EXPERIMENT 5; Acid-Base TitrationA.Trial 1Trial2Trial 35.210 gMass of KHP5.210 g5.210 g(C3H,O4)Mole of KHPInitial Burette0.000.505.00Volume NaOH (mL)Final Burette31.0026.2529.95Volume NaOH (mL)Total Volume ofNaOH (mL)Moles of NaOHMolarity of NaOH(mol/L)Calculate the average Molarity of NaOHB. Titration of Vinegar (HC,H3O2)A 2.352 g sample of Vinegar is titrated with 0.08751MNaOH, and it requires 22.31 mL of NaOH to reach theendpoint. Calculate the mass percent of acetic acid(HC,H;O,) in the vinegar sample?Hint for Calculation of Mass percent of Vinegar(HC,H;O2)1. Calculate mole of HC,H;O2 from concentration(M) and volume of NaOH (L)needed to titrate Vinegar2. Calculate mass (g) of Vinegar from mole ofVinegar3. Mass % Vinegar = (calculated mass Vinegar/2.352 g) X 100C. A solution of malonic acid, H;C;H¿O4 ¸wasstandardized by titration with 0.100 M NaOH solution.If 20.76 mL of the NaOH solution were required toneutralize completely 13.15 mL of the malonic acidsolution, what is the molarity of the malonic acidsolution?

Hii there, As there are multiple question posted. we are answering first question. If you need…

EXPERIMENT 5; Acid-Base Titration A. Trial 1 Trial 2 Trial 3 Mass of KHP 5.210 g 5.210 g 5.210 g (C3H;O4) Mole of KHP ---------- Initial Burette 0.00 0.50 5.00 Volume NaOH (mL) Final Burette 31.00 26.25 29.95 Volume NaOH (mL) Total Volume of NaOH (mL) Moles of NaOH Molarity of NaOH (mol/L) Calculate the average Molarity of NAOH

EXPERIMENT 5; Acid-Base Titration A. Trial 1 Trial 2 Trial 3 Mass of KHP 5.210 g 5.210 g 5.210 g (C3H;O4) Mole of KHP ---------- Initial Burette 0.00 0.50 5.00 Volume NaOH (mL) Final Burette 31.00 26.25 29.95 Volume NaOH (mL) Total Volume of NaOH (mL) Moles of NaOH Molarity of NaOH (mol/L) Calculate the average Molarity of NAOH

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Questions 11-14 refer to the same strong base/strong acid (SB/SA) titration. A 15.00 mL solution of 0.100 M sodium hydroxide (NaOH) is being titrated with 0.250 M hydrobromic acid (HBr). What is the total solution volume (in mL) at the equivalence point? (Nearest whole number)
Question Completion Status: For Solution A, approximately how many mL of HCI were needed to exceed the buffering capacity of this solution? Enter the number only. Don't enter units as part of your answer and don't use commas or scientific notation. Titration curve of Solution A 14 13 12 11- 10 9 8 pH 7 6 5 4 3 2 1 5 4 3 2 1 HCI added (mL) 0 1 2 3 4 5 NaOH added (mL) 14 13 12- 11 10 9 8 pH 7 6.8 2 65 4 3 2 1 Titration curve of Solution B Click Save and Submit to save and submit. Click Save All Answers to save all answers. 54 32 1 HCI added (mL) 0123 NaOH added (mL)
Consider the titration of 80.0 mL of 0.100 M Ba (OH), by 0.400 M HCI. Calculate the pH of the resulting solution after the following volumes of HCI have been added. a. 0.0 mL pH = b. 20.0 mL pH = c. 30.0 mL pH= d. 40.0 mL PH= e. 90.0 mL pH = Submit Answer Try Another Version 1 item attempt remaining
3
Part C. Determination of Unknown Balanced chemical equation: Hcl+ Na OH -o Nacl t Hg0 Unknown number NA Trial 1 Trial 2 Trial 3 Volume of unknown HCl titrated 10.00mL NA NA Initial burette reading 0.90mL Final burette reading 30.90mL Volume NaOH added 20 00ML 50.00MLXO.144 m=43.2m moles NaOH added 4.335m (Use average molarity from part B and multiply it by the volume you added) 0.4335 M moles HCl 4.335m 04335m 10.00 mL molarity of HCl Actual Molarity (from instructor) 0.3150M percentage error
What is the pH at the equivalence point in the titration of 60.00 mL of 0.200 M HA (a generic acid with K, = (2.26x10^-5)) with 0.400 M NaOH? Enter your answer in scientific notation with 3 sig figs. Do not include any units in your answer. Do not round any intermediate calculations. Note: Your answer is assumed to be reduced to the highest power possible. Your Answer: x10 Answer
hi, can you reexplain the 0.6149 g * 1 mol KHP204.22 g= 0.003011 mol KHP Moles of KHP(KHC8H4O4) used in second titration = 0.003011 mol (Answer)
Assume that the molarity of the standardized KMnO4 solution is 0.025 M and it takes 23.00 mL to reach the endpoint when titrating a dilute sample of H2O2. How many moles of H2O2 are in the diluted sample initially? Group of answer choices
2 Match the following acid-base titration plots with the correct titrant and sample. Veq = 20.00 mL mea a. b. Hd (1) Titrant: Strong Acid; Sample: Strong Base (2) Titrant: Strong Base; Sample: Strong Acid (3) Titrant: Strong Acid; Sample: Weak Base (4) Titrant: Strong Base; Sample: Weak Acid 3219 11 10 9 8 7 6 5 4 3 2 1 0 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 0 0 5 5 10 10 15 Volume 15 Volume 20 20 25 25 30 30
Pls ans all
1 L - 1 F F D 1 F F C J 7₁ C " I F < 4:32 Question 11 of 13 A) CN and HCN B) CN- What is the primary species in solution at the halfway point in a titration of HCN with KOH? C) HCN and OH- D) OH- all? Tap here or pull up for additional resources Submit
A student performed three H2SO4 titrations in Part 2 using their 0.1048 M NaOH solution which they previously standardized in Part 1. They reported the following measurements: Determination #1 Determination #2 Determination #3 Final NaOH Burette Reading 37.65 mL 28.35 mL 26.23 mL Initial NaOH Burette Reading 12.80 mL 4.05 mL 1.73 mL For every 1 mole of H2SO4, how many moles of NaOH are needed to reach equivalence in the titration? Report the integer number only (no units).