What is the pH at the equivalence point when 25.0 mL of a 0.225 M solution of acetic acid (CH3COOH) is titrated with 0.100 M NaOH to its end point?
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- Consider the nanoscale-level representations for Question 111 of the titration of the aqueous strong acid HA with aqueous NaOH, the titrant. Water molecules and Na+ ions are omitted for clarity. Which diagram corresponds to the situation: (a) After a very small volume of titrant has been added to the initial HA solution? (b) Halfway to the equivalence point? (c) When enough titrant has been added to take the solution just past the equivalence point? (d) At the equivalence point? Nanoscale representations for Question 111.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?When a diprotic acid, H2A, is titrated with NaOH, the protons on the diprotic acid are generally removed one at a time, resulting in a pH curve that has the following generic shape: a. Notice that the plot has essentially two titration curves. If the first equivalence point occurs at 100.0 mL NaOH added, what volume of NaOH added corresponds to the second equivalence point? b. For the following volumes of NaOH added, list the major species present after the OH reacts completely. i. 0 mL NaOH added ii. between 0 and 100.0 mL NaOH added iii. 100.0 mL NaOH added iv. between 100.0 and 200.0 mL NaOH added v. 200.0 mL NaOH added vi. after 200.0 mL NaOH added c. If the pH at 50.0 mL NaOH added is 4.0, and the pH at 150.0 mL NaOH added is 8.0, determine the values Ka1, and Ka2 for the diprotic acid.
- Does the pH of the solution increase, decrease, or stay the same when you (a) Add solid sodium oxalate, Na2C2O4, to 50.0 mL of 0.015-M oxalic acid? (b) Add solid ammonium chloride to 100. mL of 0.016-M HCl? (c) Add 20.0 g NaCl to 1.0 L of 0.012-M sodium acetate, NaCH3COO?A solution is prepared by dissolving 0.350 g of benzoic acid, HC7H5O2, in water to make 100.0 mL of solution. A 30.00-mL sample of the solution is titrated with 0.272 M KOH. Calculate the pH of the solution (a) before titration. (b) halfway to the equivalence point. (c) at the equivalence point.Phenol, C6H5OH, is a weak organic acid. Suppose 0.515 g of the compound is dissolved in enough water to make 125 mL of solution. The resulting solution is titrated with 0.123 M NaOH. C6H5OH(aq) + OH(aq) C6H5O(aq) + H2O() (a) What is the pH of the original solution of phenol? (b) What are the concentrations of all of the following ions at the equivalence point: Na+, H3O+, OH, and C6H5O? (c) What is the pH of the solution at the equivalence point?
- The titration of 0.100 M acetic acid with 0.100 M NaOH is described in the text. What is the pH of the solution when 35.0 mL of the base has been added to 100.0 mL of 0.100 M acetic acid?The weak base ethanolamine. HOCH2CH2NH2, can be titrated with HCl. HOCH2CH2NH2(aq)+H3O+(aq)HOCH2CH2NH3+(aq)+H2O(l) Assume you have 25.0 mL of a 0.010 M solution of ethanolamine and titrate it with 0.0095 M HCl. (Kb for ethanolamine is 3.2 107.) (a) What is the pH of the ethanolamine solution before the titration begins? (b) What is the pH at the equivalence point? (c) What is the pH at the halfway point of the titration? (d) Which indicator in Figure 17.11 would be the best choice to detect the equivalence point? (e) Calculate the pH of the solution after adding 5.00, 10.0, 20.0, and 30.0 mL of the acid. (f) Combine the information in parts (a), (b), (c), and (e), and plot an approximate titration curve.Aniline hydrochloride, (C6H5NH3)Cl, is a weak acid. (Its conjugate base is the weak base aniline, C6H5NH2.) The acid can be titrated with a strong base such as NaOH. C6H5NH3+(aq)+OH(aq)C6H5NH2(aq)+H2O(l) Assume 50.0 mL of 0.100 M aniline hydrochloride is titrated with 0.185 M NaOH. (Ka for aniline hydrochloride is 2.4 105.) (a) What is the pH of the (C6H5NH3) solution before the titration begins? (b) What is the pH at the equivalence point? (c) What is the pH at the halfway point of the titration? (d) Which indicator in Figure 17.11 could be used to detect the equivalence point? (e) Calculate the pH of the solution after adding 10.0, 20.0, and 30.0 mL of base. (f) Combine the information in parts (a), (b), (c), and (e), and plot an approximate titration curve.
- A buffer solution is prepared by dissolving 1.50 g each of benzoic acid, C6H5CO2H, and sodium benzoate, NaC6H5CO2, in 150.0 mL of solution. (a) What is the pH of this buffer solution? (b) Which buffer component must be added, and in what quantity, to change the pH to 4.00? (c) What quantity of 2.0 M NaOH or 2.0 M HCl must be added to the buffer to change the pH to 4.00?You are given the following acidbase titration data, where each point on the graph represents the pH after adding a given volume of titrant (the substance being added during the titration). a What substance is being titrated, a strong acid, strong base, weak acid, or weak base? b What is the pH at the equivalence point of the tiration? c What indicator might you use to perform this titration? Explain.Three students titrate different samples of the same solution of HCI to obtain its molarity. Below are their data. Student A: 20.00mLHCl+20.00mLH2O 0.100 M NaOH used to titrate to the equivalence point Student B: 20.00mLHCl+40.00mLH2O 0.100 M NaOH used to titrate to the equivalence point Student C: 20.00mLHCl+20.00mLH2O 0.100 M Ba(OH)2 used to titrate to the equivalence point. All the students calculated the molarities correctly. Which (if any) of the following statements are true? (a) The molarity calculated by A is half that calculated by B. (b) The molarity calculated by A is equal to that calculated by C. (c) The molarity calculated by B is twice that calculated by C. (d) The molarity calculated by A is twice that calculated by B. (e) The molarity calculated by A is equal to that calculated by B.
