Lab2 Sophie Rahman PDF

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Lab report for lab 2: how much KHP on unknown...

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Report Sheet For Lab 2: How much potassium hydrogen phthalate is in this unknown?

Name: ___Sophie Rahman___________________________________ Partner Name: ______No partner________________________________ All answers must be typed in black ink. Increase the space between questions to insert your answers. The grading rubric is provided separately. Prelab Question: 1. What does hygroscopic mean? Readily taking up and retaining moisture Reference: https://www.merriam-webster.com/dictionary/hygroscopic 2. Write the balanced chemical reaction between KHP and NaOH? HKC8H4O4(aq) + NaOH(aq) → NaKC8H4O4(aq) +H2O(l) 3. Calculate the molecular weight (using the atomic weights of the individual atoms on the front cover of your textbook) of NaOH and KHP. Show your work with handwritten answers in notebook for pre-lab check; when you do report, then use the equation tools provided under Insert: Equation in Word to show the work. KHP(HKC8H4O4): g g ×5=5.0395 H: 1.0079 mol mol g g ×1=39.0983 K: 39.0983 mol mol g g × 8=96.0880 C: 12.0110 mol mol g g × 4=63.9976 O: 15.9994 mol mol g KHP molecular weight: 204.2234 mol g g ×1=22.9897 mol mol g g ×1=15.9897 O: 15.9897 mol mol g g ×1=1.0079 H: 1.0079 mol mol Na: 22.9897

NaOH molecular weight: 39.9970

g mol

4. Calculate the amount of NaOH needed to make a 1 L solution that is ~0.1 M NaOH. Show your work with handwritten answers in notebook for pre-lab check; when you do report, then use the equation tools provided under Insert: Equation in Word to show the work. 1

x mol NaOH 1 L solution 0.1 M ×1 L solution=0.1 mol NaOH g 39.9970 mol 0.1 mol NaOH × =3.9970 g NaOH 1 mol NaOH 1.1 M =

5. Read the Safety Data Sheet for Solid Sodium Hydroxide. Describe the safety precautions you will follow while in the lab using 25 to 50 words. Attach the Safety Data Sheet of Solid Sodium Hydroxide to your Lab 3 report. When you handle NaOH(s), you must be very careful not to let the chemical touch your skin because it’s an irritant. It can cause 2 nd or 3rd degree burns. Also avoid ingesting the chemical and contact with eyes. Reference: https://betastatic.fishersci.com/content/dam/fishersci/en_US/documents/programs/education/regula tory-documents/sds/chemicals/chemicals-s/S25881.pdf

Experimental and Results NaOH Standardization 1. Mass of NaOH transferred to the beaker ___4.1903 g__________ (don’t forget units) Table 1: Data for the Standardization of NaOH Mass of KHP Initial Final (g) Volume of Volume Base (mL) of Base (mL) Sample 1 .6787 g .99 mL 33.15

Sample 2

.6233 g

.97 mL

Sample 3

.6238 g

.99 mL

Volume of Base delievered (mL)

Molarity of NaOH (M)

Notes, comments about sample

.1033 M

mL

32.16 mL

31.23 mL 31.10 mL

30.26 mL 30.11 mL

.1009 M

Light to medium pink (on the lighter side) Light pink Light to medium pink (similar to sample

.1014 M

2

Sample 4

.6222 g

.99 mL

31.11 mL

30.12 mL

.1012 M

1) Light pink

2. Provide one sample calculation of the molarity of NaOH (use equation tools under Insert: Equation in Word): Molarity for trial 1: 1000 mL 1 mol NaOH 1 mol KHP 1 =.1033 M NaOH × × .6787 g KHP × × 1L g 1mol KHP 32.16 mL KHP 204.2234 mol 3. Use your calculator or Excel to find the average molarity of the NaOH solution and give the answer here: Average Molarity: .1017 M NaOH 4. Use your calculator or Excel to find the standard deviation of the molarity of the NaOH solution and give the answer here: Standard Deviation of molarity: .001 5. Using the correct number of significant figures, report the molarity and standard deviation of the NaOH solution and give the answer here: Average molarity: .1017 M NaOH Standard deviation: .001 Solution of NaOH is .1017 M±.001 Unknown Analysis: Use the average molarity of your NaOH for the calculations of your unknown. Table 2: Data for the Determination of %w/w of KHP in an Unknown Sample Volume of % w/w of Notes, Final Mass of Initial Unknown Base KHP in comments Volume unknown Volume Number of Base delivered unknown about of Base _____527__ (g) (mL) (mL) sample (mL) _ Sample 1 12.15 27.97 % Faint .9010 g .98 mL 13.13 Sample 2

