Question 13 of 14 A student performs a titration of 37.0 mL of a phosphoric acid (H3PO4) solution of unknown concentration with a standardized 1.25 M NaOH solution. The titration requires 26.2 mL of base to reach the third equivalence point. What is the concentration of the H3PO4 solution? 0.4425 M 4 6. 7 8 9. +/- x 10 10:50 PM 4/15/2021 P Type here to search End 19 PgUp PgDn VI Home PrtScn F7 3. 1. 近

Question 13 of 14 A student performs a titration of 37.0 mL of a phosphoric acid (H3PO4) solution of unknown concentration with a standardized 1.25 M NaOH solution. The titration requires 26.2 mL of base to reach the third equivalence point. What is the concentration of the H3PO4 solution? 0.4425 M 4 6. 7 8 9. +/- x 10 10:50 PM 4/15/2021 P Type here to search End 19 PgUp PgDn VI Home PrtScn F7 3. 1. 近

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...

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Questions 5-10 refer to the same weak acid/strong base (WA/SB) titration. The K₂ of hypobromous acid is 2.8 x 10-⁹.5 A 30.00 mL solution of 0.300 M hypobromous acid (HBrO) is being titrated with a solution that is 0.600 M in lithium hydroxide (LiOH). What is the solution pH when 25.00 mL of LiOH have been added? (Two decimal places) Type your answer...
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Question 6 Type numbers in the boxes. 45.0 mL of o.650 M HNO, is titrated by 0.650 M KOH. aby Part 1: 10 points Calculate the pH of the acid solution before any titrant is aby Part 2: 10 points added. aby Part 3: 1 points pH = aby Part 4: 10 points Calculate the pH after 1.47 mL of o.650 M KOH is added 31 points to 45.0 mL of 0.650 M HNO,. pH = Calculate the pH after 45.00 mL of o.650 M KOH is added to 45.0 mL of o.650 ΜΗΝΟ. pH = Calculate the pH after 89.5 mL of o.650 M KOH is added to 45.0 mL of 0.650 M HNO,- pH =
An HCl solution has a concentration of 0.09714 M. Then 10.00 mL of this solution was then diluted to 250.00 mL in a volumetric flask. The diluted solution was then used to titrate 250.0 mL of a saturated AgOH solution using methyl orange indicator to reach the endpoint. 1. What is the concentration of the diluted HCl solution? 2. If 7.93 mL of the diluted HCl solution was required to reach the endpoint, what is the concentration of OH− in solution? 3. What is the concentration of Ag+ in solution?
Suppose a student performed a similar standardization titration experiment. Below is the calculated concentration of NaOH from each of their titrations. Titration #1 Titration #2 Titration #3 The concentration of NaOH (M) 0.9554 0.9540 0.9551 Using the 3 concentrations, calculate the average concentration of the student's NaOH solution and calculate the ± error in parts per thousand (ppt). Show the complete calculation. Express your final answer in the form of average (with units) ± error in ppt.
of 15 > The identity of an unknown monoprotic organic acid is determined by titration. A 0.300 g sample of the acid is titrated with 0.103 M NaOH. What is the molar mass of the compound if 16.2 mL of the NaOH solution is required to neutralize the sample? molar mass: The compound is 60.0% C, 4.48% H, and 35.5% O. What is the molecular formula of the organic acid? formula: $ 4 C R F LL dº 5 % V T G 6 MacBook Pro B Y H & 7 N * 00 8 J 1 M ( 9 K O 1 L command P A. : i { option [ + 11 = ? mol 1
1. How do you prepare a 100mL of 0.1 M phosphate buffer?To make 100 mL of 0.1 M phosphate buffer: Calculate the amount of sodium phosphate needed. moles of sodium phosphate = (0.1 mol/L) x (0.1 L) = 0.01 moles mass of sodium phosphate = moles x molar mass = 0.01 moles x 142 g/mol = 1.42 g Dissolve 1.42 g of sodium phosphate in distilled water in a 100 mL volumetric flask. Adjust the pH of the solution to the desired value (usually around 7.4) using a strong acid or strong base. Bring the solution to the final volume (100 mL) with distilled water. 2. From the anterior buffer, how do you make 100mL of 0.05 M? To make 100 mL of 0.05 M phosphate buffer from the 0.1 M stock solution: Calculate the amount of the 0.1 M phosphate buffer needed. moles of phosphate buffer = (0.05 mol/L) x (0.1 L) = 0.005 moles Calculate the volume of the 0.1 M phosphate buffer needed. moles = concentration x volume (in liters) volume = moles / concentration = 0.005 moles / 0.1 mol/L = 0.05 L or 50 mL Measure…
Select the answer that best describes the experimental setup of the titration curve shown in the image below 12 Hd H] 8 4 0 10 volume of titrant (mL) 20 20 strong base (beaker) and strong acid (burette) weak base (beaker) and strong acid (burette) weak acid (beaker) and strong base (burette) strong acid (beaker) and strong base (burette)
40 mL of LiOH are titrated to the equivalence point with 65 mL of 1.2 x 10-2 M HNO3. Find the concentration (molarity) of the LiOH solution.
5 6 7 8 Questions 5-10 refer to the same weak acid/strong base (WA/SB) titration. The K, of hypobromous acid is 2.8 x 10¹⁹. A 30.00 mL solution of 0.300 M hypobromous acid (HBrO) is being titrated with a solution that is 0.600 M in lithium hydroxide (LiOH). What is the initial solution pH (i.e., when 0.00 mL of titrant have been added)? (Two decimal places) Type your answer... Questions 5-10 refer to the same weak acid/strong base (WA/SB) titration. The K, of hypobromous acid is 2.8 x 10¹⁹. A 30.00 mL solution of 0.300 M hypobromous acid (HBrO) is being titrated with a solution that is 0.600 M in lithium hydroxide (LiOH). What volume (in mL) of titrant will have been added at the half-equivalence point? (One decimal place) Type your answer... Questions 5-10 refer to the same weak acid/strong base (WA/SB) titration. The K₂ of hypobromous acid is 2.8 x 10.⁹. A 30.00 mL solution of 0.300 M hypobromous acid (HBrO) is being titrated with a solution that is 0.600 M in lithium hydroxide…
Determine the molarity of a 7.00 ml HCl solution if it takes 48.35 mL of a0.1315M NaOH solution to neutralize in a titration. HCl+NaOH-->NaCl +H2O
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