Oxalic acid, H2C2O4 , is a diprotic acid. Write a chemical equilibrium expression for each ionization step.
Added by Marina C.
Step 1
Step 1:** The first ionization step of oxalic acid can be represented as follows: \[H_2C_2O_4 \rightleftharpoons H^+ + HC_2O_4^-\] ** Show more…
Show all steps
Your feedback will help us improve your experience
Shaiju T and 86 other Chemistry 101 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Oxalic acid (C2H2O4) is a diprotic acid. Write the two ionization reactions for C2H2O4 to C2O42-.
Ivan K.
Enter a chemical equation for the first ionization step of oxalic acid. Express your answer as a chemical equation including phases.
Lijeesh K.
Oxalic acid is a relatively weak diprotic acid. Calculate the equilibrium constant for the reaction shown below from $K_{a 1}$ and $K_{a 2}$. (See Appendix H for the required $K_{a}$ values.) $\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(\mathrm{aq})+2 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})$
Recommended Textbooks
Chemistry: Structure and Properties
Chemistry The Central Science
Chemistry
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD