Question

What is the pH at the equivalence point in the titration of 0.20 L of 0.66...

What is the pH at the equivalence point in the titration of 0.20 L of 0.66 M HF with 0.33 M NaOH? (F- Kb = 1.4 x 10-11)

a. 7.00

b. 5.76

c. 8.24

d. 3.25

e. 10.75

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

NaOH + HF ----> Na+ + F- + H2O

At equivalence point all HF is converted to F-, NaF being salt of a strong base and a weak acid. It is completely dissociated.

So pH is controlled by F-

F- + H2O <-----> HF + OH-

Kb = [HF][OH-] / [F-]

use ICE table we can write: Kb = x^2 / (0.66-x)

we know HF is a weak acid and hence 0.66-x = 0.66.

x = ((1.4*10^-11)*0.66)^0.5 = (0.924*10^-11)^0.5 = 3.039*10^-6 M

Thus, pOH = -log(3.039*10^-6) = 5.5

pH = 14-5.5 = 8.5

We have used approximation at the point 0.66-x = 0.66.

Hence our answer should be 8.24.

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Rank the following titrations in order of increasing pH at the equivalence point of the titration...
Rank the following titrations in order of increasing pH at the equivalence point of the titration (1 = lowest pH and 5 = highest pH). 100.0 mL of 0.100 M C2H5NH2 (Kb = 5.6 x 10-4) by 0.100 M HCl 100.0 mL of 0.100 M HF (Ka = 7.2 x 10-4) by 0.100 M NaOH 100.0 mL of 0.100 M HC3H5O2 (Ka = 1.3 x 10-5) by 0.100 M NaOH 200.0 mL of 0.100 M H2NNH2 (Kb = 3.0 x...
Rank the following titrations in order of increasing pH at the equivalence point of the titration...
Rank the following titrations in order of increasing pH at the equivalence point of the titration (1 = lowest pH and 5 = highest pH). 100.0 mL of 0.100 M HC3H5O2 (Ka = 1.3 x 10-5) by 0.100 M NaOH 100.0 mL of 0.100 M C2H5NH2 (Kb = 5.6 x 10-4) by 0.100 M HCl 100.0 mL of 0.100 M KOH by 0.100 M HCl 100.0 mL of 0.100 M HF (Ka = 7.2 x 10-4) by 0.100 M NaOH...
Which of the following titrations will have the highest pH at the equivalence point? A) A...
Which of the following titrations will have the highest pH at the equivalence point? A) A titration of 0.25 M CH3N2H (Kb = 4.4x10-4) with 0.25 M HNO3 B) A titration of 0.25 M HClO (Ka = 3.5x10-8) with 0.25 M LiOH C) A titration of 0.25 M HF (Ka = 7.4x10-4) with 0.25 M LiOH D) A titration of 0.25 M LiOH with 0.25 M HNO3 E) A titration of 0.25 M C6H5NH2 (Kb = 4.6x10-10) with 0.25 M...
Determine the pH at the equivalence point for the titration of 25.0 mL of 0.50M HF...
Determine the pH at the equivalence point for the titration of 25.0 mL of 0.50M HF solution with 0.40 M NaOH solution. This is what I have so far: First determine Vb = Veq Ma * Va = Mb * Vb 0.5 * 25 = 0.4 * Vb Vb = 31.25 mL Moles of HF = 0.5 M * 0.025 L = 0.0125 mol HF Moles of NaOH = 0.4 M * 0.03125 L = 0.0125 mol NaOH Total Volume...
What is the pH at the equivalence point of a titration of the weak base NH3(aq)...
What is the pH at the equivalence point of a titration of the weak base NH3(aq) with the strong acid HBr(aq) if 30.00 mL of 0.200 M NH3 solution requires 30.00 mL of 0.200 M HBr to reach the equivalence point? Kb = 1.8 x 10–5 for NH3.
2)Rank the following titrations in order of increasing pH at the equivalence point of the titration...
2)Rank the following titrations in order of increasing pH at the equivalence point of the titration (1 = lowest pH and 5 = highest pH). 100.0 mL of 0.100 M HC3H5O2 (Ka = 1.3 x 10-5) by 0.100 M NaOH 200.0 mL of 0.100 M (C2H5)2NH (Kb = 1.3 x 10-3) by 0.100 M HCl 100.0 mL of 0.100 M NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCl 100.0 mL of 0.100 M KOH by 0.100 M HCl...
What is the pH at the equivalence point for the titration of a 25.0mL sample of...
What is the pH at the equivalence point for the titration of a 25.0mL sample of 0.175 M CH3NH2 with 0.150M HBr (Kb = 4.4x10-4)
Calculate the pH at the equivalence point in the titration of the weak base, B, having...
Calculate the pH at the equivalence point in the titration of the weak base, B, having a Kb of 1.00 x 10−8 when the initial formal concentration of the B titrand is 0.01000F and its initial volume is 0.1000 L. The titrant is standardized HCl with a concentration of 0.1000 M.
Calculate the pH at the equivalence point for the titration of 0.180 M methylamine (CH3NH2) with...
Calculate the pH at the equivalence point for the titration of 0.180 M methylamine (CH3NH2) with 0.180 M HCl. The Kb of methylamine is 5.0× 10–4. pH=?
1. Determine the pH at the equivalence (stoichiometric) point in the titration of 28 mL of...
1. Determine the pH at the equivalence (stoichiometric) point in the titration of 28 mL of 0.17 M H3BO3(aq) with 0.11 M NaOH(aq). The Ka of H3BO3 is 5.8 x 10-10.
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT