H⁺. To produce x mol of Br^-, we must have consumed x mol of BrO^-₃ and 2x mol of Cr³⁺.

(a) Write the equilibrium constant expression that you would use to solve for x to find the concentrations at equilibrium. Do not try to solve the equation.

(b) Because K = 1 × 10¹¹, we suppose that the reaction will go nearly “to completion.” That is, we expect both the concentration of Br^- and Cr₂O₇^2- to be close to 0.005 00 M at equilibrium. (Why?)

That is, x ≈ 0.005 00 M. With this value of x, [H⁺] = 1.00 + 8x = 1.04 M and [BrO^-₃] = 0.010 0 – x = 0.005 0 M. However, we cannot say [Cr³⁺] = 0.0100 – 2x = 0, because there must be some small concentration of Cr³⁺ at equilibrium. Write [Cr³⁺] for the concentration of Cr³⁺and solve for [Cr³⁺]. The limiting reagent in this example is Cr³⁺. The reaction uses up Cr³⁺ before consuming BrO^-₃.