Question

The Ka value for acetic acid, CH3COOH(aq), is 1.8x10^-5. Calculate the ph of a 2.80 M...

The Ka value for acetic acid, CH3COOH(aq), is 1.8x10^-5. Calculate the ph of a 2.80 M acetic acid solution.


PH=


Calculate the ph of the resulting solution when 3.00 mL of the 2.80 M acetic acid is diluted to make a 250.0 mL solution.


PH=


Answers are not 4.6 or 3.8

0 0
Add a comment Improve this question Transcribed image text
Answer #1
Concepts and reason

Power of hydrogen ions (pH)

The pH is defined as the negative logarithmic function of the concentration of hydrogen () ion in an aqueous solution.

Molar concentration:

The molar concentration is the number of moles of a solute that will dissolve in a liter of solution. It is expressed as mol/L or M (molarity).

Fundamentals

Molar concentration:

Molar concentration = number of moles of solute in mol
volume of solution in L

Dilution equation:

M,V, %3 М,V,
2

Acid dissociation constant:

НА +Н,О —>н,0+A
к,
HA
The acid dissociation constant is K.

Equation for the pH of the solution:

pH logH
Concentration of hydrogen ion is H*].

Henderson–Hasseslbalch Equation:

[A]
pH pK,+logTHA
Acid dissociation constant of HA is pK.
Concentration of HA is [HA]
Concentration of A is| A |.

An ICE table is used to find the concentrations or moles of individual reactants or products at equilibrium.

The expansion of an ICE table is as follows:

The initial quantities is denoted as I.
The change in each quantity is denoted as C.
The equilibrium quantities is denoted as

All quantities are expressed in terms of concentrations or moles.

(1)

Given data:

Initial concentration of acetic acid, НА
= 2.80 M

НА +Н,О — >Н,О+А

[но ТА]
Н.о
к,
[НА]

Н,о — | Н,о
A
Н.О
ICE
НА
1 (М)
С (М)
Е (М) | 2.80-х
2.80
+x
+x
-х
+x
+x

At equilibrium: The change in concentration is x.

(x)(x)
к,
(2.80-x)
Substitute
1.8x10=(x)(x)
(2.80-x)

Solve x
x 7.09x103
So,
H0*]-7.09x 10 M
or
H*7.09x 10 M

H*7.09x10 M
pH-log[H
Substitute
pH -log (7.09x103)
pH 2.15

(2)

Given:

М, 3 2.80М
V 3.00mL
V2 250.0mL

Using dilution equation
M2 V2 M,V

Rearrange
М, - М,у
V,
м,
2.80 Мx3.00 mL
м,
250.0mL
М, 3 0.0336 M

Concentration of acetic acid, = 0.0336 M

+| Н,о
| — |
Н,О| +| A
ICE
НА
1(М)
0.0336
С (М)
Е (М) | 0.0336-х
-X
+x
+x
+x
+x

At equilibrium: The change in concentration is x.

(x)(x)
к,
(0.0336-x)
Substitute
(x)(x)
1.8x10 (0.0336-x)

Solvex
x 7.77x10
So,
HO*]-7.77x10 M
or
H*7.77x 10 M

H*7
pH log[H]
7.77x104 M
Substitute
pH=-log(7.77x10)
pH 3.11

Ans: Part 1

The pH of a 2.80 M acetic acid solution is

pH 2.15

Part 2

The pH of the resulting solution is

pH 3.11

Add a comment
Know the answer?
Add Answer to:
The Ka value for acetic acid, CH3COOH(aq), is 1.8x10^-5. Calculate the ph of a 2.80 M...
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