Given that the expected amount of ascorbic acid in Vitamin C tablets should be 1000 mg, using stoichiometry, back calculate the expected volume in mL of ~0.20 M NaOH (you will find the exact concentration in lab close to this value and later calculate % error based on a new number) you would need to titrate the Vitamin C tablets. This calculation will help you predict where to slow down your base addition in the titration as you approach your endpoint. Here are a few hints: We assume a 1:1 mole ratio of ascorbic acid to NaOH o If you're given mass of ascorbic acid, you should calculate the moles of ascorbic acid present, convert this into moles of NaOH and then using the concentration of NaOH, you may calculate the volume required of NaOH to neutralize the acid present in the tablet

Given that the expected amount of ascorbic acid in Vitamin C tablets should be 1000 mg, using stoichiometry, back calculate the expected volume in mL of ~0.20 M NaOH (you will find the exact concentration in lab close to this value and later calculate % error based on a new number) you would need to titrate the Vitamin C tablets. This calculation will help you predict where to slow down your base addition in the titration as you approach your endpoint. Here are a few hints: We assume a 1:1 mole ratio of ascorbic acid to NaOH o If you're given mass of ascorbic acid, you should calculate the moles of ascorbic acid present, convert this into moles of NaOH and then using the concentration of NaOH, you may calculate the volume required of NaOH to neutralize the acid present in the tablet

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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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.
During the titration of a polyprotic acid, neutralization of its ionizable protons happens... ...in a stepwise manner. ...only if a weak base is added. ...occasionally. ...simultaneously.
A 1 liter solution contains 0.589 M acetic acid and 0.442 M potassium acetate. Addition of 0.221 moles of hydroiodic acid will:(Assume that the volume does not change upon the addition of hydroiodic acid.) CHECK ALL THAT APPLY - Raise the pH slightly - Lower the pH slightly - Raise the pH by several units - Lower the pH by several units - Not change the pH - Exceed the buffer capacity
Based on that experimental titration curve of the IU unknown sample, can you sketch out the titration curve if both the starting weak acid solution and the NaOH solution are diluted 10 times? What about diluting another 10 times on top of that? Particularly, you should pay attention to the size of the pH jumps around each equivalence point, as well as the relative positions of the flat regions before or after the pH jumps. For the sake of simplicity, no need to consider activity in this exercise.
Determine the pH at the point in the titration of 40.0 mL of 0.200 M H2NNH2 with 0.100 M HNO3 after 10.0 mL of the strong acid has been added. The value of Kb for H2NNH2 is 3.0 x 10-6. 1 2 4 NEXT Use the table below to determine the moles of reactant and product after the reaction of the acid and base. H2NNH2(aq) H*(aq) H2NNH;*(aq) + Before (mol) Change (mol) After (mol) 5 RESET 0.200 0.100 +x 0.200 + x 0.200 - x -x 0.100 + x 0.100 - x 1.00 x 103 -1.00 x 10-3 2.00 x 10-3 -2.00 x 10-3 6.00 x 103 -6.00 x 10-3 7.00 x 10-3 -7.00 x 103 8.00 x 103 -8.00 x 103
6. Open the bags and add 3 drops of bromothymol blue to each bottle. The indicator turns yellow if vinegar is still present and is blue-green when vinegar is not present. Record the color of each solution in the table provided. Data Table 1: Reaction Data Sample 7. Add a 4 teaspoon of baking soda to each bag. Effervescence of CO₂ occurs if excess vinegar is present. Record your observations in the table provided. Moles of vinegar (mol) Moles of baking soda (mol) mistry Relative size of bag Was excess vinegar present? Color with indicator? Reacts with additional baking soda? Moles of gas produced (mol) Watch the limiting reagent! #1 0.042 0.014 чем yes yellow #2 0.042 0.028 #3 yes yellow 0.042 0.042 Experiment 11 yes yellow #4 Stoichiometry 0.042 0.88 more small big bigger bigger No 0.071 #5 NO Geen 0.042 Blue Calculations 1. Use Table 1 to determine which reactant was limiting in each trial. Use the amount of limiting reactant to calculate the moles of carbon dioxide gas that were…
If during the titration a drop of the standardized NaOH adhered to the side of the Erlenmeyer flask and did not enter the vinegar solution would this result in a percent acetic acid that is too high, too low or would it have no effect? Explain
Thank you! Data included in pictures! No hand drawn graphs!! Determine the equivalence point and the endpoint of the sudsy ammonia/citric acid reaction by graphing the titration data (pH vs volume citric acid) from Table 3: Titration of Sudsy Ammonia with 0.25 M Citric Acid Data with the Chosen Indicator. Draw a green arrow (use ‘Insert Shape’ feature to do this once you have copied your graph from Excel) on your graph indicating the data point that represents your equivalence point. Draw a blue arrow (use ‘Insert Shape’ feature to do this once you have copied your graph from Excel) on your graph indicating the data point that represents your endpoint. Use a graphing software to create the x-y scatter plot showing the pH of the experimental solution as a function of the volume of citric acid (the titrant) added for the titration data in Question 9. On your graph, the x-axis should be the volume of citric acid added (mL), and the y-axis should be the pH of the experimental…
1) A 1 liter solution contains 0.363 M nitrous acid and 0.484 M sodium nitrite.Addition of 0.399 moles of potassium hydroxide will:(Assume that the volume does not change upon the addition of potassium hydroxide.) Raise the pH slightly Lower the pH slightly Raise the pH by several units Lower the pH by several units Not change the pH Exceed the buffer capacity 2) A 1 liter solution contains 0.364 M hydrocyanic acid and 0.485 M sodium cyanide.Addition of 0.182 moles of potassium hydroxide will:(Assume that the volume does not change upon the addition of potassium hydroxide.) Raise the pH slightly Lower the pH slightly Raise the pH by several units Lower the pH by several units Not change the pH Exceed the buffer capacity
rial Number Total Volume of Titrant Added (mL) Post Lab Questions 34.7 b) 34.7 Burette contains 0.70M of KOH and 100mL of Mystery CH3COOH 34.7 All part of same question. 1. Using data collected from the titration, determine the concentration of the assigned unknown solution. a) Using your answer from question 1, determine the pH of the unknown acetic acid solution BEFORE the titration began (i.e., when no base has been added) if Ka of acetic acid is 1.8x10-5. Determine the pH of the solution AT EQUIVALENCE POINT of the titration.
I've used the buffer in question 4 and need to modify the last 250 mL so that it has a total osmotic concentration of 220 mM. How much NaCl should I add to get this solution?
Determine the resulting pH when 0.040 mol of solid NaOH is added to a 200.0 mL buffer containing 0.100 mol C6H5NH3CI and 0.500 M C6H5NH2. The value of Kb for C6H5NH2 is 4.3 x 101°. 1 2 3 4 NEXT Use the table below to determine the moles of reactant and product after the reaction of the acid and base. You can ignore the amount of liquid water in the reaction. C6H;NH3*(aq) Он (aq) H20(1) C6H;NH2(aq) + + Before (mol) Change (mol) After (mol) 5 RESET 0.040 -0.040 0.500 -0.500 0.100 -0.100 +x 0.040 + x 0.040 - x 0.100 + x 0.100 - x 0.060 -0.060 -x 0.140 -0.140 0.020 -0.020