From the following equilibrium constants, calculate the equilibrium constant for the reaction ${\mathbf{HO}}_{\mathbf{2}}{\mathbf{CCO}}_{\mathbf{2}}\mathbf{H}\mathbf{\rightleftharpoons }\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{+}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}$

When an uncharged weak acid is given to water, a homogeneous equilibrium is established in which aqueous acid molecules,$\mathbf{HA}\mathbf{}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$ react with liquid water to create aqueous hydronium ions and aqueous anions,$\mathbf{A}\mathbf{-}\mathbf{}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$ .

When acid molecules lose H+ ions to water, the latter is generated.

The ratio of the reactant and product concentrations in an equilibrium reaction is known as the equilibrium constant (K). If K is less than 1, the reaction should be moved to the left, and if K is greater than 1, it should be moved to the right.

Given information

$K_1=5.6\times {10}^{-2}$

_{${\mathrm{K}}_2=5.4\times {10}^{-5}$}

Equilibrium constant (K) for given reaction

Now add both the equation we get

_{$$\begin{gathered} {}_{{\mathrm{HO}}_2{\mathrm{CCO}}_2}\mathrm{H}\rightleftharpoons {\mathrm{H}}^{+}+\left(\mathrm{COOH}\right){\mathrm{COO}}^{-}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{K}}_1=5.6\times {10}^{-2} \\ \left(\mathrm{COOH}\right){\mathrm{COO}}^{-}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{K}}_2=5.4\times {10}^{-5} \\ {\mathrm{HO}}_2{\mathrm{CCO}}_2\mathrm{H}\rightleftharpoons 2{\mathrm{H}}^{+}+{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\mathrm{K}={\mathrm{K}}_1•{\mathrm{K}}_2 \\ \hspace{0.33em}\mathrm{K}={\mathrm{K}}_1•{\mathrm{K}}_2=3.0\times {10}^{-6} \end{gathered}$$}

Thus, Equilibrium constant (K) for given reaction was calculated as$3.0\times {10}^{-6}$