Acids and Bases

Review - 11 (Acids and bases) • • • K a, K b, K w Strong acids, weak acids, polyprotic acids Titrations PROBLEMS 1. A 0.1-molar solution of acetic acid -5 (Ka = 1.8 x 10 ) has a pH of about a. 1 b. 3 c. 7 d. 10 e. 14 2. 3. Ka, the acid dissociation constant, -4 for an acid is 9 x 10 at room temperature. At this temperature, what is the approximate percent dissociation of the acid in a 1.0 M solution? a. 0.03% b. 0.09% c. 3% d. 5% e. 9% a. b. 5.- c. 4.- d. 6.- e. 5. 7. Using the information below, choose the best answer for preparing a pH = 8 buffer. Acid Ka value -3 H3PO4 7.2 x 10 -8 H2PO4 6.3 x 10 2-13 HPO4 4.2 x 10 a. K2HPO4 + KH2PO4 b. H3PO4 c. K2HPO4 + K3PO4 d. K3PO4 e. K2HPO4 + H3PO4 For 4-6, use the following: -4 Ka = 1.8 x 10 for HCOOH -4 Kb = 4.4 x 10 for CH3NH2 -2 Ka1 = 3 x 10 for H3PO2 -7 Ka2 = 1.7 x 10 for H2PO2 4. 6. a solution with pH = 7 a solution with pH < 7, which is not a buffer solution a solution with pH < 7, which is a buffer a solution with a pH > 7, which is not a buffer solution a solution with pH > 7, which is a buffer A solution with an initial KCOOH concentration of 1 M, and an initial K2HPO2 concentration of 1 M is… A solution with an initial H3PO2 concentration of 1 M, and an initial KH2PO2 concentration of 1 M. Name: ____________________ • • • Buffers Salts Acid base equilibria A solution with an initial CH3NH2 concentration of 1 M, and an initial CH3NH3Cl concentration of 1 M. A solution of a weak base is titrated with a solution of a standard strong acid. The progress of the titration is followed with a pH meter. Which of the followin observations would occur? a. The pH of the solution gradually decreases throughout the experiment. b. Initially the pH of the solution drops slowly, and then it drops much more rapidly. c. At the equivalence point the pH is 7. d. After the equivalence point, the pH becomes constant because this is the buffer region. e. The pOH at the equivalence point equals the pKb of the base. d. e. The solution is acidic because of hydrolysis of the NO2 ion. The solution is basic because of hydrolysis of the NO2 ion. For 12-15, use the following: 15.- a. 12.- b. 14.- c. d. 13.- e. H2C2O4 (oxalic acid) and KHC2O4 (potassium hydrogen oxalate) KNO3 (potassium nitrate) and HNO3 (nitric acid) NH3 (ammonia) and NH4NO3 (ammonium nitrate) C2H5NH2 (ethylamine) and KOH (potassium hydroxide) CH3NH2 (methylamine) and HC2H3O2 (acetic acid) 12. The most acidic solution (lowest pH). 13. The solution with a pH nearest 7. 14. A buffer with a pH > 7. 8. 9. What is the ionization constant, Ka, for a weak monoprotic acid if a 0.30molar solution has a pH of 4.0? -10 a. 9.7 x 10 -2 b. 4.7 x 10 -6 c. 1.7 x 10 -4 d. 3.0 x 10 -8 e. 3.3 x 10 Phenol, C6H5OH, has a Ka = 1.0 x -10 10 . What is the pH of a 0.010 M solution of phenol? a. Between 3 and 7 b. 10 c. 2 d. Between 7 and 10 e. 7 10. You are given equimolar solutions of each of the following. Which has the lowest pH? a. NH4Cl b. NaCl c. K3PO4 d. Na2CO3 e. KNO3 11. When sodium nitrite is dissolved in water a. The solution is acidic because of hydrolysis of the sodium ion. b. The solution is neutral. c. The solution is basic because of hydrolysis of the sodium ion. 15. A buffer with a pH < 7. - 16. Determine the OH (aq) concentration in 1.0 M aniline (C6H5NH2) solution. (Kb for aniline is -10 4.0 x 10 .) -5 a. 2.0 x 10 M -10 b. 4.0 x 10 M -6 c. 3.0 x 10 M -7 d. 5.0 x 10 M 0 e. 1.0 x 10 M - 17. For IO3 + HC2H3O2 ⇌ HIO3 + C2H3O2 , the above equation has an equilibrium constant less than 1. What are the relative strengths of the acids and bases? Acids Bases a. HIO3 < HC2H3O2 IO3 < C2H3O2 b. HIO3 < HC2H3O2 IO3 > C2H3O2 c. HIO3 > HC2H3O2 IO3 > C2H3O2 d. HIO3 > HC2H3O2 IO3 < C2H3O2 e. HIO3 = HC2H3O2 IO3 = C2H3O2 + 2+ 18. For: ZnS(s) + 2H (aq) ⇌ Zn (aq) + H2S(aq), what is the equilibrium constant for the above reaction? The successive acid dissociation -8 constants for H2S are 9.5 x 10 (Ka1) -19 and 1 x 10 (Ka2). Ksp for ZnS -24 equals 1.6 x 10 . -24 -8 a. 1.6 x 10 / 9.5 x 10 -79 -24 b. 1 x 10 / 1.6 x 10 c. d. e. -27 -24 9.5 x 10 / 1.6 x 10 -8 -24 9.5 x 10 / 1.6 x 10 -24 -27 1.6 x 10 / 9.5 x 10 19. In the equilibrium below, which species behave as bases? + 2H2PO4 + H2O ⇌ H3O + HPO4 2i. HPO4 ii. H2PO4 iii. H2O a. i only b. i and ii c. ii and iii d. i and iii e. iii only + 20. H2C3H2O4 + 2H2O ⇌ 2H3O + 2C3H2O4 . As shown above, malonic acid is a diprotic acid. The successive equilibrium constants are -3 -6 1.5 x 10 (Ka1) and 2.0 x 10 (Ka2). What is the equilibrium constant for the above reaction? -14 a. 1.0 x 10 -6 b. 2.0 x 10 -12 c. 4.0 x 10 -9 d. 3.0 x 10 -3 e. 1.5 x 10 - 21. HC3H5O2(aq) + HCOO (aq) ⇌ HCOOH(aq) + C3H5O2 (aq). The equilibrium constant, K, for the -2 above equilibrium is 7.2 x 10 . This value implies which of the following? a. A solution with equimolar amounts of HC3H5O2(aq) and HCOO (aq) is neutral. b. C3H5O2 (aq) is a stronger base than HCOO (aq). c. HC3H5O2(aq) is a stronger acid than HCOOH(aq). d. HCOO (aq) is a stronger base than C3H5O2 (aq). e. The value of the equilibrium does not depend on the temperature.