Question

Consider an experiment where 35.0 ml of 0.175 M acetic acetic acid, HC2H3O3, is titrated with...

Consider an experiment where 35.0 ml of 0.175 M acetic acetic acid, HC2H3O3, is titrated with

0.25 M NaOH. What is the pH at the equivalence point of this titration? The Ka for acetic acid is 1.8 x 10-5.  

Thanks!

Homework Answers

Answer #1

[OH-] = 7.56 x 10-6 M

pOH = -log(7.56 x 10-6) = 5.12

pH = 14.00 - 5.12 = 8.88

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
A student performs the following experiment: 25.0 mL of 0.1 M acetic acid is titrated to...
A student performs the following experiment: 25.0 mL of 0.1 M acetic acid is titrated to the equivalence point with 25.0 mL of 0.1 M NaOH. Calculate the pH at the equivalence point. The equilibrium constant Ka for acetic acid is 1.8*10^-5. Compare this value with the experimental value. Experimental value for acetic acid is pH = 8.8 and volume of base = 24 mL at equivalence point.
A 25.0 mL sample of an acetic acid solution in titrated with a 0.09984 M NaOH...
A 25.0 mL sample of an acetic acid solution in titrated with a 0.09984 M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. Calculate the concentration of acetic acid in the 25.0 mL sample. Calculate the pH at the equivalence point (Ka acetic acid = 1.75x10^-5)
In an acid base titration experiment, 50.0 ml of a 0.0500 m solution of acetic acid...
In an acid base titration experiment, 50.0 ml of a 0.0500 m solution of acetic acid ( ka =7.5 x 10^-5) was titrated with a 0.0500 M solution of NaOH at 25 C. The system will acquire this pH after addition of 18.75 mL of the titrant: answer is 4.535
A 20.00-mL sample of formic acid (HCO2H) is titrated with a 0.100 M solution of NaOH....
A 20.00-mL sample of formic acid (HCO2H) is titrated with a 0.100 M solution of NaOH. To reach the endpoint of the titration, 30.00 mL of NaOH solution is required. Ka = 1.8 x 10-4 What is the pH of the solution after the addition of 10.00 mL of NaOH solution? What is the pH at the midpoint of the titration? What is the pH at the equivalence point?
25.00 mL of 0.1 M acetic acid are titrated with 0.05 M NaOH. Calculate the pH...
25.00 mL of 0.1 M acetic acid are titrated with 0.05 M NaOH. Calculate the pH of the solution after addition of 25 mL of NaOH. Ka(CH3COOH) = 1.8 x 10-5.
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 22.5 mL sample of an acetic acid solution is titrated with a 0.175M NaOH solution....
A 22.5 mL sample of an acetic acid solution is titrated with a 0.175M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. What was the concentration of acetic acid in the original (22.5mL) sample? What is the pH of the equivalence point? Ka acetic acid= 1.75E-5
16.5 mL of 0.168 M diprotic acid (H2A) was titrated with 0.118 M KOH. The acid...
16.5 mL of 0.168 M diprotic acid (H2A) was titrated with 0.118 M KOH. The acid ionization constants for the acid are Ka1=5.2×10−5 and Ka2=3.4×10−10. At what added volume (mL) of base does the first equivalence point occur? 20.8 mL of 0.146 M diprotic acid (H2A) was titrated with 0.14 M KOH. The acid ionization constants for the acid are Ka1=5.2×10−5 and Ka2=3.4×10−10. At what added volume (mL) of base does the second equivalence point occur? Consider the titration of...
What is the pH at equivalence point when 100 mL of a 0.175 M solution of...
What is the pH at equivalence point when 100 mL of a 0.175 M solution of acetic acid (CH3COOH) is titrated with 0.10 M NaOH to its end point?
A 20.00-mL sample of formic acid (HCO2H) is titrated with a 0.100 M solution of NaOH....
A 20.00-mL sample of formic acid (HCO2H) is titrated with a 0.100 M solution of NaOH. To reach the endpoint of the titration, 30.00 mL of NaOH solution is required. Ka = 1.8 x 10-4 What is the concentration of formic acid in the original solution? What is the pH of the formic acid solution before the titration begins (before the addition of any NaOH)?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT