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What would be the most efficient way to make up a HEPES buffer at pH 8.5? And what starting compounds/reagants would be cohsen to use.
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- How much monobasic and dibasic phosophate is needed to make 100 mL buffer containing 100 moles of each phosphate?1. Investigate what an acid-base extraction consists of and what functional groups in substances can present a reaction of this type.thank you, I would like to have a follow-up question. Would you get an answer of 4.05 pH if the value of your formic acid is 0.050 M while your sodium formate is 0.10 M? and still the same pKa which is 3.75?
- 3. The ranking of the acids in terms of increasing acid strength. 4. The ranking of the conjugate bases in terms of increasing base strength. 5.A justification for the rankings based on the factors that influence the relative stability of the different conjugate bases. I need to answer these for C3H5ClO2, ClCH2CO2, and CH3COOH. (Acetic acid, 3-chloropropanic, and chloroacetic acid). Pleaseeee help I will definitely rateAccording to the Evan's pKa table, phenol has a pKa of 9.95. p-Nitro and m-nitrophenol have pKas of 7.14 and 8.35 respectively. Using resonance structures, explain the observed trends in the acidity of these molecules.When preparing a soap it is always required that the pH be verified, if it comes out too high, how can it be lowered from 9.7 to 9.2? a.With citric acid b. Adding borax c. with dilute HCI d. with more NaOH e.a and c are correct Please can you explain why, thank you
- When water is the solvent, the pKa of acetic acid (CH3CO2H) is 4.75, but when DMSO is the solvent, the pKa is 12.6.Explain why. Hint: Review Section 2.9 and consider the ability of each solvent to solvate cations and anions.You wish to prepare an HC2H3O2 buffer with a pH of 5.14. If the pKa of is 4.74, what ratio of C2H3O2⁻/HC2H3O2 must you use?Using the information in the first picture, solve the problem in the second. I am not good at drawing molecules when determining which structure to use in relationship of pH and pKA. Please explain your answer. Thank you very much!