Question

Given the following information: 1.6g of an unknown monoprotic acid (HA) required 50.80mL of a .35M...

Given the following information: 1.6g of an unknown monoprotic acid (HA) required 50.80mL of a .35M NaOH solution to reach the equivalence point and the pH was 3.86 at 25.40mL, calculate the molar mass of the acid and the Ka of the unknown acid.

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
A solution was made by dissolving 0.580 g of an unknown monoprotic acid in water, and...
A solution was made by dissolving 0.580 g of an unknown monoprotic acid in water, and diluting the solution to a final volume of 25.00 mL. This solution was then titrated with 0.100 M NaOH. It took 36.80 mL of NaOH to reach the equivalence point, at which point the pH was 10.42. a. Determine the molar mass of the unknown acid. b. Calculate what the pH was after 18.40 mL of NaOH was added during the titration. Hint: the...
1. A 0.312 g of an unknown acid (monoprotic) was dissolved in 26.5 mL of water...
1. A 0.312 g of an unknown acid (monoprotic) was dissolved in 26.5 mL of water and titrated with 0.0850 M NaOH. The acid required 28.5 mL of base to reach the equivalence point. What is the molar mass of the acid? After 16.0 mL of base had been added in the titration, the pH was found to be 6.45. What is the Ka for the unknown acid?
The following data shows the titration of an unknown weak acid (called HA now) created by...
The following data shows the titration of an unknown weak acid (called HA now) created by dissolving 1.78 g of this acid in distilled water. Volume of 0.600 M NaOH added (1st measurement) pH of titrated solution (2nd #) 0.00 mL - ? 10.0 - 4.40 20.0 - ? 30.0 - 5.35 40.0 - 8.55 50.0 - 12.25 (a) The titration equivalence point occurred at the 40.0 mL data point. How does the data verify that HA is a weak...
A sample of 0.2140 g of an unknown monoprotic acid was dissolved in 25.0 mL of...
A sample of 0.2140 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0950MNaOH. The acid required 27.4 mL of base to reach the equivalence point. What is the molar mass of the acid?
A sample of 0.1387 g of an unknown monoprotic acid was dissolved in 25.0 mL of...
A sample of 0.1387 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.1150 MNaOH. The acid required 15.5 mL of base to reach the equivalence point. Part B After 7.25 mL of base had been added in the titration, the pH was found to be 2.45. What is the Ka for the unknown acid? (Do not ignore the change, x, in the equation for Ka.)
A 1.26 gram sample of an unknown monoprotic acid is dissolved in 50.0 mL of water...
A 1.26 gram sample of an unknown monoprotic acid is dissolved in 50.0 mL of water and titrated with a a 0.306 M aqueous potassium hydroxide solution. It is observed that after 12.7 milliliters of potassium hydroxide have been added, the pH is 3.193 and that an additional 19.3 mL of the potassium hydroxide solution is required to reach the equivalence point. (1) What is the molecular weight of the acid? ____ g/mol (2) What is the value of Ka...
A 0.872 gram sample of an unknown monoprotic acid is dissolved in 50.0 mL of water...
A 0.872 gram sample of an unknown monoprotic acid is dissolved in 50.0 mL of water and titrated with a a 0.424 M aqueous sodium hydroxide solution. It is observed that after 9.40 milliliters of sodium hydroxide have been added, the pH is 3.350 and that an additional 5.50 mL of the sodium hydroxide solution is required to reach the equivalence point. (1) What is the molecular weight of the acid? g/mol (2) What is the value of Ka for...
A 1.54 gram sample of an unknown monoprotic acid is dissolved in 30.0 mL of water...
A 1.54 gram sample of an unknown monoprotic acid is dissolved in 30.0 mL of water and titrated with a a 0.225 M aqueous potassium hydroxide solution. It is observed that after 19.3 milliliters of potassium hydroxide have been added, the pH is 7.171 and that an additional 34.5 mL of the potassium hydroxide solution is required to reach the equivalence point. (1) What is the molecular weight of the acid?  g/mol (2) What is the value of Ka for the...
A 1.22 gram sample of an unknown monoprotic acid is dissolved in 25.0 mL of water...
A 1.22 gram sample of an unknown monoprotic acid is dissolved in 25.0 mL of water and titrated with a a 0.299 M aqueous potassium hydroxide solution. It is observed that after 10.4 milliliters of potassium hydroxide have been added, the pH is 3.039 and that an additional 19.4 mL of the potassium hydroxide solution is required to reach the equivalence point. (1) What is the molecular weight of the acid? g/mol (2) What is the value of Ka for...
You have a 1.153 g sample of an unknown solid acid, HA, dissolved in enough water...
You have a 1.153 g sample of an unknown solid acid, HA, dissolved in enough water to make 20.00 mL of solution. HA reacts with KOH(aq) according to the following balanced chemical equation: HA(aq)+KOH(aq)--->KA(aq)+H2O(l) If 13.40 mL of 0.715 M KOH is required to titrate the unknown acid to the equivalence point, what is the concentration of the unknown acid? What is the molar mass of HA?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT