What is the average molarity of NaOH between the three trials? Titration Lab Trial 1 Trial 2 Trial 3 Volume of Acid Used 20.0 mL 20.2 mL 20.8 mL Molarity of Acid Used 0.05 M 0.05 M 0.05 M Initial Buret Reading 0.00 mL 0.00 mL 0.00 mL Final Buret Reading 20.7 mL 21.0 mL 21.5 mL Volume of NaOH Used A В C Molarity of NaOH Solution D E F
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- A 10.00-mL aliquot of a known HCl solution (3.800 g/L) was titrated with a 4.100 g/L NaOH solution. Calculate the range of the blank corrected titration volumes using the titration data below: Blank Trial 1 Trial 2 Trial 3 Initial Volume (mL) 12.50 0.50 12.30 34.75 Final Volume (mL) 13.10 10.45 22.35 45.273 How to Titrate: Lab Techniques Titration Titration Procedures: Add about 100mL of a 0.1 M NAOH solution to a 250mL beaker. 2. Obtain a buret and rinse with about 10 ml of the NaOH solution. 3. Fill the buret with NaOH and record starting volume to the nearest 1m" of a milliliter. 4. Measure out exactly 10.0 ml of HC in a graduated cylinder and add it to a 250mL Erlenmeyer flask along with 1 drop of phenolphthalein indicator. 5. Slowly add the NaOH to the flask with the acid and swirl. Continue adding the base until a color change (faint pink) occurs and remains for at least 10 seconds. This is the endpoint of this neutralization eaction. 6. Read and record the ending volume of NaOH in the buret, 7. Repeat steps 1 through 6 two more times for Trials 2 and 3. 8. Calculate the volume of base added in each trial. 9. Use the volume of base added and the volume of acid added to calculate the molarity of the acid for each trial. 10. Calculate the average molarity of the acid. 11. Calculate…13,How many milliliters of 0.0854 M NaOH are required to titrate 25.00 mL of 0.1250M HI to the equivalence point? Group of answer choices A, 17.08 B, 17.1 C, 36.6 D, no correct answer E, 36.593
- Standardization of 1N H2SO4 Titration 1 Titration 2 Titration 3 Titration 4 Mass of flask and Sodium Carbonate 1.5067 g 1.5098 g 1.5076 g 1.5077 Final buret reading (mL) 22 mL 22mL 21.5 mL 22.5 mL Volume of Sulphuric acid used (mL) 30 mL 30 mL 30 mL 30 mL What is the result of my titration?What is the volume of NaOH used in each trial? Titration Lab Trial 1 Trial2 Trial 3 Volume of Acid Used 20.0 mL 20.2 mL 20.8 mL Molarity of Acid Used 0.05 M 0.05 M 0.05 M Initial Buret Reading 0.00 mL 0.00 mL 0.00 mL Final Buret Reading 20.7 mL 21.0 mL 21.5 mL Volume of NaOH Used Molarity of NaOH SolutionTitle: Potentiometric Acid Titration fitulo:Tabla de ph volumen NaOH (ml) ph Titulación NaOH 20 4.25 7.39 20.1 6.24 7.03 7.21 6.83 6.59 20.2 6.59 6.24 20.3 6.83 6 20.4 7.03 4.25 20.5 7.21 20.6 7.39 1 19.9 20 20.1 20.2 20.3 20.4 20.5 20.6 20.7 V NAOH (ml) With the previous graph determine the following: a. Determine the bufer zone on the graph b. Determine the volume at the stoichiometric point of the acid titration c. Determine the average volume d. Use the medium volume to get the pka e. From pka, get the value of ka
- A 50.00 (±0.02) mL portion of an HCl solution required 29.71(±0.02) mL of 0.01963(±0.0032) M Ba(OH)2 to reach an end point with bromocresol green indicator. ? of HCL = 29.71?? ? 0.01963 ???? ??(??)2 ?? ? 2 ???? ??? ???? ??(??)2/ 50.00?? = 0.02333 ? Calculate the uncertainty of the result (absolute error).Calculate the coefficient of variation for the result.Part A: Standardization of a Sodium Hydroxide Solution Titration 1 Titration 2 Titration 3 Mass of 125 mL flask 45.849g 46.715g 44.953g Mass of flask and KHP 46.849g 47.745g 46.003g Initial buret reading (mL) 0.5 ml 0.5 ml 0.5 ml Final buret reading (mL) 27.8 ml 26.5 ml 26.7 ml Volume of NaOH used (mL) 45.11 ml 45.06 ml 45.14 ml Calculations Titration 1 Titration 2 Titration 3 Moles of KHP Moles of NaOH Molarity of NaOH Average Molarity of NaOH: _______________QUESTION 68 For the titration of HF with NaOH, which location on the titration curve corresponds to the particulate diagram below? 14 IV.- Key 13 12 11 10 I. I. 2. Volume of titrant added (ml.) OI. O II. O II. O IV. %3D Hd
- how would i find the average titre value in ml if I have 10% ethanol 1.3 ml (of water used) as titre value 1 and 1.1 ml (of water used) as titre value 2 ?How many milliliters of 8.50x10-2 M NaOH are required to titrate each of the following solutions to the equivalence point? Show Transcribed Text V = 45.0 mL of 9.00x10-² M HNO3 Submit ▾ Part B V = —| ΑΣΦ 35.0 mL of 8.50x10-2 M CH3COOH Show Transcribed Text V = Request Answer IVE ΑΣΦ P Pearson Submit C IVE ΑΣΦ Ĉ Request Answer 55.0 mL of a solution that contains 1.80 g of HCl per liter www ? SW ? ? mL mL mLHow many milliliters of water have to be added to 143.8 mL of 0.25 M HCI to reduce the concentration to 0.10 M HCI? Vwater = mL eTextbook and Media Hint Save for Later Attempts: 0 of 2 used Submit Ans O Type here to search 10 insert @ %23 $ & * 4 7 8. 10 %3D Q W R PI A G K C V B NM pause alt alt ctri