Chapter 12, Problem 12.24PAE

Interpretation Introduction

Interpretation: From the two reactions with different equilibrium constants ${\text{K}}_{\text{1}}$ and ${\text{K}}_2$, the equilibrium constant for another reaction has to be determined.

Concept introduction:

• For a reaction $\text{aA}\text{+ bB}\to \text{cC + dD}$, the equilibrium constant is given by ${\text{K}}_{\text{eq}}\text{ = }\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}$

  • Equilibrium constant is directly proportional to products and inversely proportional to the reactants.

  For a reversible reaction, ${\text{K}}_{\text{eq}}{}^{forwardreaction}\text{ = }\frac{1}{{\text{K}}_{\text{eq}}{}^{backwardreaction}}$

  • If a reaction is multiplied by n, then the ${\text{K}}_{\text{eq}}$ value changes as ${\text{K}}_{\text{eq}}{}^{\text{new}}\text{ = }{\left[{\text{K}}_{\text{eq}}{}^{\text{old}}\right]}^{\text{n}}$

  Given:

  ${\text{K}}_1=3\times {10}^{-13}$ for the reaction ${\text{HAsO}}_4{}^{2-}\left(aq\right)\rightleftarrows {\text{AsO}}_4{}^{\text{3-}}\left(aq\right){\text{+H}}^{\text{+}}\left(aq\right)$

  ${\text{K}}_2=1.8\times {10}^7$ for the reaction ${\text{HAsO}}_4{}^{2-}\left(aq\right){\text{+H}}^{\text{+}}\left(aq\right)\rightleftarrows {\text{H}}_{\text{2}}{\text{AsO}}_{\text{4}}{}^{\text{-}}\left(aq\right)$