Question

A volume of 25.0 mL OF 0.140 M HCl is titrated against a 0.140 M CH3NH2...

A volume of 25.0 mL OF 0.140 M HCl is titrated against a 0.140 M CH3NH2 solution added to it from a buret.

A.) Calculate the pH value of the solution after 10.0 mL of CH3NH2 solution has been added.

B.) Calculate the pH value of the solution after 25.0 mL of CH3NH2 solution have been added.

C.) Calculate the pH value of the solution after 35.0 mL of CH3NH2 solution have been added.

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A 25.0 mL sample of a 0.0500 M solution of aqueous trimethylamine is titrated with a...
A 25.0 mL sample of a 0.0500 M solution of aqueous trimethylamine is titrated with a 0.0625 M solution of HCl. Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added; pKb of (CH3)3N = 4.19 at 25°C.
A 25.0 mL sample of a 0.290 M solution of aqueous trimethylamine is titrated with a...
A 25.0 mL sample of a 0.290 M solution of aqueous trimethylamine is titrated with a 0.363 M solution of HCl. Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added; pKb of (CH3)3N=4.19 at 25 degrees C
A 25.0-mL sample of a 0.070 M solution of aqeous trimethylamine is titrated with a 0.088...
A 25.0-mL sample of a 0.070 M solution of aqeous trimethylamine is titrated with a 0.088 M solution of HCl. Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added; pKb of (CH3)3N = 4.19 at 25 C.
A 25.0-mL sample of a 0.350 M solution of aqueous trimelhylamine is titrated with a 0.438...
A 25.0-mL sample of a 0.350 M solution of aqueous trimelhylamine is titrated with a 0.438 M solution of HCI. Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added; pK_b of (CH_3)_3N = 4.19 at 25 degree C. pH after 10.0 mL of acid have been added: pH after 20.0 mL of acid have been added: pH after 30.0 mL of acid have been added:
A 25.0 mL sample of a 0.115 M solution of acetic acid is titrated with a...
A 25.0 mL sample of a 0.115 M solution of acetic acid is titrated with a 0.144 M solution of NaOH. Calculate the pH of the titration mixture after 10.0, 20.0, and 30.0 mL of base have been added. (The Ka for acetic acid is 1.76 x 10^-5). 10.0 mL of base = 20.0 mL of base = 30.0 mL of base =
a) 125 mL of a 0.098 M aniline (pKb = 9.40) solution is titrated with 0.217...
a) 125 mL of a 0.098 M aniline (pKb = 9.40) solution is titrated with 0.217 M HCl. What is the pH after 75 ml of HCl have been added? b) 125 mL of a 0.098 M aniline (pKb = 9.40) solution is titrated with 0.217 M HCl. What is the pH after 25 ml of HCl have been added? c) 125 mL of a 0.098 M aniline (pKb = 9.40) solution is titrated with 0.217 M HCl. What is...
A 25.0 mL of a weak acid is titrated with a strong base (0.1 M). Calculate...
A 25.0 mL of a weak acid is titrated with a strong base (0.1 M). Calculate the pH of the solution during the titration if the weak acid concentration is 0.10 M and its Ka = 1.8 x 10-5 and 10.0 mL of base has been added.
Consider the titration of 100.0 mL of 0.200 M CH3NH2 by 0.100 M HCl. For each...
Consider the titration of 100.0 mL of 0.200 M CH3NH2 by 0.100 M HCl. For each volume of HCl added, decide which of the components is a major species after the HCl has reacted completely. Kb for CH3NH2 = 4.4 x 10-4. 1) Calculate the pH at the equivalence point for this titration 2) At what volume of HCl added in this titration does the pH = 10.64? Express your answer in mL and include the units in your answer.
a. You titrate 25.0 mL of 0.60 M NH3 with 0.60 M HCl. Calculate the pH...
a. You titrate 25.0 mL of 0.60 M NH3 with 0.60 M HCl. Calculate the pH of the solution after adding 5.00, 15.0, 22.0, and 30.0 mL of the acid. Ka = 5.6×10-10 pH(5.00 mL added) ------------------------------ pH(15.0 mL added)------------------------------ pH(22.0 mL added) ---------------------------- pH(30.0 mL added)---------------------------- b. itration of 27.9 mL of a solution of the weak base aniline, C6H5NH2, requires 28.64 mL of 0.160 M HCl to reach the equivalence point. C6H5NH2(aq) + H3O+(aq) ⇆ C6H5NH3+(aq) + H2O(ℓ)...
A 25.0 ml sample of 0.090 M solution of aqueous trimethylamine is nitrated with a 0.11...
A 25.0 ml sample of 0.090 M solution of aqueous trimethylamine is nitrated with a 0.11 M solution of HCl. Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added, pkb of (ch3)3n = 4.19 (part 2, after adding 20.0 mL)
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT