Question

In: Chemistry

PART A- Ethylamine, an organic amine, C2H5NH2, a weak base with Kb of 8.3×10−5, reacts with...

PART A-

Ethylamine, an organic amine, C2H5NH2, a weak base with Kb of 8.3×10−5, reacts with water to form its conjugate acid, C2H5NH+3.

1) Write the equation of dissociation for ethylamine.

Express your answer as a chemical equation. Identify all of the phases in your answer.

2)What is the expression constant dissociation for ethylamine?

PART B - Phosphorous acid dissociates to form dihydrogen phosphite and hydronium ion. Phosphorous acid has Kaof 5.0×10−2.

1)Write the equation dissociation for phosphorous acid.

Express your answer as a chemical equation. Identify all of the phases in your answer.

2)What is the expression constant dissociation for phosphorous acid?

Solutions

Expert Solution


Related Solutions

Ethylamine, C2H5NH2, is a weak base. 250.0 mL of a 0.160 M C2H5NH2  solution is titrated with...
Ethylamine, C2H5NH2, is a weak base. 250.0 mL of a 0.160 M C2H5NH2  solution is titrated with 0.500 M HCl (aq). Find the KbKb value in a lecture slide. Calculate the pH at the following points. 30. mL of HCl solution added. At the halfway point. At the equivalence point. Verify any assumption you made. 90. mL of HCl solution added.
Find the [OH−] in a 0.400 M solution of ethylamine (C2H5NH2). For ethylamine, Kb=5.6×10−4.
Find the [OH−] in a 0.400 M solution of ethylamine (C2H5NH2). For ethylamine, Kb=5.6×10−4.
Ammonia, NH3, is a weak base with a Kb value of 1.8×10−5. Part A Part complete...
Ammonia, NH3, is a weak base with a Kb value of 1.8×10−5. Part A Part complete What is the pH of a 0.100 M ammonia solution? Express your answer numerically to two decimal places. View Available Hint(s) pH = 11.13 SubmitPrevious Answers Correct Part B What is the percent ionization of ammonia at this concentration? Express your answer with the appropriate units.
Ammonia, NH3, is a weak base with a Kb value of 1.8×10−5. Part A What is...
Ammonia, NH3, is a weak base with a Kb value of 1.8×10−5. Part A What is the pH of a 0.270 M ammonia solution? Part B What is the percent ionization of ammonia at this concentration?
Saccharin is a weak organic base with a Kb of 4.80 × 10–3. A 0.297-g sample...
Saccharin is a weak organic base with a Kb of 4.80 × 10–3. A 0.297-g sample of saccharin dissolved in 25.0 mL of water has a pH of 12.190. What is the molar mass of saccharin? A sample of ammonia (Kb = 1.8 × 10–5) is titrated with 0.1 M HCl. At the equivalence point, what is the approximate pH of the solution? What is the equilibrium concentration of ammonium ion in a 0.33 M solution of ammonia (NH3, Kb...
The Kb for an amine is 2.105 × 10-5. What percentage of the amine is protonated...
The Kb for an amine is 2.105 × 10-5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.788? (Assume that all OH– came from the reaction of B with H2O.)
The Kb for an amine is 8.695 × 10-5. What percentage of the amine is protonated...
The Kb for an amine is 8.695 × 10-5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.207? (Assume that all OH– came from the reaction of B with H2O.)
Consider the titration of 70.0 mL of 0.0300 M C2H5NH2 (a weak base; Kb = 0.000640)...
Consider the titration of 70.0 mL of 0.0300 M C2H5NH2 (a weak base; Kb = 0.000640) with 0.100 M HClO4. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL pH =   (b) 5.3 mL pH =   (c) 10.5 mL pH =   (d) 15.8 mL pH =   (e) 21.0 mL pH =   (f) 29.4 mL pH =  
Consider the titration of 70.0 mL of 0.0300 M C2H5NH2 (a weak base; Kb = 0.000640)...
Consider the titration of 70.0 mL of 0.0300 M C2H5NH2 (a weak base; Kb = 0.000640) with 0.100 M HClO4. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL pH =   (b) 5.3 mL pH =   (c) 10.5 mL pH =   (d) 15.8 mL pH =   (e) 21.0 mL pH =   (f) 29.4 mL pH =  
Consider the titration of 50.0 mL of 0.0500 M C2H5NH2 (a weak base; Kb = 0.000640)...
Consider the titration of 50.0 mL of 0.0500 M C2H5NH2 (a weak base; Kb = 0.000640) with 0.100 M HIO4. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL pH =   (b) 6.3 mL pH =   (c) 12.5 mL pH =   (d) 18.8 mL pH =   (e) 25.0 mL pH =   (f) 30.0 mL pH =  
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT