Question

a) 30.0 ml of an HCl solution of unknown molarity was titrated with .200 M NaOH....

a) 30.0 ml of an HCl solution of unknown molarity was titrated with .200 M NaOH. Phenolphthalien was used as the indicator. The titrated solution turned a very pale pink after 21.8 mL of NaOH was added. What was the initial molarity of the HCl?

b)Consider the following three titrations:

100.0 mL of 0.100 M CH3NH2 (Kb = 4.4x10-4) titrated with 0.100 M HCl

100.0 mL of 0.100 M NH3 (Kb = 1.8x10-5) titrated with 0.100 M HCl

100.0 mL of 0.100 M HF (Ka = 3.5x10-4) titrated with 0.100 M NaOH

Which of the titrations will have the lowest pH at the halfway point?

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

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
What is the molarity of an HCl solution if 13.0 mL HCl solution is titrated with...
What is the molarity of an HCl solution if 13.0 mL HCl solution is titrated with 26.6 mL of 0.175 M NaOH solution? HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
1) Rank the following titrations in order of increasing pH at the halfway point to equivalence...
1) Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 = highest pH). 100.0 mL of 0.100 M KOH by 0.100 M HCl 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 HF (Ka = 7.2 x 10-4) by 0.100 M NaOH...
If a 30.0 mL solution of 0.40M barium hydroxide is titrated with 0.25M HCl using an...
If a 30.0 mL solution of 0.40M barium hydroxide is titrated with 0.25M HCl using an indicator, HIn, with a Ka of 3.4x10^-6. What is the total volume of the solution when the indicator first changes color?
1) Rank the following titrations in order of increasing pH at the halfway point to equivalence...
1) Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 = highest pH). 12345 100.0 mL of 0.100 M HC3H5O2 (Ka = 1.3 x 10-5) by 0.100 M NaOH 12345 100.0 mL of 0.100 M HCl by 0.100 M NaOH 12345 100.0 mL of 0.100 M KOH by 0.100 M HCl 12345 100.0 mL of 0.100 M NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCl 12345...
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...
A 600.0 mL sample of 0.20 M HF is titrated with 0.20 M NaOH. Determine the...
A 600.0 mL sample of 0.20 M HF is titrated with 0.20 M NaOH. Determine the pH of the solution after the addition of 100.0 mL of NaOH. The Ka of HF is 3.5 * 10^-4
A 25 mL aliquot of an HCl solution is titrated with 0.100 M NaOH. The equivalence...
A 25 mL aliquot of an HCl solution is titrated with 0.100 M NaOH. The equivalence point is reached after 21.27 mL of the base were added. Calculate the concentration of the acid in the original solution, the pH of the original HCl solution and the original NaOH solution
1. A25.0 mL sample of 0.100 M lactic acid (hc3h5o5, pka= 3.86) is titrated with a...
1. A25.0 mL sample of 0.100 M lactic acid (hc3h5o5, pka= 3.86) is titrated with a 0.100 M NaOH solution. Calculate the pH after the addition of 0.0, 4.0, 8.0, 12.5, 20.0, 24.0, 24.5, 24.9, 25.0, 25.1, 26.0, 28.0, and 30.0 of the NaOH. Plot the results of your calculation, as a pH versus mililiters of NaOH added. 2. Repeat the procedure in problem 1, but the titration of 25.0 mL 0.100 M NH3 (kb= 1.8*10^-5) with 0.100 M HCl....
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...
A 100.0 mL sample of 0.20 M HF is titrated with 0.10M KOH. Determine the pH...
A 100.0 mL sample of 0.20 M HF is titrated with 0.10M KOH. Determine the pH of the solution before teh addition of any KOH. The Ka of HF is 3.5x10^-4
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT