144 13- 12. 11 Titration of Weak Acid 10- 9 86 7. Equivalence point pH, 8.72 65 A 4 3- 210 0001-0 to nojuan anta 0 5 10 15 20 25 30 35 40 45 50 Volume of 0.100 M NaOH added (mL) Fig. 10.2: Titration curve of a weak acid with a strong base The shape of a titration curve depends on the ionization constants of the acid and base and their concentrations. The example given above is for a weak acid and strong base. The equivalence point is defined as the point where the amount of base added is equal to the original amount of acid in the solution. Therefore, at the equivalence point, the number of moles of base added equals the number of moles of acid originally present in the solution being titrated. After determining the equivalence point, the concentration of the basic solution can be easily calculated with the titration formula (when the stoichiometry of acid and base is 1:1): Molarity of acid (mol/L) x Volume of acid = Molarity of base (mol/L) x Volume of base Macid X Vacid = Mbase X Vbase moles of acid in original solution = moles of base added Concentration of base (mol/L) or [base] = [moles of acid in original solution] (volume in L of base at equivalence point)

144 13- 12. 11 Titration of Weak Acid 10- 9 86 7. Equivalence point pH, 8.72 65 A 4 3- 210 0001-0 to nojuan anta 0 5 10 15 20 25 30 35 40 45 50 Volume of 0.100 M NaOH added (mL) Fig. 10.2: Titration curve of a weak acid with a strong base The shape of a titration curve depends on the ionization constants of the acid and base and their concentrations. The example given above is for a weak acid and strong base. The equivalence point is defined as the point where the amount of base added is equal to the original amount of acid in the solution. Therefore, at the equivalence point, the number of moles of base added equals the number of moles of acid originally present in the solution being titrated. After determining the equivalence point, the concentration of the basic solution can be easily calculated with the titration formula (when the stoichiometry of acid and base is 1:1): Molarity of acid (mol/L) x Volume of acid = Molarity of base (mol/L) x Volume of base Macid X Vacid = Mbase X Vbase moles of acid in original solution = moles of base added Concentration of base (mol/L) or [base] = [moles of acid in original solution] (volume in L of base at equivalence point)

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter15: Acid–base Equilibria

Section: Chapter Questions

Problem 57P

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Similar questions

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.
Ammonia gas is bubbled into 275 mL of water to make an aqueous solution of ammonia. To prepare a buffer with a pH of 9.56, 15.0 g of NH4Cl are added. How many liters of NH3; at 25C and 0.981 atm should be used to prepare the buffer? Assume no volume changes and ignore the vapor pressure of water.
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?
Sketch a titration curve for the titration of potassium hydroxide with HCl, both 0.100 M. Identify three regions in which a particular chemical species or system dominates the acid-base equilibria.
Determine the dominant acid-base equilibrium that results when each of the following pairs of solutions is mixed. Indicate the equilibrium by writing 1 for a strong acid, 3 for a weak acid, 4 for an acidic buffer, 7 for a neutral solution, 10 for a basic buffer, 11 for a weak base, and 13 for a strong base. (a) 10.0 mL of 0.15 M NaOH + 15.0 mL of 0.10 M HNO3 (b) 25.0 mL of 0.10 M HCl + 10.0 mL of 0.25 M NH3 (c) 50.0 mL of 0.050 M NaOH + 50.0 mL of 0.10 M NH3 (d) 50.0 mL of 0.10 M NH3 + 50.0 mL of 0.05 M HCl
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.)