1311975310Titration of 10.0mL of an unknown diprotic acid with 1.00M NaOH.5pH = 4.0115pH = 9,2710Using the titration curve shown above:1. Calculate the concentration of the acid using the data from equivalence point #1.2025. Calculate the value of pKa2 from the data at 17.0mL.302. Calculate the concentration of the sulfurous acid using the data from equivalence point #2.3. Calculate the volumes for midpoint #1 and midpoint #2.4. Calculate the value of pKal from midpoint #1 and pKa2 from midpoint #2.5. Calculate the value of pKal using the data from 3.0mL.

13 11 9 7 5 3 1 0 Titration of 10.0mL of an unknown diprotic acid with 1.00M NaOH. 5 pH = 4.01 15 pH=9,27 10 Using the titration curve shown above: 1. Calculate the concentration of the acid using the data from equivalence point #1. 20 5. Calculate the value of pKal using the data from 3.0mL. 25 6. Calculate the value of pKa2 from the data at 17.0mL. 2. Calculate the concentration of the sulfurous acid using the data from equivalence point #2. 3. Calculate the volumes for midpoint #1 and midpoint #2. 4. Calculate the value of pKal from midpoint #1 and pKa2 from midpoint #2. 30

13 11 9 7 5 3 1 0 Titration of 10.0mL of an unknown diprotic acid with 1.00M NaOH. 5 pH = 4.01 15 pH=9,27 10 Using the titration curve shown above: 1. Calculate the concentration of the acid using the data from equivalence point #1. 20 5. Calculate the value of pKal using the data from 3.0mL. 25 6. Calculate the value of pKa2 from the data at 17.0mL. 2. Calculate the concentration of the sulfurous acid using the data from equivalence point #2. 3. Calculate the volumes for midpoint #1 and midpoint #2. 4. Calculate the value of pKal from midpoint #1 and pKa2 from midpoint #2. 30

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter14: Equilibria In Acid-base Solutions

Section: Chapter Questions

Problem 74QAP: Fifty cm3 of 1.000 M nitrous acid is titrated with 0.850 M NaOH. What is the pH of the solution (a)...

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Fifty cm3 of 1.000 M nitrous acid is titrated with 0.850 M NaOH. What is the pH of the solution (a) before any NaOH is added? (b) at half-neutralization? (c) at the equivalence point? (d) when 0.10 mL less than the volume of NaOH to reach the equivalence point is added? (e) when 0.10 mL more than the volume of NaOH to reach the equivalence point is added? (f) Use your data to construct a plot similar to that shown in Figure 14.10 (pH versus volume NaOH added).
Malonic acid (HO2CCH2CO2H) is a diprotic acid. In the titration of malonic acid w ith NaOH, stoichiometric points occur at pH = 3.9 and 8.8. A 25.00-mL sample of malonic acid of unknown concentration is titrated with 0.0984 M NaOH, requiring 31.50 mL of the NaOH solution to reach the phenolphthalein end point. Calculate the concentration of the initial malonic acid solution. (Sec Exercise 113.)
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
Consider the following two acids: pKa1 = 2.98; pKa2 = 13.40 HO2CCH2CH2CH2CH2CO2H Adipic acid pKa1 = 4.41; pKa2 = 5.28 In two separate experiments the pH was measured during the titration of 5.00 mmol of each acid with 0.200 M NaOH. Each experiment showed only one stoichiometric point when the data were plotted. In one experiment the stoichiometric point was at 25.00 mL added NaOH, and in the other experiment the stoichiometric point was at 50.00 mL NaOH. Explain these results.
Lactic acid is a common by-product of cellular respiration and is often said to cause the burn associated with strenuous activity. A 25.0-mL sample of 0.100 M lactic acid (HC3H5O3, pKa = 3.86) is titrated with 0.100 M NaOH solution. Calculate the pH after the addition of 0.0 mL, 4.0 mL, 8.0 mL, 12.5 mL, 20.0 mL, 24.0 mL, 24.5 mL, 24.9 mL, 25.0 mL, 25.1 mL, 26.0 mL, 28.0 mL, and 30.0 mL of the NaOH. Plot the results of your calculations as pH versus milliliters of NaOH added.
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
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.