Write the equilibrium constant expression for this reaction: c6h10o4 (aq) + 2oh− (aq) → 2ch3coo− (aq) + c2h6o2 (aq)

The equilibrium constant of reaction, usually denoted as K, is a unit of ratio. The ratio involves concentrations of products to reactants. But you also have to incorporate their stoichiometric coefficients in the reaction as their respective exponents. Note that substances in their aqueous state are the ones that are included only in the expression. To properly show you how it's done, consider this equilibrium reaction:

aA (aq) + bB (l) ⇆ nN (aq)

The equilibrium constant for this reaction is:
K = [N]ⁿ/[A]ᵃ
where the [] brackets denotes concentration in molarity

Now, let's apply this to the given equation:
C₆H₁₀O₄ (aq) + 2 OH⁻ (aq) → 2 CH₃COO⁻ (aq) + C₂H₆O₂ (aq)

Since all of the substances are in aqueous state, the expression for K is therefore,

K = [CH₃COO⁻]²[C₂H₆O₂] / [C₆H₁₀O₄][OH⁻]²

2.00 is the !