Problem 22: Buffer solutions
Buffer solutions are solutions which resist to changes in pH. Usually, buffer solutions consist of a weak acid and its conjugate base (for example CH3COOH/CH3COO?) or a weak base and its conjugate acid (for example NH3/NH4+). A buffer solution is formed by partial neutralization of a weak acid with a strong base or of a weak base with a strong acid. Alternatively, buffer solutions can be prepared by mixing the precalculated concentrations of each of the constituents.
The pH of a buffer solution, which is composed of a weak acid HA and its conjugate base A? is calculated by the Henderson-Hasselbalch equation:
[A?]
pH = pK+ log
a[HA]
where Ka is the acid dissociation constant of the weak acid HA and [HA] and [A?] are the concentrations of HA and A? in the buffer solution, respectively. Questions
1. Calculate the pH of a buffer solution, which contains 0.200 M formic acid (Ka = 2.1×10?4) and 0.150 M sodium formate.
2. Calculate the change in pH of the buffer solution in Question 1 when 0.01000 M of sodium hydroxide is added to the solution.
3. Calculate the volume of 0.200 M of sodium hydroxide which must be added to 100.0 cm3 of 0.150 M of acetic acid (CH3COOH, Ka = 1.8×10-5) in order to prepare a buffer solution with pH = 5.00.
4. The pH of a buffer solution containing 0.0100 M benzoic acid (C6H5COOH, Ka = 6.6×10?5) and 0.0100 M sodium benzoate is: a. 5.00, b. 4.18, c. 9.82, d. 9.00
In the problems below equal volumes of the following solutions A and B are mixed: 5. A: 0.100 M CH3COOH (Ka = 1.8×10?5), B: 0.0500 M NaOH
(i) The final solution: a. contains a weak acid, b. contains a strong base, c. is a buffer solution, d. none of the above
(ii) The pH of the final solution is: a. 3.02, b. 4.44, c. 3.17, d. 7.00 6 A: 0.100 M CH3COOH (Ka = 1.8×10?5), B: 0.150 M NaOH
(i) The final solution: a. contains a weak acid, b. contains a strong base, c. is a buffer solution, d. none of the above
(ii) The pH of the final solution is: a. 12.00, b. 12.70, c. 13.18, d. 12.40 7. A: 0.150 M CH3COOH, B: 0.100 M NaOH
(i) The final solution: a. contains a weak acid, b. contains a strong base, c. is a buffer solution, d. none of the above
(ii) The pH of the final solution is: a. 3.17, b. 3.02, c. 2.78, d. 3.22 8. A: 0.100 M CH3COOH, B: 0.100 M NaOH
(i) The final solution: a. contains a weak acid, b. contains a strong base, c. is a buffer solution, d. none of the above (ii) The pH of the final solution is: a. 7.00, b. 13.00, c. 2.87, d. 3.02
Problem 23: Titration of weak acids
Weak acids are titrated with solutions of strong bases of known concentration
(standard solutions). The solution of weak acid (analyte) is transferred into a 250-cm3 conical flask and the solution of strong base (titrant) is delivered from a burette. The equivalence point of the titration is reached when the amount of added titrant is chemically equivalent to the amount of analyte titrated. The graph, which shows the change of pH as a function of volume of titrant added, is called titration curve.
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The equivalence point of a titration is the theoretical point, which cannot be determined experimentally. It can only be estimated by observing some physical change associated with the process of the titration. In acid-base titrations, the end point is detected by using acid/base indicators. Questions
1. Construct the titration curve by calculating a few characteristic points and select indicator for the titration of 50.00 cm3 of 0.1000 M of acetic acid (CH3COOH, Ka = 1.8×10-5) with 0.1000 M of sodium hydroxide; you may consult Table 1. Table 1: Some common acid/base indicators Common name Transition range, pH Color change Methyl orange 3.2 - 4.4 red-orange Methyl red 4.2 - 6.2 red-yellow Bromothymol blue 6.0 - 7.6 yellow-blue Phenol red 6.8 - 8.2 yellow-red Phenolphthalein 8.0 - 9.8 colorless-red Thymolphthalein 9.3 - 10.5 colorless-blue
2. Ascorbic acid (vitamin C, C6H8O6) is a weak acid and undergoes the following dissociation steps:
C6H8O6 C6H7O6- + H+ Ka1 = 6.8×10-5 C6H7O6- C6H6O62- + H+ Ka2 = 2.7×10-12
Hence, ascorbic acid can be titrated with sodium hydroxide according to the first acid dissociation constant.
50.00 cm3 of 0.1000 M of ascorbic acid are titrated with 0.2000 M of sodium hydroxide.
(i) The initial pH of the solution is: a. 7.00, b. 2.58, c. 4.17, d. 1.00
(ii) The volume of titrant required for the equivalence point is: a. 50.00 cm3, b. 35.00 cm3, c. 25.00 cm3, d. 20.00 cm3
(iii) The pH of the solution after the addition of 12.5 cm3 of titrant is equal to: a. 4.17, b. 2.58, c. 7.00, d. 4.58
(iv). The pH at the equivalence point is: a. 7.00, b. 8.50, c. 8.43, d. 8.58
(v) The indicator, which must be chosen for the titration is: The appropriate indicator for the titration is (refer to Table 1):
a. bromothymol blue, b. phenol red, c. phenolphthalein, d. thymolphthalein (vi) The pH of the solution after the addition of 26.00 cm3 of titrant is: a. 13.30, b. 11.30, c. 11.00, d. 11.42
Problem 24: Separation by Extraction
Extraction is one the most common separation methods and it is based on the distribution equilibria of a substance between two immiscible liquids whose densities differ appreciably so that they are separated easily after mixing.
The more usual case is the extraction of an aqueous solution with an organic solvent, whereupon the inorganic ions and the polar organic compounds are found mainly in the aqueous phase and the polar organic compounds distribute in the organic phase. Inorganic ions may be reacted with an appropriate reagent to yield a non-polar compound that distributes in the organic phase
When a species S (solute) is distributed between two solvents 1 and, then we have the following equilibrium
KD
(S)1 (S)2
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where KD is the distribution coefficient given by
(a) KD=s2 (1)
(as)1
where (aS)1 and (aS)2 are the activities of S in phases 1 and 2. For a given system of solvents and species S, KD depends practically solely on temperature. Separations by extraction are commonly
performed with a separatory funnel (Fig. 1), a
common and easy to use glass apparatus found in any chemical laboratory.
Equation 1.1 is valid only if the solute S is present in both phases in the same form. Otherwise, if any dissociation, dimerization, complexation of the solute takes place, the distribution ratio, D, is used instead, which is given by D = (CS)2 / (CS)1 (1.2) where (CS)1 and (CS)2 are the analytical concentrations of S in phases 1 and 2 (rather than equilibrium concentrations of given species).
By convention, when one of the two solvents is water, Equation 1.2 is written with the aqueous concentration in the denominator and the
organic solvent concentration in the numerator. Figure 1. A typical separatory funnel. D is a conditional constant dependent on a
variety of experimental parameters like the concentration of S and of any other species involved in any equilibria with S in either phase and most likely on the pH of the aqueous phase (e.g. if S participates in any acid-base type equilibrium).
If W0 g of S is initially present in V1 mL of solvent 1 and S is extracted successively with equal fractions of V2 mL of solvent 2, the quantity Wn of S that remains in phase 1 after n such extractions is given by
n
or
??V1
Wn = ??DV+V??W0 (3)
21??
n
??WV1
? (4) fn = n= ???W0?DV2+V1?
where fn is the fraction of S that remains in solvent 1 after n extractions.
One can derive from Equations 1.3 and 1.4 that for a given volume of extractant, successive extractions with smaller individual volumes of extractant are more efficient than with all the volume of the extractant. 1 Prove Equation (3). 2 Substance S is distributed between chloroform and water with a distribution
ratio D = 3.2. If 50 cm3 of an aqueous solution of S is extracted with (a) one 100-cm3, and (b) four 25-cm3 portions of chloroform, calculated the percentage of S which is finally extracted in each case. 3 What is the minimum number of extractions required for the removal of at
least 99% of substance X from 100 cm3 of an aqueous solution containing
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