1) Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 =highest pH).2 200.0 mL of 0.100 M (C₂H5)2NH (Kb = 1.3 x 10-³) by 0.100 M HCI4 100.0 mL of 0.100 M HCI by 0.100 M NaOH3 100.0 mL of 0.100 M KOH by 0.100 M HCI15100.0 mL of 0.100 M NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCI200.0 mL of 0.100 M HC₂H30₂ (Ka = 1.8 x 10-5) by 0.100 M NaOHConsider the major species present at the halfway point. From this, you should be able to deduce the order of the pHwith only minimal calculationsSubmit Answer You have entered that answer before Incorrect. Tries 5/45 Previous Tries2) Rank the following titrations in order of increasing pH at the equivalence point of the titration (1= lowest pH and5 highest pH).2 100.0 mL of 0.100 M NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCI3200.0 mL of 0.100 M HC₂H30₂ (Ka = 1.8 x 10-5) by 0.100 M NaOH5 100.0 mL of 0.100 M KOH by 0.100 M HCI41100.0 mL of 0.100 M HF (Ka = 7.2 x 10-4) by 0.100 M NaOH200.0 mL of 0.100 M (C₂H5)2NH (Kb = 1.3 x 10-3) by 0.100 M HCI

Answered: Image /qna-images/answer/0599f511-3687-4830-998f-108071953dad.jpg

) Rank the following titrations in order of increasing pH at the halfway point to equ ighest pH). 2 200.0 mL of 0.100 M (C₂H5)2NH (Kb = 1.3 x 10-3) by 0.100 M HCI 100.0 mL of 0.100 M HCI by 0.100 M NaOH 3 100.0 mL of 0.100 M KOH by 0.100 M HCI 4 1 5 100.0 mL of 0.100 M NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCI 200.0 mL of 0.100 M HC₂H302 (Ka = 1.8 x 10-5) by 0.100 M NaOH Consider the major species present at the halfway point. From this, you should be ab vith only minimal calculations Submit Answer You have entered that answer before Incorrect. Tries 5/45 Previou ) Rank the following titrations in order of increasing pH at the equivalence point of = highest pH). 2 € 100.0 mL of 0.100 M NH3 (Kb: = 1.8 x1 0-5) by 0.100 M HCI 3 € 200.0 mL of 0.100 M HC₂H302 (Ka = 1.8 x 10-5) by 0.100 M NaOH 5 100.0 mL of 0.100 M KOH by 0.100 M HCI 4 € 100.0 mL of 0.100 M HF (Ka = 7.2 x 10-4) by 0.100 M NaOH 1 € 200.0 mL of 0.100 M (C₂H5)2NH (K₁ = 1.3 x 10-³) by 0.100 M HCI

) Rank the following titrations in order of increasing pH at the halfway point to equ ighest pH). 2 200.0 mL of 0.100 M (C₂H5)2NH (Kb = 1.3 x 10-3) by 0.100 M HCI 100.0 mL of 0.100 M HCI by 0.100 M NaOH 3 100.0 mL of 0.100 M KOH by 0.100 M HCI 4 1 5 100.0 mL of 0.100 M NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCI 200.0 mL of 0.100 M HC₂H302 (Ka = 1.8 x 10-5) by 0.100 M NaOH Consider the major species present at the halfway point. From this, you should be ab vith only minimal calculations Submit Answer You have entered that answer before Incorrect. Tries 5/45 Previou ) Rank the following titrations in order of increasing pH at the equivalence point of = highest pH). 2 € 100.0 mL of 0.100 M NH3 (Kb: = 1.8 x1 0-5) by 0.100 M HCI 3 € 200.0 mL of 0.100 M HC₂H302 (Ka = 1.8 x 10-5) by 0.100 M NaOH 5 100.0 mL of 0.100 M KOH by 0.100 M HCI 4 € 100.0 mL of 0.100 M HF (Ka = 7.2 x 10-4) by 0.100 M NaOH 1 € 200.0 mL of 0.100 M (C₂H5)2NH (K₁ = 1.3 x 10-³) by 0.100 M HCI

Chemistry

9th Edition

ISBN:9781133611097

Author:Steven S. Zumdahl

Publisher:Steven S. Zumdahl

Chapter15: Acid-base Equilibria

Section: Chapter Questions

Problem 14Q: Consider the following four titrations. i. 100.0 mL of 0.10 M HCl titrated by 0.10 M NaOH ii. 100.0...

