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ANSWERS TO ACID-BASE PROBLEM SET

ANSWERS TO ACID-BASE PROBLEM SET. 1. Given: 1.0 M C 6 H 5 NH 2 (aniline) ; K b = 4.0 x 10 -10 Find: pH of solution a. Identify and classify C 6 H 5 NH 2 : It is a base, a weak base. b. Write equations : CH 6 H 5 NH 2 + H 2 O W C 6 H 5 HN 3 + + OH-

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ANSWERS TO ACID-BASE PROBLEM SET

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  1. ANSWERS TO ACID-BASE PROBLEM SET

  2. 1. Given: 1.0 M C6H5NH2 (aniline) ; Kb = 4.0 x 10-10 Find: pH of solution a. Identify and classify C6H5NH2 : It is a base, a weak base. b. Write equations : CH6H5NH2 + H2O W C6H5HN3+ + OH- c. Find equilibrium concentration : 1.0 - x d. Set up Kb : Kb = [C6H5NH3+][OH-]/[C6H5NH2] = 4.0 x 10-10 x2 = 4.0 x 10-10/1.0 - x (ignore x in the denominator) e. x2 = 4.0 x 10-10/1.0 ; x = 2.0 x 10-5 M = [OH-] f. pOH = 4.70 ; pH = 14 - pOH = 9.30

  3. 2. Given : 0.10 M CH3NH3Cl ; pH = 5.86 Find : Kb for CH3NH2 a. Identify and classify CH3NH3Cl: Salt; therefore dissociates. CH3NH3Cl  CH3NH3+ + Cl- b. Which reacts further with water? CH3NH3+; it is a weak acid. c. Write equations: CH3NH3+W H+ + CH3NH2 d. Find equilibrium concentration : 0.10 - x But: pH = 5.86; so [H+] = 1.4 x 10-6 M = x = [CH3NH2] [CH3NH3+] = 0.10 - 1.4 x 10-6 = 0.10 M e. Ka = [H+][CH3NH2]/[CH3NH3+] = (1.4 x 10-6)2/ 0.10 = 2.0 x 10-11 f. Kb = Kw/Ka = 1.0 x 10-14/2.0 x 10-11 = 5.0 x 10-4

  4. 3. Given : 0.10 M HN3 ; pH = 3.0 Find: Ka for HN3 ? a. Identify and classify HN3 : weak acid ! b. Equation: HN3W H+ + N3- c. Equilibrium concentration : 0.10 - x But : pH = 3.0 [H+] = 1 x 10-3M = x = [N3-] [HN3] = 0.10 - 0.001 = 0.10 M d. Ka = [H+][N3-] = (1 x 10-3)2 = 1 x 10-5 [HN3] 0.10

  5. 4. Given : 0.10 M NaBO2 ; Ka = 5.8 x 10-10 for HBO2. Find: pH ? a. Identify and classify NaBO2 as a salt; therefore it dissociates : NaBO2 Na+ + BO2- b. Which reacts further with water? BO2- because it is basic (it accepts a proton from water) c. Equation: BO2- + H2O W HBO2 + OH- d. Equilibrium concentration 0.100-x e. Kb = [HBO2] x [OH-]/[BO2-] = x2/(0.100-x) = ?

  6. Value of Kb is not given and must be calculated : Kb = Kw/Ka = 1.0 x 10-14/5.8 x 10-10 = 1.7 x 10-5 x2 = 1.7 x 10-5/0.100-x (ignore x) x2/0.100 = 1.7 x 10-5 x2 = 1.7 x 10-6 x = 1.3 x 10-3 M = [OH-] pOH = 2.89 pH = 14 - 2.89 = 11.11

  7. 5. Given : 0.100 mole Na3AsO4 in 1.00 L solution; pH 11.5. Find : Ka of HAsO42- a. Start by identifying Na3AsO4as a salt. Salts dissociate completely (when soluble in water) Na3AsO4 3Na+ + AsO43- in H2O. b. Look to see what reacts further with H2O!! Na+ will not react with water!! AsO42- will react with water because it will accept a proton from water!! Write the equation: AsO43- + H2O W HAsO42- + OH- (Notice that AsO43- is basic because it is the anion of a weak acid and will react with water and accept a proton from water) c. Set up Kb : Kb = [HAsO42-][OH-]/[AsO43-]

  8. d. At Equilibrium : AsO43- + H2O W HAsO42- + OH- Initial : 0.100 M - 0 0 Change : -x - +x +x Equilibrium : 0.100-x - x x

  9. But : pH = 11.50 ; [H+] = 3.2 x 10-12 M ; [OH-] = Kw/[H+] [OH-] = 1.0 x 10-14 M/3.2 x 10-12 M = x = 3.1 x 10-3 M So : x = [OH-] = [HAsO42-] [AsO43-] = 0.100 - x = 0.097 M Kb = (3.1 x 10-3)2/0.097 = 9.9 x 10-5 f. Notice that : HAsO42- is the conjugate acid of AsO43- g. Ka x Kb = Kw Ka = Kw/Kb = 1.0 x 10-14/1.0 x 10-10 = 9.9 x 10-5

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