Question
a 20.0 ml sample of 0.10 mol... CAN YOU EXPLAIN How to do B,C,D plz. i need it asap!! thank you so much

A 20.0 mL sample of 0.10 mol/L CH3COOH was titrated with 0.20 mol/L. NaOH. Ka for CH3COOH is 1.8 x 10-6 a. What volume of NaO
0 0
Add a comment Improve this question Transcribed image text
Answer #1

a)

find the volume of NaOH used to reach equivalence point

M(CH3COOH)*V(CH3COOH) =M(NaOH)*V(NaOH)

0.1 M *20.0 mL = 0.2M *V(NaOH)

V(NaOH) = 10 mL

Answer: 10 mL

b)

At half equivalence point, pH = pKa

use:

pKa = -log Ka

= -log (1.8*10^-5)

= 4.7447

So, pH = 4.7447

Answer: 4.74

c)

Given:

M(CH3COOH) = 0.1 M

V(CH3COOH) = 20 mL

M(NaOH) = 0.2 M

V(NaOH) = 10 mL

mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)

mol(CH3COOH) = 0.1 M * 20 mL = 2 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.2 M * 10 mL = 2 mmol

We have:

mol(CH3COOH) = 2 mmol

mol(NaOH) = 2 mmol

2 mmol of both will react to form CH3COO- and H2O

CH3COO- here is strong base

CH3COO- formed = 2 mmol

Volume of Solution = 20 + 10 = 30 mL

Kb of CH3COO- = Kw/Ka = 1*10^-14/1.8*10^-5 = 5.556*10^-10

concentration ofCH3COO-,c = 2 mmol/30 mL = 0.0667M

CH3COO- dissociates as

CH3COO- + H2O -----> CH3COOH + OH-

0.0667 0 0

0.0667-x x x

Kb = [CH3COOH][OH-]/[CH3COO-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((5.556*10^-10)*6.667*10^-2) = 6.086*10^-6

since c is much greater than x, our assumption is correct

so, x = 6.086*10^-6 M

[OH-] = x = 6.086*10^-6 M

use:

pOH = -log [OH-]

= -log (6.086*10^-6)

= 5.2157

use:

PH = 14 - pOH

= 14 - 5.2157

= 8.7843

Answer: 8.78

d)

Given:

M(CH3COOH) = 0.1 M

V(CH3COOH) = 20 mL

M(NaOH) = 0.2 M

V(NaOH) = 15 mL

mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)

mol(CH3COOH) = 0.1 M * 20 mL = 2 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.2 M * 15 mL = 3 mmol

We have:

mol(CH3COOH) = 2 mmol

mol(NaOH) = 3 mmol

2 mmol of both will react

excess NaOH remaining = 1 mmol

Volume of Solution = 20 + 15 = 35 mL

[OH-] = 1 mmol/35 mL = 0.0286 M

use:

pOH = -log [OH-]

= -log (2.857*10^-2)

= 1.5441

use:

PH = 14 - pOH

= 14 - 1.5441

= 12.4559

Answer: 12.46

Add a comment
Know the answer?
Add Answer to:
a 20.0 ml sample of 0.10 mol... CAN YOU EXPLAIN How to do B,C,D plz. i...
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. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT