Question

Consider a 2×10−2 M  solution of the weak acid butanoic acid , for which pKa =  4.83 ....

Consider a 2×10−2 M  solution of the weak acid butanoic acid , for which pKa =  4.83 .

You can search for the structure of this compound online, although precise knowledge of the structure is not needed.

Your answers need to be within 5% of the correct answer to be considered correct. Don't round or you will fall outside the 5% margin of error.

Part A: Calculate  [H+] (units in M)

Part B: Calculate the concentration of butanoate ions (units in M)

Part C: Calculate the concentration of butanoic acid in equilibrium. (units in M)

Part D: Calculate [OH−] (units in M)

Part E: Calculate the pH

Homework Answers

Answer #1

pKa = 4.83

Ka = 10^-pKa

Ka = 1.48 x 10^-5

C = 2 x 10^-2 M

part A)

[H+] = sqrt (Ka x C)

[H+]   = sqrt (1.48 x 10^-5 x 2 x 10^-2)

[H+] = 5.44 x 10^-4 M

part B)

concentration of butanoate ion = 5.44 x 10^-4 M

part C)

concentration of butanoic acid in equilibrium = 2 x 10^-2 - (5.44 x 10^-4)

                                                                      = 0.0195 M

part D)

[H+] [OH- ] = 1.0 x 10^-14

5.44 x 10^-4 [OH-] = 1.0 x 10^-14

[OH-] = 1.84 x 10^-11 M

part E)

pH = -log [H+]

pH = -log (5.44 x 10^-4)

pH = 3.26

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Consider a 0.2 M  solution of   dimethylammonium chloride. The ion   dimethylammonium is the conjugate acid of the...
Consider a 0.2 M  solution of   dimethylammonium chloride. The ion   dimethylammonium is the conjugate acid of the the weak base dimethylamine.The pKb  of dimethylamine is 3.27   You can search for the structure of this compound online, although precise knowledge of the structure is not needed. Don't round to avoid being outside the 5% margin of error. Part A: Calculate  [H+] (units in M) Part B: Calculate the concentration of dimethylammonium ions. (units in M) [Answer was 0.200 M] Part C: Calculate the concentration...
A 1.45 L buffer solution consists of 0.134 M butanoic acid and 0.344 M sodium butanoate....
A 1.45 L buffer solution consists of 0.134 M butanoic acid and 0.344 M sodium butanoate. Calculate the pH of the solution following the addition of 0.063 moles of NaOH. Assume that any contribution of the NaOH to the volume of the solution is negligible. The Ka of butanoic acid is 1.52 × 10-5.
A 1.49 L buffer solution consists of 0.186 M butanoic acid and 0.258 M sodium butanoate....
A 1.49 L buffer solution consists of 0.186 M butanoic acid and 0.258 M sodium butanoate. Calculate the pH of the solution following the addition of 0.062 moles of NaOH. Assume that any contribution of the NaOH to the volume of the solution is negligible. The Ka of butanoic acid is 1.52 × 10-5.
A 1.35 L buffer solution consists of 0.173 M butanoic acid and 0.309 M sodium butanoate....
A 1.35 L buffer solution consists of 0.173 M butanoic acid and 0.309 M sodium butanoate. Calculate the pH of the solution following the addition of 0.076 moles of NaOH. Assume that any contribution of the NaOH to the volume of the solution is negligible. The Ka of butanoic acid is 1.52 × 10-5.
A 1.36 L buffer solution consists of 0.184 M butanoic acid and 0.331 M sodium butanoate....
A 1.36 L buffer solution consists of 0.184 M butanoic acid and 0.331 M sodium butanoate. Calculate the pH of the solution following the addition of 0.068 moles of NaOH. Assume that any contribution of the NaOH to the volume of the solution is negligible. The Ka of butanoic acid is 1.52 × 10^-5.
A 1.37 L buffer solution consists of 0.136 M butanoic acid and 0.337 M sodium butanoate....
A 1.37 L buffer solution consists of 0.136 M butanoic acid and 0.337 M sodium butanoate. Calculate the pH of the solution following the addition of 0.070 moles of NaOH. Assume that any contribution of the NaOH to the volume of the solution is negligible. The Ka of butanoic acid is 1.52 × 10-5. pH =
A. Calculate the percent ionization of 8.0×10−3 M butanoic acid (Ka=1.5×10−5) B. Calculate the percent ionization...
A. Calculate the percent ionization of 8.0×10−3 M butanoic acid (Ka=1.5×10−5) B. Calculate the percent ionization of 8.0×10−3 M butanoic acid in a solution containing 8.0×10−2 M sodium butanoate.
The pH of a 6.92×10-4 M solution of a weak monoprotic acid is 5.16. Calculate pKa...
The pH of a 6.92×10-4 M solution of a weak monoprotic acid is 5.16. Calculate pKa for this monoprotic acid to two decimal places.
1). Consider a 0.4 M solution of   anilinium chloride. The ion   anilinium is the conjugate acid...
1). Consider a 0.4 M solution of   anilinium chloride. The ion   anilinium is the conjugate acid of the the weak base aniline.The pKb   of aniline is 9.13   Part a).   Calculate  [H+] __________ M Part b). Calculate the concentration of   anilinium ions. __________ M Part c). Calculate the concentration of   aniline in equilibrium. _________ M Part d). Calculate [OH−] ________ M Part e). Calculate [Cl-] ________ M Part f). Calculate the pH _______ M 2). Calculate the pH and the concentration [Na+] =of all...
0.1 M solution of weak acid has pH=4.0. Calculate pKa. 20 mL of 0.1 M solution...
0.1 M solution of weak acid has pH=4.0. Calculate pKa. 20 mL of 0.1 M solution of weak acid was mixed with 8 mL 0.1 M solution of NaOH. Measured pH was 5.12. Calculate pKa. Calculate the pH of a 0.2M solution of ammonia. Ka=5.62×10-10. Calculate pH of 0.01 M aniline hydrochloride. Aniline pKb=9.4.
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT