Is equilibrium point for a coupled equilibria of arbitary number of coupled reactions is a unique point?

I was solving a question of salt hydrolysis for a mixture of Weak Acid-Base Salts, given the initial amount of salt and Equilibrium constants. I was trying to approach the problem in two ways, Assuming initially that we have all reactants and no products and move the reaction forward OR assuming all the reactants have been converted into product instantaneously and moving the reaction backward for equilibrium.

As concentrations of various species in equilibrium depends on equilibrium constant powers of their coefficients, which does not seem like any linear or straightforward relation. I wonder if both approaches are equivalent.

I.e. What guarantees the uniqueness of an equilibrium point (that is, specific concentrations of various species at eq.), given just the specific values of initial variables (initial concentrations)?

The paper https://aip.scitation.org/doi/10.1063/1.1669753 mentions something like that but the math went over my head. I wonder if there is any simple explanation based on the first principles of thermodynamics, like the curve for free energy for the whole system, does have some restrictions (it should be concave or something like that).

• Both approaches are equivalent if you do not apply any simplifying assumption. E.g. the notorious formula pH=0.5*(pKa - log c ) for weak acid pH applies 2 such assumptions ( c >> [H+] and [H+] >> [OH-]). Commented Jun 9, 2021 at 7:18
• My question is more on theoretical side...how can we prove that both approaches are equivalent. Commented Jun 9, 2021 at 8:52
• Such task is rather for mathematicians. Ask yourself, does the solution of the given set of equations for tthe given set of variables depend on the way how you solve it ? The existence of multiple solutions and the proof of it is matter on mathematics. Fortunately, usually just one of solutions makes scientific sense, e.g. concentrations cannot be negative. Commented Jun 9, 2021 at 9:48
• Usually, resolving the equation set leads to a polynomial function f([H+])=0 and only one value makes sense. Then, given [H+] leads to unique concentrations of various compounds forms. Commented Jun 9, 2021 at 9:51

One of the first experimental demonstrations that the same equilibrium point is reached if the reaction starts with all reactants or all products was $$\ce{2HI<=>I_2 + H_2}$$ made by Bodenstein in $$\approx 1894$$. I have redrawn the data.
figure. $$\ce{ 2HI=I_2 + H_2}$$. $$x$$ is the fraction of HI at $$721$$ K. $$x=[HI]/([HI]+2[I_2])$$
In the reaction $$\ce{A<=>B}$$ the equilibrium constant is determined by the ratio of rate constants forward and back, $$k_1,\;k_{-1}$$ respectively. If another reaction is coupled to the first one $$\ce{B<=>C}$$ then this has a different equilibrium constant with rate constants $$k_2,\;k_{-2}$$ but these are separate reactions and so the rate constants and hence equilibrium constants are unchanged when A, B, C are together in solution. The concentrations are changed, of course, because when C is added it changes the concentration of B and hence A.