Question

Experiment #13—Acid-Base Stoichiometry Experimental data Step 1: Standardization of NaOH titrant | | mass of sulfamic acid used | burette reading for NaOH solution | | :--- | :--- | :--- | :--- | | Trial #1 | m = 0.3293 g | Vinit = 1.32 mL | Vfinal = 32.66 mL | | Trial #2 | m = 0.3516 g | Vinit = 3.74 mL | Vfinal = 37.12 mL | | Trial #3 | m = 0.4081 g | Vinit = 2.69 mL | Vfinal = 41.52 mL | Step 2: Determination of unknown sulfamic acid solution Vsulfamic acid = 20.00 mL | | burette reading for NaOH solution | | :--- | :--- | :--- | | Trial #1 | Vinit = 2.08 mL | Vfinal = 31.39 mL | | Trial #2 | Vinit = 0.65 mL | Vfinal = 29.82 mL | | Trial #3 | Vinit = 1.09 mL | Vfinal = 30.37 mL | Step 3: Analysis of experimental data Calculate the concentration of NaOH solution for each standardization trial and determine the average value for [NaOH]. Calculate the concentration if sulfamic acid solution for each trial and report the average and standard deviation of your measurements. Answer all questions listed at the end of the procedure in the lab manual as part of your lab report.

          Experiment #13—Acid-Base Stoichiometry

Experimental data

Step 1: Standardization of NaOH titrant

| | mass of sulfamic acid used | burette reading for NaOH solution |
| :--- | :--- | :--- | :--- |
| Trial #1 | m = 0.3293 g | Vinit = 1.32 mL | Vfinal = 32.66 mL |
| Trial #2 | m = 0.3516 g | Vinit = 3.74 mL | Vfinal = 37.12 mL |
| Trial #3 | m = 0.4081 g | Vinit = 2.69 mL | Vfinal = 41.52 mL |

Step 2: Determination of unknown sulfamic acid solution

Vsulfamic acid = 20.00 mL

| | burette reading for NaOH solution |
| :--- | :--- | :--- |
| Trial #1 | Vinit = 2.08 mL | Vfinal = 31.39 mL |
| Trial #2 | Vinit = 0.65 mL | Vfinal = 29.82 mL |
| Trial #3 | Vinit = 1.09 mL | Vfinal = 30.37 mL |

Step 3: Analysis of experimental data

Calculate the concentration of NaOH solution for each standardization trial and determine the average value for [NaOH].

Calculate the concentration if sulfamic acid solution for each trial and report the average and standard deviation of your measurements.

Answer all questions listed at the end of the procedure in the lab manual as part of your lab report.

Show more…

Experiment #13—Acid-Base Stoichiometry

Experimental data

Step 1: Standardization of NaOH titrant

| | mass of sulfamic acid used | burette reading for NaOH solution |
| :— | :— | :— | :— |
| Trial #1 | m = 0.3293 g | Vinit = 1.32 mL | Vfinal = 32.66 mL |
| Trial #2 | m = 0.3516 g | Vinit = 3.74 mL | Vfinal = 37.12 mL |
| Trial #3 | m = 0.4081 g | Vinit = 2.69 mL | Vfinal = 41.52 mL |

Step 2: Determination of unknown sulfamic acid solution

Vsulfamic acid = 20.00 mL

| | burette reading for NaOH solution |
| :— | :— | :— |
| Trial #1 | Vinit = 2.08 mL | Vfinal = 31.39 mL |
| Trial #2 | Vinit = 0.65 mL | Vfinal = 29.82 mL |
| Trial #3 | Vinit = 1.09 mL | Vfinal = 30.37 mL |

Step 3: Analysis of experimental data

Calculate the concentration of NaOH solution for each standardization trial and determine the average value for [NaOH].

Calculate the concentration if sulfamic acid solution for each trial and report the average and standard deviation of your measurements.

Answer all questions listed at the end of the procedure in the lab manual as part of your lab report.

Added by Cynthia R.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Madhur L

05/07/2023

Step 1

Molar mass of H3NSO3 = 1(3) + 14 + 32 + 16(3) = 97 g/mol Trial #1: moles of H3NSO3 = (0.3293 g) / (97 g/mol) = 0.003396 mol Trial #2: moles of H3NSO3 = (0.3516 g) / (97 g/mol) = 0.003622 mol Trial #3: moles of H3NSO3 = (0.4081 g) / (97 g/mol) = 0.004207 mol Show more…

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Experiment #13—Acid-Base Stoichiometry Experimental data Step 1: Standardization of NaOH titrant Mass of sulfamic acid used Burette reading for NaOH solution Trial #1 m = 0.3293 g Vinit = 1.32 mL Vfinal = 32.66 mL Trial #2 m = 0.3516 g Vinit = 3.74 mL Vfinal = 37.12 mL Trial #3 m = 0.4081 g Vinit = 2.69 mL Vfinal = 41.52 mL Step 2: Determination of unknown sulfamic acid solution Vsulfamic acid = 20.00 mL Burette reading for NaOH solution Trial #1 Vinit = 2.08 mL Vfinal = 31.39 mL Trial #2 Vinit = 0.65 mL Vfinal = 29.82 mL Trial #3 Vinit = 1.09 mL Vfinal = 30.37 mL Step 3: Analysis of experimental data Calculate the concentration of NaOH solution for each standardization trial and determine the average value for [NaOH]. Calculate the concentration of sulfamic acid solution for each trial and report the average and standard deviation of your measurements. Answer all questions listed at the end of the procedure in the lab manual as part of your lab report.

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Transcript

00:01 Hello students, in this question basically we have done as the three trials trivalent triad 2 and trial 3 for that we have to do some calculations let's say for calculating the number of moles for the sulfamic acid in this case we know that sulfamic acid is represented as the h3nso3 so we can say that here the number of moles of sno3 for the three trials is number of moles for h3nso3 that is a sulfonic acid if we calculate this because mass and molar mass molar mass for this h3nso3 is the 97 gram per mole and given mass is given so we know that number of moles is equal to given mass upon my molar mass so we can easily write it as this so that should be for trial one it should be 0 .329 divided by 97 which is almost equal to 0 .003396 moles and for trial two it should be 0 .3516 and divided by 97 is the molar mass so that should be equal to the almost 0 .003622 moles and for this trial three that is 0 .4081 divided by 97 so that should be almost equal to the 0 .004207 moles so that should be here so we have this sulfamic acid as we know this is a sulfamic acid this is the moles for sulfanic acid for trial one trial two and trial three and now we have to calculate the volume for nmh for particular titrations so volume for nmh as we know that volume here should be the v is equal to v2 minus v1 v2 here is final volume and v1 here is initial volume so volume of nmh for trial one is 3266 minus 1 .32 so that should be almost equal to the 31 .34 ml and for trial two the volume should be the 37 .12 minus 3 .74 so that should be almost equal to 33 .38 ml and the volume for the trial three that should be 41 .52 minus 2 .69 that is initial volume so this volume should be 38 .83 ml and we can we know that the in this according to the stoichiometry if we can say that the number of moles for nmh should be equal to the almost the should be equal to the number of moles for h3 and so3 sulfamic acid because it is in the one ratio one of stoichiometry in the particular reaction so we can say that number of moles is equal to number of the moles for the nmh so this is also known as the number of moles for nmh so now we…

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