Solution Found!
The following equilibrium constant have been determined
Chapter 14, Problem 30P(choose chapter or problem)
The following equilibrium constants have been determined for oxalic acid at \(25^{\circ} \mathrm{C}\):
\(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)\ \leftrightharpoons\ H^{+}(a q)+\mathrm{HC}_{2} \mathrm{O}_{4}^{-}(a q)\)
\(K_{c}^{\prime}=6.5 \times 10^{-2}\)
\(\mathrm{HC}_{2} \mathrm{O}_{4}^{-}(a q)\ \leftrightharpoons\ H^{+}(a q)+\mathrm{C}_{2} O_{4}^{2-}(a q)\)
\(K_{c}^{\prime \prime}=6.1 \times 10^{-5}\)
Calculate the equilibrium constant for the following reaction at the same temperature:
\(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)\ \leftrightharpoons\ 2 \mathrm{H}^{+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q)\)
Questions & Answers
QUESTION:
The following equilibrium constants have been determined for oxalic acid at \(25^{\circ} \mathrm{C}\):
\(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)\ \leftrightharpoons\ H^{+}(a q)+\mathrm{HC}_{2} \mathrm{O}_{4}^{-}(a q)\)
\(K_{c}^{\prime}=6.5 \times 10^{-2}\)
\(\mathrm{HC}_{2} \mathrm{O}_{4}^{-}(a q)\ \leftrightharpoons\ H^{+}(a q)+\mathrm{C}_{2} O_{4}^{2-}(a q)\)
\(K_{c}^{\prime \prime}=6.1 \times 10^{-5}\)
Calculate the equilibrium constant for the following reaction at the same temperature:
\(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)\ \leftrightharpoons\ 2 \mathrm{H}^{+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q)\)
ANSWER:Step 1 of 2
The equilibrium constant for the oxalic acid is given as follows.
