Question
What is the $\mathrm{pH}$ of a buffer solution containing $0.005 \mathrm{M}$ oxalic acid and $0.05 \mathrm{M}$ potassium oxalate solution in one-liter? The dissociation constant of oxalic acid is $6.5 \times 10^{-2}$ at $25^{\circ} \mathrm{C}$.
Step 1
Step 1: Write the chemical equation for the dissociation of oxalic acid (H2C2O4) in water: \[H_{2}C_{2}O_{4} \rightleftharpoons H^{+} + HC_{2}O_{4}^{-}\] Show more…
Show all steps
Your feedback will help us improve your experience
Ahmed Ali and 52 other 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
What is the pH of a solution that is 0.25 M in oxalic acid (H2C2O4) and 0.55 M in sodium oxalate (Na2C2O4)? The Ka for oxalic acid is 5.6 x 10^-2. H2C2O4 (aq) + H2O (l) = HC2O4^- (aq) + H3O+ (aq)
Calculate the $\mathrm{pH}$ at $25^{\circ} \mathrm{C}$ of a $0.25 \mathrm{M}$ aqueous solution of oxalic acid $\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right) .\left(K_{\mathrm{a}_{1}} \text { and } K_{\mathrm{a}_{2}} \text { for oxalic acid are } 6.5 \times 10^{-2}\right.$ and $6.1 \times 10^{-5}$, respectively.)
What is the pH of a solution that is 0.25 M in oxalic acid (H2C2O4) and 0.55 M in sodium oxalate (Na2C2O4)? The Ka for oxalic acid is 5.6 x 10-2. H2C2O4 (aq) + H2O (l) ⇌ HC2O4- (aq) + H3O+ (aq)
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD