Question

In a titration of .4534g of an unknown monoprotic acid, 45.95 mL of 0.1172M of NaOH...

In a titration of .4534g of an unknown monoprotic acid, 45.95 mL of 0.1172M of NaOH was required to reach the endpoint. what is the formula mass of the unknown acid? Plesse show steps thank you

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
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?
A 0.3012 g sample of an unknown monoprotic acid requires 24.13 mL of 0.0944 M NaOH...
A 0.3012 g sample of an unknown monoprotic acid requires 24.13 mL of 0.0944 M NaOH for neutralization to a phenolphthalein end point. There are 0.32 mL of 0.0997 M HCl used for back-titration. How many moles of OH- are used? How many moles of H+ from HCl? How many moles of H+ are there in the solid acid? (Use Eq. 5.) moles H+ in solid iacid =moles OH- in NaOH soln. - moles H+ in HCl soln. What is...
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...
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 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?
Titration of 0.01500 grams of diprotic acid required 36.22ml of 0.0120 M NaOH to reach the...
Titration of 0.01500 grams of diprotic acid required 36.22ml of 0.0120 M NaOH to reach the endpoint. What is the mass of a mole of the acid?
A solution of an unknown monoprotic acid was prepared by dissolving 72.9 mg of the acid...
A solution of an unknown monoprotic acid was prepared by dissolving 72.9 mg of the acid into 50. mL of water. 10.0 mL of 0.10 M NaOH is needed to titrate a 25 mL aliquot of the unknown monoprotic acid. From this info, calculate the molar mass of the unknown acid (g/mol).
a) If your titration solution is 0.409 M in NaOH, and the endpoint occues at 13.00...
a) If your titration solution is 0.409 M in NaOH, and the endpoint occues at 13.00 mL of titrant, how many mmol of NaOH are required to reach the endpoint? ____ mmol NaOH b) How many mmol of acetic acid (HC2H3O2) are required to react with the NaOH? ____ mmol HC2H3O2 c) How many grams of acetic acid is this? ____ grams HC2H3O2 d) If the mass of analyte is 10.15 grams, what is the mass % of acetic acid...
(a) If your titration solution is 0.410 M in NaOH, and the endpoint occurs at 13.50...
(a) If your titration solution is 0.410 M in NaOH, and the endpoint occurs at 13.50 mL of titrant, how many mmol of NaOH are required to reach the endpoint? __?__mmol NaOH (b) How many mmol of acetic acid (HC2H3O2) are required to react with the NaOH? __?__ mmol HC2H3O2 (c) How many grams of acetic acid is this? __?__ grams HC2H3O2 (d) If the mass of analyte is 10.05 grams, what is the mass % of acetic acid in...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT