In the titration of 10.00 mL sample of vinegar solution with 0.35 M NaOH as titrant, the initial burette reading was 2.50 mL and the final burette reading was 45.30 mL. Calculate the mass percent of acetic acid (Molar mass = 60.05 g/mol) in the vinegar sample. (The density of the vinegar is 1.00 g/mL) % 5.0 .a O % 8.0 .b O % 6.0 .c O % 9.0 .d O % 7.0 .e O
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- A 250.0-mg sample of an organic weak acid is dissolved in an appropriate solvent and titrated with 0.0556 M NaOH, requiring 32.58 mL to reach the end point. Determine the compound’s equivalent weight.Na2CO3 served as the primary standard in a titration experiment. Find the molarity of the titrant given the following data in 3 decimal places. Show solutions Primary Standard Used: Na2CO3Formula Mass of 1º standard: 105.99 g/mol% purity of 1º standard: 95% Trial 1 2 3 1º Standard weight, g 0.1005 0.1001 0.0997 Net volume of HCl, mL 9.30 9.00 8.90 Molarity of HCl X1 X2 X3What mass of Ba(OH)2 is present in a sample if it is titrated to its equivalence point with 44.20 mL of 0.1000 N H2SO4? Note: Present complete solutions for the following problem. Express your final answers up to two (2) decimal places.
- Express the concentration of acetic acid in both samples as % by mass of acetic per 100mL of solution given : %= mass/ volumex100% Given: NaOH vs CH3COOH Burette solution is NaOH and the pipette solution is 5.0 mL of Vinegar Titration Initial burette reading Final burette reading Volume of NaOH consumed Average volume of NaOH Approximate 0.0 mL 20.1 mL 20.1 mL 20.3 mL Titration 1 0.0 mL 20.9 mL 20.9 mL Titration 2 0.0 mL 20.0 mL 20.0 mL To find the average volume Average volume = 20.1 mL + 20.9 mL + 20.0 mL320.1 mL + 20.9 mL + 20.0 mL3 = 20.3 mL Concentration of Vinegar = Volume of NaOH * Concentration of NaOHVolume of vinegarVolume of NaOH * Concentration of NaOHVolume of vinegar = 20.3 mL * 0.0647 M5.0 mL20.3 mL * 0.0647 M5.0 mL = 0.2627 M Concentration of NaOH = Volume of H2SO4 * Concentration of H2SO4Volume of NaOHVolume of H2SO4 * Concentration of…100.0 cm3 water is titrated with 0.0049 M ascorbic acid solution. What is the analytical concentrationof the weak acid at 12.73 cm3 added titrant volume?Calculate the grams of Acetic Acid ( M. W. 60.05g/ mole) in a unknown sample. A sample of Acetic Acid was titrated to the end point using 23.76M NaOH solution. The initial buret reading was 0.6559mL and the final reading was 0.02435mL.
- A mixture of NaOH (M.wt = 40) and Na2CO3 (M.wt = 106) is titrated with 0.3 M HCl, requiring 30 mL for phenolphthalein end point and an additional 15 mL to reach the modified methyl orange end point, How many milligrams of NaOH and Na2CO3 are in the mixture *0.0585g of Na2 C2O4 10 mL of distilled water, 2 M H2SO4 were added to adjust the KMnO4 solution prepared as 0.1M, heating was performed and 8.4mL titrant was spent as a result of titration. Calculate the true concentration of potassium permanganate accordingly.cedric and astrid titrated a 15.00 ml aliquot of grapefruit juice with a 0.134 M NaOH solution to the end point. the initial buret reading was 1.04 ml and the final buret reading was 24.83ml. H3C6H5O7(aq) + 3 NaOH(aq) yeilds Na3C6H5O7(aq) + 3 H2O)(l) The volume of NaOH titrated is 23.79ml 24.83ml - 01.04ml = 23.79ml of NaOH ***What is the mass of citric acid in the juice sample? 0.204g of H3C6H5O7 ( can you please show how to calculate this answer)
- A different titration experiment using a 0.127M standardized NaOH solution to titrate a 27.67 mL solution with an unknown Molarity concentration (M) of sulfuric acid (H2SO4) gave the following molarities for 3 trials. Initial Burette Reading (mL) Final Burette Reading (mL) Delivered vol (mL) Acid Concentration (M) Trial 1 0.0358 Trial 2 0.0341 Trial 3 0.0331 From the 3 trials, determine the average Molarity concentration of the H2SO4 to 3 significant digits. Don't include a unit. From the 3 trials, determine the standard deviation of the Molarity concentration of the H2SO4 to 3 significant digits. Don't include a unit.From the 3 trials, determine the relative standard deviation of the Molarity concentration of the H2SO4 to 3 significant digits. Don't include a unit. Do you think the above experiment was accurate? Precise? Explain your answer using supporting values.A soda ash sample was contaminated with NaOH. It was made up of 0.00069 moles of NaOH and 0.00115 moles of Na2CO3. What would be the expected volumes of Vphth and Vmo if the said sample was titrated with 0.1M HCl?4) A 25.00-mL sample of vinegar is diluted with deionized to 100.0 mL. Then a 10.00-mL aliquot (portion) of the dilute vinegar solutionis titrated with 0.1055 M NaOH solution using phenolphthalein as theindicator. (a) If 19.60 mL of the base solution is required to reach the end-point, calculate the molar concentration of acetic in dilute vinegar solution. (b) What is the molar concentration of acetic acid in the original (undiluted) vinegar? The acid-base reaction occurs as follows: HC2H3O2(aq) + NaOH(aq) --> H2O(l) + NaC2H3O2(aq)