An analytical chermist is titrating 150.6 ml. of a 0.6600 M solution of diethylamine (C,H,), NH) with a 0.1500 M solution of HIO, The pky of dethylamine is 2.89. Calculate the ptt of the base solution after the chemist has added 777.3 ml. of the HIO, solution to t. Note for advanced students: you may assume the final volume equals the wntial volume of the solution plus the volume of HIO, solution added Round your answer to 2 decimal places. pH =

An analytical chermist is titrating 150.6 ml. of a 0.6600 M solution of diethylamine (C,H,), NH) with a 0.1500 M solution of HIO, The pky of dethylamine is 2.89. Calculate the ptt of the base solution after the chemist has added 777.3 ml. of the HIO, solution to t. Note for advanced students: you may assume the final volume equals the wntial volume of the solution plus the volume of HIO, solution added Round your answer to 2 decimal places. pH =

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter15: Additional Aqueous Equilibria

Section: Chapter Questions

Problem 93QRT: When 40.00 mL of a weak monoprotic acid solution is titrated with 0.100-M NaOH, the equivalence...

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When 40.00 mL of a weak monoprotic acid solution is titrated with 0.100-M NaOH, the equivalence point is reached when 35.00 mL base has been added. After 20.00 mL NaOH solution has been added, the titration mixture has a pH of 5.75. Calculate the ionization constant of the acid.
A diprotic acid, H2B(MM=126g/moL), is determined to be a hydrate, H2B xH2O. A 10.00-g sample of this hydrate is dissolved in enough water to make 150.0 mL of solution. Twenty-five milliliters of this solution requires 48.5 mL of 0.425 M NaOH to reach the equivalence point. What is x?
A 10.00-g sample of the ionic compound NaA, where A is the anion of a weak acid, was dissolved in enough water to make 100.0 mL of solution and was then titrated with 0.100 M HCl. After 500.0 mL HCl was added, the pH was 5.00. The experimenter found that 1.00 L of 0.100 M HCl was required to reach the stoichiometric point of the titration. a. What is the molar mass of NaA? b. Calculate the pH of the solution at the stoichiometric point of the titration.
A 0.400-M solution of ammonia was titrated with hydrochloric acid to the equivalence point, where the total volume was 1.50 times the original volume. At what pH does the equivalence point occur?
A monoprotic organic acid that has a molar mass of 176.1 g/mol is synthesized. Unfortunately, the acid produced is not completely pure. In addition, it is not soluble in water. A chemist weighs a 1.8451-g sample of the impure acid and adds it to 100.0 mL of 0.1050 M NaOH. The acid is soluble in the NaOH solution and reacts to consume most of the NaOH. The amount of excess NaOH is determined by titration: It takes 3.28 mL of 0.0970 M HCl to neutralize the excess NaOH. What is the purity of the original acid, in percent?
A solution is made by diluting 25.0 mL of concentrated HCl (37% by weight; density = 1.19 g/mL) to exactly 500 mL. Calculate the pH of the resulting solution.
Consider die titration of 50.0 mL of 0.10 M H3A (Ka1 = 5.0 104, Ka2 = 1.0 108, Ka3 = 1.0 1012) titrated by 0.10 M KOH. a. Calculate the pH of the resulting solution at 125 mL of KOH added. b. At what volume of KOH added does pH = 3.30? c. At 75.0 mL of KOH added, is the solution acidic or basic?
Repeat the procedure in Exercise 61, but for the titration of 25.0 mL of 0.100 M pyridine with 0.100 M hydrochloric acid (Kb for pyridine is 1.7 109). Do not calculate the points at 24.9 and 25.1 mL.
A solution consisting of 25.00 g NH4Cl in 178 mL of water is titrated with 0.114 M KOH. (a) How many milliliters of KOH are required to reach the equivalence point? (b) Calculate [Cl-], [K+], [NH3], and [OH-] at the equivalence point. (Assume that volumes are additive.) (c) What is the pH at the equivalence point?
A friend asks the following: Consider a buffered solution made up of the weak acid HA and its salt NaA. If a strong base like NaOH is added, the HA reacts with the OH to form A. Thus the amount of acid (HA) is decreased, and the amount of base (A) is increased. Analogously, adding HCI to the buffered solution forms more of the acid (HA) by reacting with the base (A). Thus how can we claim that a buffered solution resists changes in the pH of the solution? How would you explain buffering to this friend?