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For the titration of 50.0 mL of 0.100-M HCl with 0.100-M NaOH, calculate the pH when these volumes of NaOH have been added: (a) 10.0 mL (b) 25.00 mL (c) 45.0 mL (d) 50.5 mL
The titration curves for two acids with the same base are shown on this graph. (a) Which is the curve for the weaker acid? Explain your choice. (b) Give the approximate pH at the equivalence point for the titration of each acid. (c) Explain why the pH at the equivalence point differs for each acid. (d) Explain why the starting pH values of the two acids differ. (e) Which indicator or indicators, phenolphthalein, bromthymol blue, or methyl red, could be used for the titration of Acid 1? For the titration of Acid 2? Explain your choices.
Calculate the volume of 0.150-M HCl required to titrate to the equivalence point for each of these samples. (a) 25.0 mL of 0.175-M KOH (b) 15.0 mL of 6.00-M NH3 (c) 15.0 mL of propylamine, CH3CH2CH2NH2, which has a density of 0.712 g/mL (d) 40.0 mL of 0.0050-M Ba(OH)2
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.
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.
Calculate the pH change when 10.0 mL of 0.100-M NaOH is added to 90.0 mL pure water, and compare the pH change with that when the same amount of NaOH solution is added to 90.0 mL of a buffer consisting of 1.00-M NH3 and 1.00-M NH4Cl. Assume that the volumes are additive. Kb of NH3 = 1.8 × 10-5.
Consider the following four titrations. i. 100.0 mL of 0.10 M HCl titrated by 0.10 M NaOH ii. 100.0 mL of 0.10 M NaOH titrated by 0.10 M HCl iii. 100.0 mL of 0.10 M CH3NH2 titrated by 0.10 M HCl iv. 100.0 mL of 0.10 M HF titrated by 0.10 M NaOH Rank the titrations in order of: a. increasing volume of titrant added to reach the equivalence point. b. increasing pH initially before any titrant has been added. c. increasing pH at the halfway point in equivalence. d. increasing pH at the equivalence point. How would the rankings change if C5H5N replaced CH3NH2 and if HOC6H5 replaced HF?
Using Figure 17.11, suggest an indicator to use in each of the following titrations: (a) The weak base pyridine is titrated with HCl. (b) Formic acid is titrated with NaOH. (c) Ethylenediamine, a weak diprotic base, is titrated with HCl. Figure 17.11 Common acid-base indicators. The color changes occur over a range of pH values. Notice that o few indicators hove color changes over two different pH ranges.
Given three acid-base indicatorsâ€”methyl orange (end point at pH 4), bromthymol blue (end point at pH 7), and phenolphthalein (end point at pH 9)â€”which would you select for the following acid-base titrations? (a) perchloric acid with an aqueous solution of ammonia (b) nitrous acid with lithium hydroxide (c) hydrobromic acid with strontium hydroxide (d) sodium fluoride with nitric acid
What is the pH of a buffer that is 0.175 M in a weak acid and 0.200 M in the acids conjugate base? The acids ionization constant is 5.7 104.
Chloropropionic acid, ClCH2CH2COOH, is a weak monoprotic acid with Ka = 7.94 105. Calculate the pH at the equivalence point in a titration of 10.00 mL of 0.100 M chloropropionic acid with 0.100 M KOH. Choose an indicator from Table 16.4 for the titration. Explain your choice. TABLE 16.5 Properties of Several Indicators