6.E: Acid-Base Equilibria (Exercises)
Polyprotic Acids
1. Which of the following concentrations would be practically equal in a calculation of the equilibrium concentrations in a 0.134- M solution of H 2 CO 3 , a diprotic acid:
- \(\ce{[H3O+]}\),
- \([OH^−]\)
- \([H_2CO_3]\)
- \(\ce{[HCO3- ]}\)
- \(\ce{[CO3^2- ]}\)
No calculations are needed to answer this question.
- Answer
- [H 3 O + ] and \(\ce{[HCO3- ]}\) are equal, H 3 O + and \(\ce{HCO3-}\) are practically equal
2. Calculate the concentration of each species present in a 0.050- M solution of H 2 S.
3. Calculate the concentration of each species present in a 0.010- M solution of phthalic acid, C 6 H 4 (CO 2 H) 2 .
- Answer
- \(\ce{C6H4(CO2H)2}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{C6H4(CO2H)(CO2)-}(aq) \hspace{20px} K_\ce{a}=1.1×10^{−3}\)
- \(\ce{C6H4(CO2H)(CO2)}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{C6H4(CO2)2^2-}(aq) \hspace{20px} K_\ce{a}=3.9×10^{−6}\)
- [C 6 H 4 (CO 2 H) 2 ] 7.2 × 10 −3 M , [C 6 H 4 (CO 2 H)(CO 2 ) − ] = [H 3 O + ] 2.8 × 10 −3 M , \(\ce{[C6H4(CO2)2^2- ]}\)3.9 × 10 −6 M , [OH − ] 3.6 × 10 −12 M
4. Salicylic acid, HOC 6 H 4 CO 2 H, and its derivatives have been used as pain relievers for a long time. Salicylic acid occurs in small amounts in the leaves, bark, and roots of some vegetation (most notably historically in the bark of the willow tree). Extracts of these plants have been used as medications for centuries. The acid was first isolated in the laboratory in 1838.
a. Both functional groups of salicylic acid ionize in water, with K a = 1.0 × 10 −3 for the—CO 2 H group and 4.2 × 10 −13 for the −OH group. What is the pH of a saturated solution of the acid (solubility = 1.8 g/L)?
b. Aspirin was discovered as a result of efforts to produce a derivative of salicylic acid that would not be irritating to the stomach lining. Aspirin is acetylsalicylic acid, CH 3 CO 2 C 6 H 4 CO 2 H. The −CO 2 H functional group is still present, but its acidity is reduced, K a = 3.0 × 10 −4 . What is the pH of a solution of aspirin with the same concentration as a saturated solution of salicylic acid (See Part a)?
c. Under some conditions, aspirin reacts with water and forms a solution of salicylic acid and acetic acid:
\[\ce{CH3CO2C6H4CO2H}(aq)+\ce{H2O}(l)⟶\ce{HOC6H4CO2H}(aq)+\ce{CH3CO2H}(aq)\]
d. Which of the acids salicylic acid or acetic acid produces more hydronium ions in solution such a solution?
e. What are the concentrations of molecules and ions in a solution produced by the hydrolysis of 0.50 g of aspirin dissolved in enough water to give 75 mL of solution?
5. The ion HTe − is an amphiprotic species; it can act as either an acid or a base.
- Wh at is K a for the acid reaction of HTe − with H 2 O?
- What is K b for the reaction in which HTe − functions as a base in water?
- Demonstrate whether or not the second ionization of H 2 Te can be neglected in the calculation of [HTe − ] in a 0.10 M solution of H 2 Te.
- Answer
- a. (K_{\ce a2}=1×10^{−5};\)
- b. \(K_\ce{b}=4.3×10^{−12};\)
- c. \(\ce{\dfrac{[Te^2- ][H3O+]}{[HTe- ]}}=\dfrac{(x)(0.0141+x)}{(0.0141−x)}≈\dfrac{(x)(0.0141)}{0.0141}=1×10^{−5}\). Solving for x gives 1 × 10 −5 M . Therefore, compared with 0.014 M , this value is negligible (0.071%).
Buffers
6. Explain why a buffer can be prepared from a mixture of NH 4 Cl and NaOH but not from NH 3 and NaOH.
7. Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the acid H 3 PO 4 and a salt of its conjugate base NaH 2 PO 4 .
- Answer
- Excess H 3 O + is removed primarily by the reaction:
- \(\ce{H3O+}(aq)+\ce{H2PO4-}(aq)⟶\ce{H3PO4}(aq)+\ce{H2O}(l)\) Excess base is removed by the reaction: \(\ce{OH-}(aq)+\ce{H3PO4}(aq)⟶\ce{H2PO4-}(aq)+\ce{H2O}(l)\)
8. Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the base NH 3 and a salt of its conjugate acid NH 4 Cl.
9. What is [H 3 O + ] in a solution of 0.25 M CH 3 CO 2 H and 0.030 M NaCH 3 CO 2 ?
\(\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−5}\)
- Answer
- [H 3 O + ] = 1.5 × 10 −4 M
10. What is [H 3 O + ] in a solution of 0.075 M HNO 2 and 0.030 M NaNO 2 ?
- Answer
- \(\ce{HNO2}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{NO2-}(aq) \hspace{20px} K_\ce{a}=4.5×10^{−5}\)
11. What is [OH − ] in a solution of 0.125 M CH 3 NH 2 and 0.130 M CH 3 NH 3 Cl?
- Answer
- \(\ce{CH3NH2}(aq)+\ce{H2O}(l)⇌\ce{CH3NH3+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=4.4×10^{−4}\)
- [OH − ] = 4.2 × 10 −4 M
12. What is [OH − ] in a solution of 1.25 M NH 3 and 0.78 M NH 4 NO 3 ?
- Answer
- \(\ce{NH3}(aq)+\ce{H2O}(l)⇌\ce{NH4+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=1.8×10^{−5}\)
13. What concentration of NH 4 NO 3 is required to make [OH − ] = 1.0 × 10 −5 in a 0.200- M solution of NH 3 ?
- Answer
- [NH 4 NO 3 ] = 0.36 M
14. What concentration of NaF is required to make [H 3 O + ] = 2.3 × 10 −4 in a 0.300- M solution of HF?
15. What is the effect on the concentration of acetic acid, hydronium ion, and acetate ion when the following are added to an acidic buffer solution of equal concentrations of acetic acid and sodium acetate:
- HCl
- KCH 3 CO 2
- NaCl
- KOH
- CH 3 CO 2 H
- Answer
- a. The added HCl will increase the concentration of H 3 O + slightly, which will react with \(\ce{CH3CO2-}\) and produce CH 3 CO 2 H in the process. Thus, \(\ce{[CH3CO2- ]}\) decreases and [CH 3 CO 2 H] increases.
- b. The added KCH 3 CO 2 will increase the concentration of \(\ce{[CH3CO2- ]}\) which will react with H 3 O + and produce CH 3 CO 2 H in the process. Thus, [H 3 O + ] decreases slightly and [CH 3 CO 2 H] increases.
- c. The added NaCl will have no effect on the concentration of the ions.
- d. The added KOH will produce OH − ions, which will react with the H 3 O + , thus reducing [H 3 O + ]. Some additional CH 3 CO 2 H will dissociate, producing \(\ce{[CH3CO2- ]}\) ions in the process. Thus, [CH 3 CO 2 H] decreases slightly and \(\ce{[CH3CO2- ]}\) increases.
- e. The added CH 3 CO 2 H will increase its concentration, causing more of it to dissociate and producing more \(\ce{[CH3CO2- ]}\) and H 3 O + in the process. Thus, [H 3 O + ] increases slightly and \(\ce{[CH3CO2- ]}\) increases.
16. What is the effect on the concentration of ammonia, hydroxide ion, and ammonium ion when the following are added to a basic buffer solution of equal concentrations of ammonia and ammonium nitrate:
- KI
- NH 3
- HI
- NaOH
- NH 4 Cl
17. What will be the pH of a buffer solution prepared from 0.20 mol NH 3 , 0.40 mol NH 4 NO 3 , and just enough water to give 1.00 L of solution?
- Answer
- pH = 8.95
18. Calculate the pH of a buffer solution prepared from 0.155 mol of phosphoric acid, 0.250 mole of KH 2 PO 4 , and enough water to make 0.500 L of solution.
19. How much solid NaCH 3 CO 2 •3H 2 O must be added to 0.300 L of a 0.50- M acetic acid solution to give a buffer with a pH of 5.00? (Hint: Assume a negligible change in volume as the solid is added.)
- Answer
- 37 g (0.27 mol)
20. What mass of NH 4 Cl must be added to 0.750 L of a 0.100- M solution of NH 3 to give a buffer solution with a pH of 9.26? (Hint: Assume a negligible change in volume as the solid is added.)
21. A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate. Use 1.80 × 10 −5 as K a for acetic acid.
a. What is the pH of the solution?
b. Is the solution acidic or basic?
22. What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to 0.200 L of the original buffer?
- pH = 5.222;
- The solution is acidic. (c) pH = 5.221
23. A 5.36–g sample of NH 4 Cl was added to 25.0 mL of 1.00 M NaOH and the resulting solution diluted to 0.100 L.
- What is the pH of this buffer solution?
- Is the solution acidic or basic?
- What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to the solution?
24. Which acid in [link] is most appropriate for preparation of a buffer solution with a pH of 3.1? Explain your choice.
- Answer
- To prepare the best buffer for a weak acid HA and its salt, the ratio \(\dfrac{\ce{[H3O+]}}{K_\ce{a}}\) should be as close to 1 as possible for effective buffer action. The [H 3 O + ] concentration in a buffer of pH 3.1 is [H 3 O + ] = 10 −3.1 = 7.94 × 10 −4 M
- We can now solve for K a of the best acid as follows:
- \(\dfrac{\ce{[H3O+]}}{K_\ce{a}}=1\)
- \(K_\ce{a}=\dfrac{\ce{[H3O+]}}{1}=7.94×10^{−4}\)
- In [link] , the acid with the closest K a to 7.94 × 10 −4 is HF, with a K a of 7.2 × 10 −4 .
25. Which acid in [link] is most appropriate for preparation of a buffer solution with a pH of 3.7? Explain your choice.
26. Which base in [link] is most appropriate for preparation of a buffer solution with a pH of 10.65? Explain your choice.
- Answer
- For buffers with pHs > 7, you should use a weak base and its salt. The most effective buffer will have a ratio \(\dfrac{\ce{[OH- ]}}{K_\ce{b}}\) that is as close to 1 as possible. The pOH of the buffer is 14.00 − 10.65 = 3.35. Therefore, [OH − ] is [OH − ] = 10 −pOH = 10 −3.35 = 4.467 × 10 −4 M .
- We can now solve for K b of the best base as follows: \(\dfrac{\ce{[OH- ]}}{K_\ce{b}}=1\) K b = [OH − ] = 4.47 × 10 −4 In [link] , the base with the closest K b to 4.47 × 10 −4 is CH 3 NH 2 , with a K b = 4.4 × 10 −4 .
27. Which base in [link] is most appropriate for preparation of a buffer solution with a pH of 9.20? Explain your choice.
28. Saccharin, C 7 H 4 NSO 3 H, is a weak acid ( K a = 2.1 × 10 −2 ). If 0.250 L of diet cola with a buffered pH of 5.48 was prepared from 2.00 × 10 −3 g of sodium saccharide, Na(C 7 H 4 NSO 3 ), what are the final concentrations of saccharine and sodium saccharide in the solution?
- Answer
- The molar mass of sodium saccharide is 205.169 g/mol. Using the abbreviations HA for saccharin and NaA for sodium saccharide the number of moles of NaA in the solution is:
- 9.75 × 10 −6 mol. This ionizes initially to form saccharin ions, A − , with: [A − ] = 3.9 × 10 −5 M
29. What is the pH of 1.000 L of a solution of 100.0 g of glutamic acid (C 5 H 9 NO 4 , a diprotic acid; K 1 = 8.5 × 10 −5 , K 2 = 3.39 × 10 −10 ) to which has been added 20.0 g of NaOH during the preparation of monosodium glutamate, the flavoring agent? What is the pH when exactly 1 mol of NaOH per mole of acid has been added?
Acid-Base Titrations
30. Explain how to choose the appropriate acid-base indicator for the titration of a weak base with a strong acid.
- Answer
- At the equivalence point in the titration of a weak base with a strong acid, the resulting solution is slightly acidic due to the presence of the conjugate acid. Thus, pick an indicator that changes color in the acidic range and brackets the pH at the equivalence point. Methyl orange is a good example.
31. Explain why an acid-base indicator changes color over a range of pH values rather than at a specific pH.
32. Why can we ignore the contribution of water to the concentrations of H 3 O + in the solutions of following acids:
- 0.0092 M HClO, a weak acid
- 0.0810 M HCN, a weak acid
- 0.120 M \(\ce{Fe(H2O)6^2+}\) a weak acid, K a = 1.6 × 10 −7
but not the contribution of water to the concentration of OH − ?
- Answer
- In an acid solution, the only source of OH − ions is water. We use K w to calculate the concentration. If the contribution from water was neglected, the concentration of OH − would be zero.
33. Why can we ignore the contribution of water to the concentration of OH − in a solution of the following bases:
0.0784 M C 6 H 5 NH 2 , a weak base
0.11 M (CH 3 ) 3 N, a weak base
but not the contribution of water to the concentration of H 3 O + ?
34. Draw a curve for a series of solutions of HF. Plot [H 3 O + ] total on the vertical axis and the total concentration of HF (the sum of the concentrations of both the ionized and nonionized HF molecules) on the horizontal axis. Let the total concentration of HF vary from 1 × 10 −10 M to 1 × 10 −2 M .
- Answer
35. Draw a curve similar to that shown in Figure for a series of solutions of NH 3 . Plot [OH − ] on the vertical axis and the total concentration of NH 3 (both ionized and nonionized NH 3 molecules) on the horizontal axis. Let the total concentration of NH 3 vary from 1 × 10 −10 M to 1 × 10 −2 M .
36. Calculate the pH at the following points in a titration of 40 mL (0.040 L) of 0.100 M barbituric acid ( K a = 9.8 × 10 −5 ) with 0.100 M KOH.
a. no KOH added
b. 20 mL of KOH solution added
c. 39 mL of KOH solution added
d. 40 mL of KOH solution added
e. 41 mL of KOH solution added
- Answer
- a. pH = 2.50; b. pH = 4.01; c. pH = 5.60; d. pH = 8.35; e. pH = 11.08
37. The indicator dinitrophenol is an acid with a K a of 1.1 × 10 −4 . In a 1.0 × 10 −4 - M solution, it is colorless in acid and yellow in base. Calculate the pH range over which it goes from 10% ionized (colorless) to 90% ionized (yellow).