ZuoTi.Pro
Question

(617.5 ml of 9.00 M NHAOH were added to 25.0 ml of H3PO4 to reach the endpoint. Find the molarity of H3PO4 (Balance the react
science   chemistry   
0 0
Add a comment Improve this question Transcribed image text
Next > < Previous
Sort answers by oldest

Homework Answers

Answer #1

Balanee equatton Poy 3My014 mal of Poy reaes wth 3 moeo NHuD (N3Poy t3 2o Let M a malacthg 3 py Males of poy = 025 L M 0175 L

0 0
Add a comment
Know the answer?
Add Answer to:
(617.5 ml of 9.00 M NHAOH were added to 25.0 ml of H3PO4 to reach the...
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. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions

  • urgent!!! The neutralization reaction between ammonium hydroxide (NH4OH, MM = 35.04 g/mol) and phosphoric acid (H3PO4,...

    urgent!!! The neutralization reaction between ammonium hydroxide (NH4OH, MM = 35.04 g/mol) and phosphoric acid (H3PO4, MM = 98.00 g/mol) proceeds according to the following balanced eqation: 3NH4OH(aq) + H3PO4(aq) ->3H20(1) + (NH4)3PO4(aq) What is the final molarity of the ammonium ion (NH4+, MM = 18.05 g/mol) when 25.0 mL of 1.20 molar ammonium hydroxide is reacted with 25.0 mL of 2.00 molar phosphoric acid?

  • final molarity of ammonium The neutralization reaction between ammonium hydroxide (NH4OH, MM = 35.04 g/mol) and...

    final molarity of ammonium The neutralization reaction between ammonium hydroxide (NH4OH, MM = 35.04 g/mol) and phosphoric acid (H3PO4, MM 98.00 g/mol) proceeds according to the following balanced eqation: 3NH2OH(aq) + H3PO4(aq)->3H20(1) + (NH4)3PO4(aq) What is the final molarity of the ammonium ion (NH4+. MM = 18.05 g/mol) when 25.0 mL of 1.20 molar ammonium hydroxide is reacted with 25.0 mL of 2.00 molar phosphoric acid?

  • An aqueous solution of Ca(OH)2with a concentration of 0.143 M was used to titrate 25.00 mL...

    An aqueous solution of Ca(OH)2with a concentration of 0.143 M was used to titrate 25.00 mL of aqueous HCl. 14.73 mL of the Ca(OH)2was required to reach the endpoint of the titration. How many moles of base were required to react completely with the acid in this reaction? How many moles of HCl were present in the original 25.00 mL of acid? What is the molarity of the original HCl solution? 04 Question (a points) aSee page 166 Watch the...

  • → X COD Data: & p 029 A 25.0 mL volume of 0.235 M Sr(NO3)2 is...

    → X COD Data: & p 029 A 25.0 mL volume of 0.235 M Sr(NO3)2 is added to 20.0 mL 0.300 M (NH4)3PO4. What is the maximum amount of solid product that can be collected? XI FIRE, ALT+F10 (PC) ER ALT+FN+F10 (Mac). TT TT Arial 3 (12pt) E.E. T. % DO Q E SE T' T, mq - - - O fx Mashups - 16 @ @ % :p 105 10 A 25.0 ml volume of 0.235 M Sr(NO3)2 is...

  • 9. a If 25.0 mL of KOH solution is neutralized by 13.31 mL of 0.111 M...

    9. a If 25.0 mL of KOH solution is neutralized by 13.31 mL of 0.111 M H3PO4, what is the molarity of the KOH? State formula, of course you were going to do that, right? b) How many mL of 0.123 M H2SO4 solution is needed to neutralize 25,0 mL of 0.250 M KOH solution?

  • Suppose you have 50.5 mL of a solution of H3PO4. You titrate it with a 0.10...

    Suppose you have 50.5 mL of a solution of H3PO4. You titrate it with a 0.10 M solution of Sr(OH)2 and use 15.3 mL to reach the endpoint. What is the concentration of H3PO4? What was the pH of the original H3PO4 solution?

  • An aqueous solution of Ca(OH)2with a concentration of 0.161 M was used to titrate 25.00 mL...

    An aqueous solution of Ca(OH)2with a concentration of 0.161 M was used to titrate 25.00 mL of aqueous HCl. 18.63 mL of the Ca(OH)2was required to reach the endpoint of the titration. Part 1 (1 point) How many moles of base were required to react completely with the acid in this reaction? x 10 mol Ca(OH)2 -- Part 2 (1 point) How many moles of HCl were present in the original 25.00 mL of acid? x 10 mol HCl +...

  • A 15.44 mL sample of acetic acid requires 32.10 mL of 0.0500 M Ba(OH)2 to reach...

    A 15.44 mL sample of acetic acid requires 32.10 mL of 0.0500 M Ba(OH)2 to reach the endpoint. a) Calculate the molarity of the unknown acid b) what is the theoretical pH when 15.8 mL of Ba(OH)2 is added? c) What is the theoretical pH at the point when a total of 32.10 mL Ba(OH)2 is added?

  • Calculate the molarity of H2SO4 that results after 10.0 mL of H2O are added to 25.0...

    Calculate the molarity of H2SO4 that results after 10.0 mL of H2O are added to 25.0 mL of 0.500 M H2SO4 in preparation for the following reaction. Assume the volumes are additive. H2SO4(aq)+ 2NaOH—> Na2SO4(aq) + 2H2O(l) answer is: 0.357 M

  • You are instructed to create 400. mL of a 0.40 M phosphate buffer with a pH...

    You are instructed to create 400. mL of a 0.40 M phosphate buffer with a pH of 6.4. You have phosphoric acid and the sodium salts NaH2PO4, Na2HPO4, and Na3PO4 available. (Enter all numerical answers to three significant figures.) H3PO4(s) + H2O(l) H3O+(aq) + H2PO4−(aq)     Ka1 = 6.9 ✕ 10−3 H2PO4−(aq) + H2O(l) H3O+(aq) + HPO42−(aq)     Ka2 = 6.2 ✕ 10−8 HPO42−(aq) + H2O(l) H3O+(aq) + PO43−(aq) Ka3 = 4.8 ✕ 10−13 Which of the available chemicals will you use...

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