Question

The titration curve for a 25.0 ml sample of benzoic acid shows that 10.0 ml of...

The titration curve for a 25.0 ml sample of benzoic acid shows that 10.0 ml of 0.100M NaOH is required to reach the equivalence point.

1. What is the concentration of the original benzoic acid sample?

2. What is the pH at the equivalence point?

3. Calculate the pH after addition of 5.00 ml of NaOH and the pOH after addition of 20.0 ml NaOH.

Homework Answers

Answer #1

In equivalence point:

HB + NaOH ---> H2O + Na+ and B-

B- and H2O <----< HB+ and OH-

there will be more OH- ions, expect basic solution

Kb = [HB+][OH-] / [B-]

NOTE, volume has changed...

VT = 25 ml + V2... Since no notes on the NaOH solution is made, assume the same volume

Vt = 25 ml

Kb = [HB+][OH-] / [B-]

[HB+] = [OH-] = x

[B-] = 0.15

Substitute and solve for x

Kb = [HB+][OH-] / [B-]

Kb = Kw/Ka =(10^-14)/(4.5*10^-4) = 2.22*10^-11

2.22*10^-11 = x*x / (0.15)

3.32*10^-12 = x^2

X = 1.825*10^-6

[OH-] = x = 1.825*10^-6

pOH = -log(1.825*10^-6) = 5.74

pH = 14- pOH = 14-5.74 = 8.26

which is basic as expected

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
The titration curve for a 25.0 ml sample of benzoic acid shows that 10.0 ml of...
The titration curve for a 25.0 ml sample of benzoic acid shows that 10.0 ml of 0.100M NaOH is required to reach the equivalence point. (See #3 for Ka info.) i) What is the concentration of the original benzoic acid sample? ii) What is the pH at the equivalence point? iii) Calculate the pH after addition of 5.00 ml of NaOH and the pOH after addition of 20.0 ml NaOH.
Calculate the pH during the titration of 25.0 ml of 0.150 M benzoic acid with 0.150...
Calculate the pH during the titration of 25.0 ml of 0.150 M benzoic acid with 0.150 M NaOH after the following additions of titrant: 15.0 ml; 25.0 ml; 40.0 ml.
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 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)
You titrate a 20.0 mL acid sample containing 1.104 g of ascorbic acid (molar mass=176.12 g/mol...
You titrate a 20.0 mL acid sample containing 1.104 g of ascorbic acid (molar mass=176.12 g/mol and pKa=4.10) with 0.200 M NaOH. Calculate the following 1. The pH at the begining of the titration, before any base was added 2. The volume of base (NaOH) required to reach the equivalence point 3. The pH at the equivalence point and the half-equivalence point 4. The pH after 35.5 mL of base have beedn added
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?
Determine the pH during the titration of 74.5 mL of 0.345 M benzoic acid (Ka =...
Determine the pH during the titration of 74.5 mL of 0.345 M benzoic acid (Ka = 6.3×10-5) by 0.345 M KOH at the following points. (a) Before the addition of any KOH (b) After the addition of 17.0 mL of KOH (c) At the half-equivalence point (the titration midpoint) (d) At the equivalence point (e) After the addition of 112 mL of KOH
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)...
Consider the titration of a 25.0 −mL sample of 0.180 M CH3NH2 with 0.155 M HBr....
Consider the titration of a 25.0 −mL sample of 0.180 M CH3NH2 with 0.155 M HBr. Determine each of the following. a. the pH at one-half of the equivalence point b. the pH at the equivalence point c. the pH after adding 6.0 mL of acid beyond the equivalence point i already found that the initial pH is 11.95, the volume of acid added to reach equivelance point is 29.0mL, and the pH og 6.0mL of added acid is 11.23
Consider the titration of a 45.0 mL sample of 0.100 M HN3 with 0.250 M KOH....
Consider the titration of a 45.0 mL sample of 0.100 M HN3 with 0.250 M KOH. (Ka of HN3 = 2.5 x 10-5) Calculate each of the following: (a) How many mL are required to reach the equivalence point? (b) What is the initial pH of the acid solution? (c) What is the pH after the addition of 5.00 mL of KOH? (d) What is the pH after the addition of 9.00 mL of KOH? (e) What is the pH...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT