Question

A 1.550-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of...

A 1.550-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 25.0 mL of this solution was titrated with 0.06307-M NaOH. The pH after the addition of 15.77 mL of base was 4.73, and the equivalence point was reached with the addition of 37.11 mL of base.

a) How many millimoles of acid are in the original solid sample? Hint: Don't forget the dilution.

mmol acid

b) What is the molar mass of the acid?

g/mol

c) What is the pKa of the acid?

pKa =

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 1.224-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of...
A 1.224-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 55.0 mL of this solution was titrated with 0.08096-M NaOH. The pH after the addition of 12.88 mL of base was 7.00, and the equivalence point was reached with the addition of 41.81 mL of base. a) How many millimoles of acid are in the original solid sample? Hint: Don't forget the dilution. mmol acid b) What is the molar mass of the...
A 2.211-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of...
A 2.211-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 30.0 mL of this solution was titrated with 0.06886-M NaOH. The pH after the addition of 25.59 mL of base was 4.22, and the equivalence point was reached with the addition of 41.47 mL of base. a) How many millimoles of acid are in the original solid sample? Hint: Don't forget the dilution. mmol acid b) What is the molar mass of the...
A 3.670-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of...
A 3.670-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 20.0 mL of this solution was titrated with 0.07662-M NaOH. The pH after the addition of 23.98 mL of base was 5.90, and the equivalence point was reached with the addition of 44.00 mL of base. a) How many millimoles of acid are in the original solid sample? Hint: Don't forget the dilution. mmol acid b) What is the molar mass of the...
A 1.074-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of...
A 1.074-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 40.0 mL of this solution was titrated with 0.06810-M NaOH. The pH after the addition of 24.91 mL of base was 4.77, and the equivalence point was reached with the addition of 37.97 mL of base. a) How many millimoles of acid are in the original solid sample? Hint: Don't forget the dilution. _______ mmol acid b) What is the molar mass of...
A 1.731-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of...
A 1.731-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 45.0 mL of this solution was titrated with 0.09322-M NaOH. The pH after the addition of 11.78 mL of base was 4.23, and the equivalence point was reached with the addition of 36.47 mL of base. a) How many millimoles of acid are in the original solid sample? Hint: Don't forget the dilution. ______mmol acid b) What is the molar mass of the...
A 5.171 g sample of a solid, weak, monoprotic acid is used to make a 100.0...
A 5.171 g sample of a solid, weak, monoprotic acid is used to make a 100.0 mL solution. 26.00 mL of the resulting acid solution is then titrated with 0.09671 M NaOH. The pH after the addition of15.00 mL of the base is 5.51, and the endpoint is reached after the addition of 47.85 mL of the base. (b) What is the molar mass of the acid? (c) What is the pKa of the acid?
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 certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH. 1)A...
A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH. 1)A solution is made by titrating 9.00 mmol (millimoles) of HA and 1.00 mmol of the strong base. What is the resulting pH? 2)More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 41.0 mL ?
A 0.6071 g sample of a weak monoprotic acid was titrated to its endpoint using 0.1325...
A 0.6071 g sample of a weak monoprotic acid was titrated to its endpoint using 0.1325 M NaOH. The initial volume of NaOH was 0.58 mL. The final volume of NaOH was 37.98 mL. What is the molar mass of the weak monoprotic acid? Show all work.
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT