Question

Assuming the mass of the weak acid used in the titration was 0.138 g, what is...

Assuming the mass of the weak acid used in the titration was 0.138 g, what is the molar mass (MM) of the weak acid? (The standardized base is 0.100 M NaOH.)

Homework Answers

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
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?
Titration of a weak acid with a strong base. the ph curve for titration of 50.0ml...
Titration of a weak acid with a strong base. the ph curve for titration of 50.0ml of a 0.100 M of acetic acid with a 0.100 M solution of NaOH (aq). For clarity, water molecules have been omitted from the molecular art. a) If the acetic acid being titrated here were replaced by hydrochloric acid, would the amount of base needed to reach the equivalence point change? b) Would the pH at the equivalence point change? -yes the pH at...
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 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...
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 component in a sleeping pill is a weak monoprotic acid (HX). A titration required 12.00...
A component in a sleeping pill is a weak monoprotic acid (HX). A titration required 12.00 mL of 0.200 M NaOH to titrate 0.4421 g of this acid. (a) What is the molar mass of this compound? (b)If this acid has the percent composition of 52.162 % C, 6.567 % H, 15.212 % N, and 26.059 % O, what is the empirical formula and molecular formula of this aicd?
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 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 titration is done using a 0.1302 M NaOH solution to determine the molar mass of...
A titration is done using a 0.1302 M NaOH solution to determine the molar mass of a monoprotic weak acid (HA). If 50.00 mL of a 1.863 g sample of the unknown acid (HA) is titrated to the equivalence point/end point with 70.11 mL of a 0.1302 M NaOH aqueous solution, what is the molar mass of the unknown acid?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT