Question

6.0.35 g FeSO4 is dissolved to give 100 mL stock solution.25 mL of this stock solution in turn is diluted to 250 mL. Calculat


solup of an unknown H2SO4 solution is titrated to the endpoint with 34.50 mL 0.0946 M NaOH. Calculate the Molarity M of the s
0 0
Add a comment Improve this question Transcribed image text
Answer #1

86: 0:35 g Fe 804 in 100 mL stock solution. Molan mass of Feeson = 151.9 g mot 15119 g FeSO4 = 1 mole 1 g Feson = 151.9gmoit• Concentration of Fe 2t in stock solution (M)=2.3x10 m volume of the stock solution taken .. 25 m2 Volume of the solution af

Add a comment
Know the answer?
Add Answer to:
6.0.35 g FeSO4 is dissolved to give 100 mL stock solution.25 mL of this stock solution...
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. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • 1. In the second step she prepared the vinegar solution for the titration in the following...

    1. In the second step she prepared the vinegar solution for the titration in the following manner 25.0 mL of vinegar were diluted to 250.0 mL in a volumetric flask, and 25.0 mL of this diluted solu- tion required 8.27 mL of the above standardized NaOH (gquestion 5a) to reach the phenolphthalein endpoint. What is the molarity of the acid as it was titrated (diluted vinegar)? a. Use the dilution equation to determine the molarity of the acid prior to...

  • 5.00 mL of a solution containing a monoprotic acid was placed in a 100-mL volumetric flask,...

    5.00 mL of a solution containing a monoprotic acid was placed in a 100-mL volumetric flask, diluted to the mark with deionized water and mixed well. Then, 25.00 mL of this diluted acid solution was titrated with 0.08765 M NaOH. 9.23 mL of NaOH was required to reach the endpoint. a. What does the term monoprotic mean? b. Determine the molar ratio between the acid and NaOH. c. Calculate the moles of NaOH used in this titration. d. Calculate the...

  • A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml. 25.00 ml of the...

    A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml. 25.00 ml of the solution is taken out and is titrated to a neutral endpoint with 0.10 M NaOH. The titrated portion is then mixed with the remaining untitrated portion and the pH of the mixture is measured. Mass of acid weighed out (grams) 0.773 Volume of NaOH required to reach endpoint: (ml) 19.0 pH of the mixture Ihalf neutralized solution 3.54 Calculate the following...

  • 1. A solution of sodium hydroxide (NaOH) was standardized against potassium hydrogen phthalate (KHP). A known...

    1. A solution of sodium hydroxide (NaOH) was standardized against potassium hydrogen phthalate (KHP). A known mass of KHP was titrated with the NaOH solution until a light pink color appeared using phenolpthalein indicator. Using the volume of NaOH required to neutralize KHP and the number of moles of KHP titrated, the concentration of the NaOH solution was calculated. Molecular formula of Potassium hydrogen phthalate: HKC8H404 Mass of KHP used for standardization (g) 0.5100 Volume of NaOH required to neutralize...

  • Standardization of NaOH solution experiment: A 0.75 g sample of pure acid KHP was titrated to...

    Standardization of NaOH solution experiment: A 0.75 g sample of pure acid KHP was titrated to phenolphthalein endpoint using 45.34 ml NaOH of unknown concentration. The formula weight of pure KHP is 204.22g/mol. Write the chemical equation for the neutralization reaction, indicate the color change of the indicator at the endpoint, and calculate molarity of the NaOH solution. Show your work. 1. Chemical equation ------------- 2. Endpoint color change ---- 3. NaOH molarity --------- Determine percent purity of impure KHP...

  • 4. Vinegar is a solution of acetic acid. Acetic acid has a 1:1 ratio when reacted...

    4. Vinegar is a solution of acetic acid. Acetic acid has a 1:1 ratio when reacted with NaOH. a. Suppose that the molarity of acetic acid in vinegar is 0.5 M, how many grams of acetic acid are present in the 5 ml of vinegar solution used in the titration? (molar mass of acetic acid = 60 g/mol). b. If vinegar has a density of 1.01 g/mol. Calculate the % of acetic acid in vinegar by mass. 5. 50 ml...

  • A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml. 25.00...

    A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml. 25.00 ml of the solution is taken out and is titrated to a neutral endpoint with 0.10 M NaOH. The titrated portion is then mixed with the remaining untitrated portion and the pH of the mixture is measured. Mass of acid weighed out (grams) 0.762 Volume of NaOH required to reach endpoint: (ml) 18.0 pH of the mixture (half neutralized solution) 3.15 Calculate the following...

  • 25.00 mL of a sulfuric acid solution was standardized by titration with 0.2500 M NaOH using...

    25.00 mL of a sulfuric acid solution was standardized by titration with 0.2500 M NaOH using phenolpthalein indicator. 30.52 ml of the NaOH was required. Find the molarity of the sulfuric acid. H2SO4 (aq) + 2NaOH (aq) ----> Na2SO4 (aq) + 2H2O (l)

  • Questions: 1. 495 mg of dried KHP is dissolved in 35 mL of distilled water and...

    Questions: 1. 495 mg of dried KHP is dissolved in 35 mL of distilled water and titrated with potassium hydroxide (KOH). If it took 22.02 mL of KOH to reach the endpoint, determine the concentration of KOH. Show calculations. If you forgot to dry the above KHP sample and it was later determined to contain 5.00 % water by weight, what would be the actual concentration of the KOH, taking into account the water content? Show calculations. 2. If you...

  • A student prepared a base solution using 7 mL of 6M NaOH diluted with distilled water...

    A student prepared a base solution using 7 mL of 6M NaOH diluted with distilled water to a final volume of approximately 400 mL. The student titrated a standardized solution of HCl to determine the exact concentration of the prepared base solution. In the first titration, she found that it took 22.47 mL of the base solution to neutralize 20.00 mL of a 0.98861 M HCl solution. In the second titration, she found that it took 20.12 mL of the...

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
ADVERTISEMENT