Question

A 0.1255 g sample containing dimethylphthalate, C6H4(COOCH3)2 and unreactive species was refluxed with 50.00 mL of...

A 0.1255 g sample containing dimethylphthalate, C6H4(COOCH3)2 and unreactive species was refluxed with 50.00 mL of 0.1031 M NaOH to hydrolyze the ester groups.

C6H4(COOCH3)2 + 2OH- = C6H4(COO)22- + 2CH3O H

After the reaction was complete, the excess NaOH was back titrated with 24.27 mL of 0.1644 M HCl. Calculate the % hydrogen in the sample.   

0 0
Add a comment Improve this question Transcribed image text
Request Professional Answer

Request Answer!

We need at least 10 more requests to produce the answer.

0 / 10 have requested this problem solution

The more requests, the faster the answer.

Request! (Login Required)


All students who have requested the answer will be notified once they are available.
Know the answer?
Add Answer to:
A 0.1255 g sample containing dimethylphthalate, C6H4(COOCH3)2 and unreactive species was refluxed with 50.00 mL of...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Similar Homework Help Questions
  • 6. (10 points) A 0.9929-g sample containing dimethylphthalate, C6H4(COOCH3)2 (194.19 g/mol), and unreactive (inert) species was...

    6. (10 points) A 0.9929-g sample containing dimethylphthalate, C6H4(COOCH3)2 (194.19 g/mol), and unreactive (inert) species was refluxed with 50.00 mL of 0.1215 M NaOH to hydrolyze the ester groups (this process is called saponification). C6H4(COOCH3)2 + 2OH-  C6H4(COO)22- + 2CH3OH. After the reaction was complete, the excess NaOH was back-titrated with 24.53 mL of 0.1613 M HCl. Calculate the percentage of dimethylphthalate in the sample.

  • O.9471 g sample dimethylphthalate C6H4(COOCH3)2 (M.M.=194.19 g/mole) and unreactive species was reflux with 50.00 ml of...

    O.9471 g sample dimethylphthalate C6H4(COOCH3)2 (M.M.=194.19 g/mole) and unreactive species was reflux with 50.00 ml of 0.1215 M NaOH to hydrolyze the ester group. After the reaction was complete, the excess NaOH was back-titrated with 24.27ml of 0.1644 M HCl. Calculate the percentage of dimethylphthalate in the sample

  • A 50.00-ml NaOH sample of unknown concentration was titrated with 0.1274 M HCI.

    Titrations 1. A 50.00-ml NaOH sample of unknown concentration was titrated with 0.1274 M HCI. if 33.61 mL of the HCl solution were required to neutralize the NaOH sample, what is the molarity of the NaOH sample? Show the steps in your calculation  2. A 25.00-ml HCl sample of unknown concentration was titrated with 0.5631 M Al(OH)3. If 37.62 mL of the Al(OH) solution were required to neutralize the HCl sample, what is the molarity of the HCl sample? (Assume Al(OH)3 is...

  • 50.00-ml. sample containing acetic acid (CH.COOH, F.W. 60.05) was transferred into a 550.0-ml. volumetric flask and...

    50.00-ml. sample containing acetic acid (CH.COOH, F.W. 60.05) was transferred into a 550.0-ml. volumetric flask and diluted to the mark with water. A 25.00-ml portion of the dilute solution was titrated with 25.47 mL of 0.1012 M NaOH solution to the end point. Calculate the "% (w/v) of acetic acid in the sample. Keep appropriate number of significant figures for the result. (10 points)

  • Assume that a 50.00 mL sample of 0.1000 M HCl in an Erlenmeyer flask is titrated...

    Assume that a 50.00 mL sample of 0.1000 M HCl in an Erlenmeyer flask is titrated with 0.1000 M NaOH. (a) What is the initial pH before any base is added? (6) What is the pH of the solution after 10.00 mL of NaOH? (c) What is the pH of the solution after 20.00 mL of NaOH? (d) What is the pH of the solution after 50.00 mL of NaOH? (e) What is the pH of the solution after 60.00...

  • A 0.2726 g sample of metal was dissolved in 50.00 mL of 0.500 M HCl. After...

    A 0.2726 g sample of metal was dissolved in 50.00 mL of 0.500 M HCl. After all the metal had dissolved, the leftover acid was titrated with 0.1054 M NaOH. If 24.36 mL of 0.1054 M NaOH were required to neutralize the leftover acid, what was the atomic mass of the metal? The metal dissolved to form M+2 ions in solution. The correct answer is 24.3g, please show all your work, thank you.

  • The Fe2+ (55.845 g/mol) content of a 2.264 g steel sample dissolved in 50.00 mL of...

    The Fe2+ (55.845 g/mol) content of a 2.264 g steel sample dissolved in 50.00 mL of an acidic solution was determined by tiration with a standardized 0.120 M potassium permanganate (KMnO4, 158.034 g/mol) solution. The titration required 44.82 mL to reach the end point. What is the concentration of iron in the steel sample? Express your answer as grams of Fe per gram of steel Mn2+5 Fe34 H20 + 5 Fe2+ MnO8 H concentration g Fe/g steel A 1.969 g...

  • Hint Check A signment Score: Resources 100/300 uestion 3 of 3 A 1.749 g sample containing...

    Hint Check A signment Score: Resources 100/300 uestion 3 of 3 A 1.749 g sample containing an unknown amount of arsenic trichloride, with the rest being inerts, NAHCO, and HCl aqueous solution. To this solution was added 1.820 g of KI and 50.00 mL of a 0.00871M KIO, solution. The excess I was titrated with 50.00 mL of a 0.02000 M Na,S,O, solution. dissolved into a was What is the mass percent of arsenic trichloride in the original sample? mass...

  • 4. A 0.5255 g sample containing an unknown mass of SnCl2 is dissolved in 50.00 mL...

    4. A 0.5255 g sample containing an unknown mass of SnCl2 is dissolved in 50.00 mL of water and titrated with 0.02050 M KMnO4 under mildly acidic conditions. The unbalanced titration reaction is shown below. MnO4 (aq) + Sn²(aq) MnO2(8) + Sn** (aq) a. Use the half-reaction method to balance the titration reaction. (5 pts) b. Given that 32.25 mL of KMnO, was required to reach the endpoint, calculate the percent SnCl2 in the sample. (5 pts)

  • A 25.00 mL sample containing an unknown amount of Al3+ and Pb2+ required 17.03 mL of...

    A 25.00 mL sample containing an unknown amount of Al3+ and Pb2+ required 17.03 mL of 0.04947 M EDTA to reach the end point. A 50.00 mL sample of the unknown was then treated with F- to mask the A13+. To the 50.00 mL sample, 25.00 mL of 0.04947 M EDTA was added. The excess EDTA was then titrated with 0.02164 M Mn2+. A total of 25.3 mL was required to reach the methylthymol blue end point. Determine pA13+ and...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
Active Questions
ADVERTISEMENT