Question

1. A 28.5 mL sample of 0.218 M ethylamine, C2H5NH2, is titrated with 0.340 M nitric...

1. A 28.5 mL sample of 0.218 M ethylamine, C2H5NH2, is titrated with 0.340 M nitric acid.

After adding 26.9 mL of nitric acid, the pH is _____.

2. A 24.7 mL sample of 0.292 M methylamine, CH3NH2, is titrated with 0.325 M perchloric acid.

At the equivalence point, the pH is _____.

Homework Answers

Answer #1

1)

millimoles of ethylamine,= 28.5 x 0.218 = 6.213

millimoles of HNO3 = 26.9 x 0.340 = 9.146

C2H5NH2 + HNO3   ----------------> C2H5NH3+ NO3-

6.213             9.146                               0    

   0    2.933    6.213

here strong acid reamins. so

[HNO3] = 2.933 / (28.5 + 26.9) = 0.0529 M

pH = -log (0.0529)

pH = 1.28

2)

at equivalence point :

millimoles of acid = millimoles of base

0.325 x V = 24.7 x 0.292

V = 22.192 mL

here salt only remians.

[salt] = 24.7 x 0.292 / (24.7 + 22.192) = 0.1538 M

pH = 7 - 1/2 (pKb + log C)

     = 7 - 1/2 (3.30 + log 0.1538)

pH = 5.76

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
2.)a.)A 24.1 mL sample of 0.370 M methylamine, CH3NH2, is titrated with 0.352 M nitric acid....
2.)a.)A 24.1 mL sample of 0.370 M methylamine, CH3NH2, is titrated with 0.352 M nitric acid. At the equivalence point, the pH is B.)A 27.8 mL sample of 0.243 M dimethylamine, (CH3)2NH, is titrated with 0.226 M nitric acid. The pH before the addition of any nitric acid is C.)A 23.2 mL sample of 0.389 M diethylamine, (C2H5)2NH, is titrated with 0.387 M hydrochloric acid. After adding 33.3 mL of hydrochloric acid, the pH is
A 20.4 mL sample of 0.325 M trimethylamine, (CH3)3N, is titrated with 0.353 M perchloric acid....
A 20.4 mL sample of 0.325 M trimethylamine, (CH3)3N, is titrated with 0.353 M perchloric acid. After adding 7.40 mL of perchloric acid, what will be the pH ?
A 28.2 mL sample of 0.397 M methylamine, CH3NH2, is titrated with 0.398 M hydroiodic acid....
A 28.2 mL sample of 0.397 M methylamine, CH3NH2, is titrated with 0.398 M hydroiodic acid. After adding 10.6 mL of hydroiodic acid, the pH is .
A 25mL sample of 0.10M C2H5NH2 (ethylamine) is titrated with 0.150M HCl. What is the pH...
A 25mL sample of 0.10M C2H5NH2 (ethylamine) is titrated with 0.150M HCl. What is the pH of the solution after 9mL of acid have been added to the amine? [Kb(C2H5NH2) = 6.5 X 10^-4]
A 21.1 mL sample of a 0.485 M aqueous hypochlorous acid solution is titrated with a...
A 21.1 mL sample of a 0.485 M aqueous hypochlorous acid solution is titrated with a 0.347 M aqueous sodium hydroxide solution. What is the pH at the start of the titration, before any sodium hydroxide has been added? 12- A 24.0 mL sample of 0.323 M ethylamine, C2H5NH2, is titrated with 0.284 M nitric acid. At the titration midpoint, the pH is :________________.
A 28.5 mL sample of 0.253 M diethylamine, (C2H5)2NH, is titrated with 0.247 M hydrobromic acid....
A 28.5 mL sample of 0.253 M diethylamine, (C2H5)2NH, is titrated with 0.247 M hydrobromic acid. After adding 13.0 mL of hydrobromic acid, the pH is_____
75.0 ml sample of 0.200 M sodium hydroxide is titrated with .200 M nitric acid. calculate...
75.0 ml sample of 0.200 M sodium hydroxide is titrated with .200 M nitric acid. calculate ph A) after adding 40.00 mL of HNO3 B) at the equivalent point
Consider the titration of a 26.0 −mL sample of 0.175 M CH3NH2 with 0.150 M HBr....
Consider the titration of a 26.0 −mL sample of 0.175 M CH3NH2 with 0.150 M HBr. Determine each of the following. 1. the pH at 4.0 mL of added acid 2.the pH at the equivalence point 3.the pH after adding 6.0 mL of acid beyond the equivalence point
1. A 100.0-mL sample of 0.500 M sodium hydroxide is titrated with 0.100 M nitric acid....
1. A 100.0-mL sample of 0.500 M sodium hydroxide is titrated with 0.100 M nitric acid. Calculate the pH before the titration begins. 2. A 1.0-L buffer solution contains 0.100 mol HCN and 0.100 mol LiCN. The value of Ka for HCN is 4.9 x 10-10. Because the initial amounts of acid and conjugate base are equal, the pH of the buffer is equal to pKa = -log (4.9 x 10-10) = 9.31. Calculate the new pH after the addition...
Consider the titration of a 26.9 −mL sample of 0.115 M RbOH with 0.100 M HCl....
Consider the titration of a 26.9 −mL sample of 0.115 M RbOH with 0.100 M HCl. Determine each of the following. C. the pH at 5.1 mL of added acid? D. the pH at the equivalence point? E. the pH after adding 4.1 mL of acid beyond the equivalence point?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT