a) Consider: A4(aq) + 2 H3O*(aq) –→ H2A² (aq) + 2 H2O(1). This is the net ionic reaction that occurs between the beginning of the titration and to the equivalence point. (Type in 1st, 2nd, 3rd or 4th.) b) After point along the titration curve, the analyte is just starting to gain the 4th proton. (Type in A, B, C...) c) The point along the titration curve where concentrations of H2A²¯ and H3A¯ are equal is (Турe in A, B, С...) d) Molecules and ions: H4A, H3A', H2A²", HA3- or A4- Of the molecule and ions shown above, the molecule or ion that has the highest concentration when 36.0 mL of 0.15 M HCI has been added is (Do not use super- or subscripts. Write H3A¯ as H3A-.) For H4A: Ka1 = 2.5 x 102, Ka2 = 7.5 x 105, Ka3 = 9.2 x 10-9, and Ka4 = 4.3 x 10-13 e) The pH at midpoint 2 is (Report your pH va to two sig figs.) f) The value of Kp3 for A4 is . (Use two sig figs.) (Do not use superscripts, subscripts or carets! Write 1.2 x 103 as 1.2x10-3 or 1.2e-3.)

How many mL of (6.22x10^-1) M HCOOH would it take to reach the equivalence point in the titration of (4.5900x10^1) mL of (4.680x10^-1) M KOH? Enter your answer in scientific notation with 3 sig figs. Do not include any units in your answer. Do not round any intermediate calculations. Note: Your answer is assumed to be reduced to the highest power possible. Your Answer: x10 Answer

Starting with 20.0 mL of 0.126 M C6H5CO₂H (pK₂ = 4.19) and using 0.063 M NaOH as a titrant, determine the pH of the solution (two decimal places) at the following points during the titration: a) Before addition of NaOH b) After addition of 10.0 mL of NaOH c) After addition of 20.0 mL of NaOH d) After addition of 30.0 mL of NaOH e) After addition of 40.0 mL of NaOH f) After addition of 50.0 mL of NaOH

Calculating the pH at equivalence of a titration A chemist titrates 50.0 mL of a 0.5851 M ethylamine (C2H5NH2) solution with 0.8192M HBr solution at 25 °C. Calculate the pH at equivalence. The pK of ethylamine is 3.19. b Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HBr solution added. pH = ] ☑

Create a BCA Table for the Equation: 2HCl(aq) + M(OH)2(aq) --> 2H2O(l) + MCl2(aq) AND FILL OUT THE FOLLOWING SUMMARY. I n this titration, 40.00 mL of 0.0521 M HCl was being neutralized by adding M(OH)2 that had a concentration of approx. 0.07 M. When a total of 11.53 mL of M(OH)2 was added, _____ moles of OH- had been added. Which reacted with the initial _____ moles of H+. After this neutralization reaction, _____ moles of H+ remained in the total volume of _____ mL. This resulted in a calculated pH of _____. The measured pH was 1.86.please answer type in word don't image upload thank you.

Really having trouble with this table for HC4H7O2 and C4H7O2, appreciate any help.

When titrating a 25 mL solution of 0.20M CH3COOH with 0.10M KOH, you realize you forgot to add an indicator initially. You've added 10 mL of KOH to the flask of acid. Ka of CH3COOH = 1.8 x 10^-5 1) Calculate the pH at this point 2) What color would your solution be if you add 40 mL KOH more. And a few drops of clorophenol blue indicator at this point. Why?

What is the pH at 90.00 mL of titrant in the titration of 50.00 mL of 0.400 M B (a generic base with Kp = (4.45x10^-10)) with 0.200 M HNO3? Enter your answer in scientific notation with 3 sig figs. Do not include any units in your answer. Do not round any intermediate calculations. Note: Your answer is assumed to be reduced to the highest power possible. Your Answer: x10 Answer

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1) Determine the pH during the titration of 74.5 mL of 0.348 M nitrous acid (Ka = 4.5×10-4) by 0.348 M NaOH at the following points.   (a) Before the addition of any NaOH ?   (b) After the addition of 18.0 mL of NaOH ?   (c) At the half-equivalence point (the titration midpoint) ?   (d) At the equivalence point ?   (e) After the addition of 112 mL of NaOH ?