The amount of ascorbic acid (vitamin C, C&HgO6) in tablets is determined by reaction with bromine and then titration of the hydrobromic acid with standard base C6H8O6 + Br2 -> C,H6O6 + 2HBR HBr + NAOH -> NaBr + H2O A certain tablet is advertised as containing 500 milligrams of vitamin C. A single tablet is dissolved in water and reacted with Br2, followed by titration with 43.20 mL of 0.1350 M NAOH. Verify if the tablet contained the advertised quantity of vitamin C.
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- JKU Portal D2 D2L ii Handshake < STARTING AMOUNT X Ch. 1 Introduction -... 4 Aktiv Chemistry 72.0 ADD FACTOR 43.2 Connect-Bio Google Docs O Quizlet A solution of phosphoric acid (H.PO4) with a known concentration of 0.250 M H3PO4 is titrated with a 0.800 M NaOH solution. How many mL of NaOH are required to reach the third equivalence point with a starting volume of 72.0 mL H3PO4 according to the following balanced chemical equation: 0.0675 mL NaOH H₂PO4 + 0.800 g NaOH 1 Question 19 of 23 3 NaOH NasPO4 + 6.75 x 104 LH₂PO4 2 mol H3PO4 → H2023 Elantra hybrid 0.250 1000 g H₂PO4 ANSWER mL H₂PO4 3 H₂O 0.001 67.5 M H3PO4 mol NaOH 2022 Subaru WRX RESET M NaOH S 7.50 22.5 LNaOH 3 2023 Kia K5 | Mid-S...help ASAP Compare the two quantities based on the given condition: A solution containing KOH and KHCO3 is titrated with 0.100 M HCl using a one-flask method. I. Volume of titrant from 0.00 mL to the phenolphthalein endpoint II. Volume of titrant from the phenolphthalein end point to the methyl red end point. (a) I < II (b) I = II (c) I > II (d) The two quantities cannot be compared due to insufficient information.(5e) Which two of the following compounds would you make a buffer of pH = 12.5. H3PO4 (MM=98.00 g/mol), NaH2PO4 (MM=119.98 g/mol), NazHPO4 (MM=141.96 g/mol), and Na3PO4 (MM = 163.94 g/mol)? [For H3PO4: pK; = 2.148; pKz=7.199; pK3=12.15]
- 3 .... 1V * 00 < (O SI Sona Psychology Research Pa X Question 5 of 9 In a titration of 49.0 mL of a 0.500 M solution of a diprotic acid H2C3H2O4 (malonic acid) with 0.255 M NaOH, how many grams of NaOH are required to reach the second equivalence point? (MW_NAOH = 39.997 g/mol) %3D 6 2 4. 9. 8. /- acBook Air psa DD F7 F3 F4 F5 F8 F10 69. F12 23 24 2 3. 4. 9. 8. 6 delete A %31 H M B. 10. command command optionBe sure to answer all parts. Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH3CH,);N (K, = 5.2 x 10-4), with 0.1000 M HCl solution after the following additions of titrant. (a) 13.00 mL: pH = 10.305 (b) 20.10 mL: pH = (c) 29.00 mL: pH = %3!Calculate the molarity of a dilute Ba(OH)2 solution if addition of 99.52 mL of the base to 0.6426 g of benzoic acid (MW=122.12g/mole) include the UNIT) required a 4.2887 mL back-titration with 0.02274 M HCI. (Answers should be written in 2 decimals place and do not
- V (acetic and CH ₂ COOH) = 10ml C = 0,271 M titrated with potassium hydroxide solution c= 0, 253 M a) pH=? (V (KOH) = 0,00, 3,00, 6,00, 9,00, 12,00 ml) 6) [H₂0], [OH-] = ? (in the solutions) pka (CH₂COOH) = 4, 15What do you call the stage of the titration in which the number of moles of the titrant is stoichiometrically less than the number of moles of the analyte? (A post-equivalence (B pre-equivalence end stage (D equivalenceA 0.1510 g KHP (MW = 204.22 g/mol) sample required a volume of 57.04 mL of NaOH solution to reach a phenolphthalein end point. Calculate the molarity of NaOH titrant. (A) 0.01012 M (B) 0.02593 M c) 0.01296 M D) 0.006481 M
- What weight of sample should be taken for analysis sos that the volume of 0.1074N NaOH used for titration equals the percentage of potassium acid phthalate (KHC8H4O4) in the sample (Answer 2,193 g)A buffer with a pH of 9.85 contains CH;NH2 and CH;NH;Cl in water. What can you conclude about the relative concentrations of CH;NH, and CH;NH;Cl in this buffer? For CH,NH2, pK, = 3.36. a) CH;NH2 > CH;NH;CI b) CH;NH2 < CH;NH,CI c) CH;NH2 = CH,NH,CI d) Nothing can be concluded about the relative concentrations of CH;NH2 and CH;NH;CI.The Ką of Benzoic acid (C6H5COOH) is 6.3 x 10-5. A 0.72 M C6H5COOH of 40.0 mL is titrated with 0.48 M NAOH. Which of the following is correct for the following two processes? (1) After add resulting solution. ml of N2OH into the above 0.72 M C6H5COOH of 40.0 mL, we have [ C6H5COOH ] = [ C6HSCO0" ] in the (2) After add the resulting solution. ml of NaOH into the above 0.72 M CGH5COOH of 40.0 mL, we have [ CgHgCOOH ] = [OH" ] in O a. The answer for (1) is 40.0 ml. The answer for (2) is 20.0 ml. O b. The answer for (1) is 20.0 ml. The answer for (2) is 40.0 ml. O. The answer for (1) is 30.0 ml. The answer for (2) is 60.0 ml. d. The answer for (1) is 60.0 ml. The answer for (2) is 30.0 ml. O e. None of the above is correct.