Propanoic acid, C3H7CO2H, is a weak acid with Ka = 1.34Ă10â»â”. (a) Calculate the value of Kb for its conjugate, the propanoate ion, C3H7CO2â». (b) Calculate the pH of a 0.10 M solution of potassium propanoate, KC3H7CO2.
Added by Thomas M.
Step 1
Given that Ka = 1.34Ă10^-5, we know that Ka * Kb = Kw (the ionization constant of water, 1.0 x 10^-14). Therefore, Kb = Kw / Ka = 1.0 x 10^-14 / 1.34 x 10^-5 = 7.46 x 10^-10. Show moreâŠ
Show all steps
Your feedback will help us improve your experience
Shaiju T and 53 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
The pH of a 0.10-M solution of propanoic acid, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH},$ a weak organic acid, is measured at equilibrium and found to be 2.93 at $25^{\circ} \mathrm{C} .$ Calculate the $K_{\mathrm{a}}$ of propanoic acid.
A 50.0 mL sample of 0.21 M propanoic acid, CH3CH2COOH, a weak monoprotic acid, is titrated with 0.11 M KOH. Ka of CH3CH2COOH = 1.4 Ă 10^-5. (a) Calculate the pH at the half-equivalence point. (b) Calculate the pH at the equivalence point.
Oluwapelumi K.
In a 0.735$M$ solution, a weak acid is 12.5$\%$ dissociated. (a) Calculate the $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right], \mathrm{pH},\left[\mathrm{OH}^{-}\right],$ and pOH of the solution. (b) Calculate $K_{\mathrm{a}}$ of the acid.
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