4. A group of students carried out a titration and graphed their results as shown in the graph below. For this graph, equivalence point is approximately 14- Acid-Base Titration Curve 13- 12- 11- 10- 9- 8- 7- 6- 5- 4- 3 2 1 0+ T 0 S 35 mls pH of 8 pH of 7 20 mls • . 10 15 20 25 30 35 40 45 50 55 60 65 70 Volume of 1.0 mol/L NaOH() (mL)

4. A group of students carried out a titration and graphed their results as shown in the graph below. For this graph, equivalence point is approximately 14- Acid-Base Titration Curve 13- 12- 11- 10- 9- 8- 7- 6- 5- 4- 3 2 1 0+ T 0 S 35 mls pH of 8 pH of 7 20 mls • . 10 15 20 25 30 35 40 45 50 55 60 65 70 Volume of 1.0 mol/L NaOH() (mL)

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter15: Complex Acid/base Systems

Section: Chapter Questions

Problem 15.9QAP

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Similar questions

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
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 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?
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.
In an acid-base titration, 21.16 mL of an NaOH solution are needed to neutralize 20.04 mL of a 0.0997 M HCl solution. To find the molarity of the NaOH solution, we can use the following procedure: First note the value of MH in the HCl solution. ____________M Find MOH- in the NaOH solution. Use Eq.3. ____________M Obtain MNaOH from MOH. ____________M
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.
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).
2. What is the pH at the equivalence point in the titration of 25.0 mL of 0.090 M phenol (Ka = 1.3 × 10−10) with 0.108 M NaOH? 7.00 5.65 8.98 11.29
K4 for acetic acid is 1.7 105 at 25C. A buffer solution is made by mixing 52.1 mL of 0.122 M acetic acid with 46.1 mL of 0.182 M sodium acetate. Calculate the pH of this solution at 25C after the addition of 5.82 mL of 0.125 M NaOH.
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?
Figure 15-3 outlines the classic scheme for separating a mixture of insoluble chloride salts from one another. Explain the chemistry involved in the various steps of the figure.