Write an equilibrium constant expression for the reaction of xylenol orange (H4Q

Question

Write an equilibrium constant expression for the reaction of xylenol orange (H4Q) with Al3+ ions: H4Q(aq) + Al3+(aq) A1Q- + 4H+. The initial concentration of H4Q is 3.0 times 10-6 M; the initial concentration of Al3+ is 4.0 times 10-6 M. Using UV-Vis spectroscopy, the concentration of A1Q- at equilibrium is determined to be 1.63 times 10 M at 65 degree C. What are the equilibrium concentrations of H4Q and Al3+? For the solution discussed in question #2, [H+] = 0.0032 M. Based on this information and your answers to question #2, what is the equilibrium constant for the reaction of xylenol orange with Al3+ at 65 degree C?

Explanation / Answer

1. Keq= ([AlQ-][H+]4 )/ [H4Q] [Al3+] (conc. of products/ conc. of reactants)

2. By stoichiometry we can say that no. of moles of AlQ produced = no. of moles of each reactant used.

[AlQ-] = 1.63*10-6

hence [H4Q] = 3*10-6 - 1.63*10-6 = 1.37*10-6 M

[Al3+] = 4*10-6 - 1.63*10-6 = 2.37*10-6 M

3. Keq= ([AlQ-][H+]4 )/ [H4Q] [Al3+]

= (1.63*10-6)(0.0032)4 / (2.37*10-6) (1.37*10-6)

= 5.26 *10-5

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