.9129 g

.91 mL

Sample 3

.9165 g

.95 mL

mL 13.11 mL

mL 12.20 mL

13.45 mL

12.50 mL

27.76 %

28.33%

pink Light pink but not as faint as sample 1 Light pink (similar 3

Sample 4

.9166 g

.97 mL

13.11 mL

12.14 mL

27.52 %

to #2) Light to medium pink

6. Provide one sample calculation of the % w/w of KHP in the unknown (use equation tools under Insert: Equation in Word): % w/w calculation for trial 1: .1017 M NaOH ×.01215 L=.001234 mol NaOH .001234 mol NaOH =.001234 mol KHP g 204.2234 mol .001234 mol KHP × =.2520 g KHP 1 mol KHP .2520 g KHP × 100 %=27.97 % .9010 g KHP 7. Use your calculator or Excel to find the average %w/w of the KHP in the unknown and give the answer here: Average %w/w: 27.90% 8. Use your calculator or Excel to find the standard deviation of the %w/w of the KHP in the unknown and give the answer here: Standard deviation of %w/w: .3 9. Using the correct number of significant figures, report the %w/w and standard deviation of KHP in the unknown and give the answer here: Average %w/w: 27.90% Standard deviation of %w/w: .3 Percent weight of KHP in unknown is 27.90%±.3 Post Lab Questions 1. Why was the NaOH not dissolved in a Volumetric Flask? A volumetric flask was not needed because to dissolve the NaOH because we only needed an approximate number, and not an exact number. Since we calculated the molarity after we didn’t need an exact volume of 1 liter. 2. What happened to the flask when you dissolved the NaOH in the beaker? An exothermic reaction occurred in the beaker because the temperature of the flask increased to a little bit above room temperature. 3. What type of reaction occurs between NaOH and phenolphthalein? An acid-base reaction occurs between the two. The phenolphthalein is the weak acid, and NaOH is the strong base. 4. Why do you only add two drops of phenolphthalein?

4

Two drops of the indicator will be the correct amount because it will provide the most accurate color change during titration. More of the given indicator would have changed the pH of the solution. 5. Were any samples not included in the calculation for molarity and if yes, why? The NaOH solution was not included to calculate the molarity; the sodium hydroxide was a hygroscopic. The fluctuating mass would have made the NaOH solution an unreliable factor. 6. Were any samples not included in the calculation for %w/w and if yes, why? All of the samples were included in the calculation for %w/w. The average molarity of NaOH was .1017 M which was used to calculate the weight percentage. The volume of base delivered of the unknown was also used in order to find the weight percentage; this got the number of moles of NaOH. Since the moles of NaOH and KHP is 1:1 ratio, # of moles of NaOH = # of moles of KHP. From that number we multiplied it by the molecular weight of KHP (204.2234 g/mol). We then used this number and divided it by the mass of the unknown for each of the four trials to find the %w/w by multiplying this number by 100% then found the average and standard deviation of the weight percentage using a calculator. 7. Explain the chemistry occurring during the titration and at the endpoint. This should be in paragraph form and contain 50 – 70 words. During the titration the base comes out of the 50 mL buret into the beaker of solution created with the indicator. With the indicator in the solution, it makes it possible to see the endpoint when the moles of hydrogen ions from the acid equal moles of the hydroxide ions from the base shown by a color change depending on the indicator used. Write a concluding statement indicating the %w/w of KHP in the unknown: When taking the weight percentage of the KHP in the unknown given in lab, it was found that the average weight percentage of KHP found in our unknown was 27.90%. The numbers and data recorded during the experiment to find the weight percentage were used. From these %w/w numbers, the average and standard deviation were found.

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