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Why can't the 250-mL Erlenmeyer flask be used to measure the vinegar needed in the titration? Briefly, 1 - 2 sentences, explain.
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- A 27.00 mL sample of sodium hydroxide was titrated (neutralized) with 31.7 mL of 0.150 M sulfuric acid. What is the concentration of the sodium hydroxide solution? Write your answer to three digits - for example 0.468 and leave of the unit M. Round off your answer to three digits - so 0.4682 is written as 0.468.Note that and answer of 0.410 will show up on Canvas as 0.41In titration, what solution is placed inside the Erlenmeyer flask? Group of answer choices diluted solution titrant analyte stock solutioni) How many moles of hydrochloric acid were there in the volume of acid added in the titration? ii) How many moles of calcium hydroxide were there in the 25 cm sample of solution that the student used in the titration? iii) What was the concentration of the calcium hydroxide solution in mol/dm?
- 3. What is an indicator and what is it used for in the concept of acid-base Titration? ExplainDuring the standardization of NaOH , a drop of titrant was present on the tip of the buret when making the final volume reading. Would this affect the final calculations of molarity of NaOH? If so, how? Explain.T1 T2 T3 Initial Volume of NaOH (mL) 0.00 3.95 7.95 Final Volume of NaOH (mL) 3.95 7.95 11.65 Volume of NaOH (mL) 3.95 4.00 3.70 Average. volume NaOH (ml) Ex .3 Data Sheet Using dilution Factor in an Acid Base Titration. HCl (aq) + NaOH (aq) NaCl (aq) + H2O (aq) HCl Solution M1 (HCl ) X V1 ( HCl ) = M2 ( HCl ) X V2 ( HCl ) 1.00 mol/L X 4.00 ml = M2 ( HCl ) X 100.0 ml Molarity of diluted HCl Solution M2 = mol / L NaOH Solution 1 - Volume of NaOH needed for dilution V1 = 10.00 ml 2 -Total volume of diluted NaOH V2 = 100.0 ml 3 - Dilution Factor = V2 / V1 =100ml / 10 ml = 4- Volume of HCl used in Titration V HCl = 10.0 ml 5- Moles of HCl = M2 (0.04 mol/L ) x VHCl 0.01 (L) = moles 6-Moles HCl =…
- Here is a video that illustrates the titration process and necessary calculations: https://www.youtube.com/watch?v=9pS7Q4MQCYI You must show all your step-by-step calculations with proper formulas, units, and significant figures (for finals answers for Molarity of Acetic Acid and % Mass of Acid in Vinegar) review Chapter 2 for multi-step calculations: addition/subtraction: fewest decimal places multiplication/division: fewest significant figures You can either type your calculations and answers (which might be time-consuming) or write them on the paper/digital tablet and insert them as pictures in the slides. Make sure they are easily read. Concentration of NaOH (from the bottle): __0.350 M___ The volume of vinegar used (that you poured in the flask): __9.50 mL____ Average volume of NaOH used (mL): Average volume of NaOH used (L): Moles of NaOH used in titration: Moles of HC2H3O2 neutralized by NaOH: Molarity of HC2H3O2 (M): Grams of HC2H3O2…Here is a video that illustrates the titration process and necessary calculations: https://www.youtube.com/watch?v=9pS7Q4MQCYI You must show all your step-by-step calculations with proper formulas, units, and significant figures (for finals answers for Molarity of Acetic Acid and % Mass of Acid in Vinegar) review Chapter 2 for multi-step calculations: addition/subtraction: fewest decimal places multiplication/division: fewest significant figures You can either type your calculations and answers (which might be time-consuming) or write them on the paper/digital tablet and insert them as pictures in the slides. Make sure they are easily read. Concentration of NaOH (from the bottle): __0.350 M___ The volume of vinegar used (that you poured in the flask): __9.50 mL____ Average volume of NaOH used (mL): Average volume of NaOH used (L): Moles of NaOH used in titration: Moles of HC2H3O2 neutralized by NaOH: Molarity of HC2H3O2 (M): Grams of HC2H3O2 (g): Percent (m/v) of HC2H3O2 in vinegar…explain the principles of acid-base titration
- Chemistry 1. (i) Why can’t we leave all acids/bases open to the atmosphere? (ii) What would happen if we leave a titrant out in the open?Volumetric Analysis Liza carefully pipets 25.0 mL of 0.525 M NaOH into a test tube. She places the test tube into a small beaker to keep it from spilling and then pipets 75.0 mL of 0.355 M HCl into another test tube. When Liza reaches to put this test tube of acid into the beaker along with test tube of base, she accidentally knocks the test tubes together hard enough to break them and their respective contents combine in the bottom of the beaker. Is the solution formed from the contents of the two test tubes acidic or basic? What is the pH of the resulting solution?For your second titration, you put 0.457 g of KHP in your flask. Pretend that the concentration of NaOH you calculated in the previous problem was 0.0976. Predict the volume (in mL) of NaOH you will use in this titration. Write your answer to two decimal places without units.