003050 x 100 060 dissociation Chem 162 2007 exam II review session 45 ET Given

# 003050 x 100 060 dissociation chem 162 2007 exam ii

This preview shows page 45 - 50 out of 84 pages.

% dissociation = (0.003/0.50) x 100 = 0.60% dissociation Chem 162-2007 exam II review session 45
ET: Given initial concentration and equilibrium concentration of weak acid, find Ka; define 1.00 x 10 -2 M as initial, and 2.77 pH as at equilibrium. 67. The pH of a 1.00 x 10 -2 M solution of cyanic acid (HOCN) is 2.77 at 25 o C. Calculate K a for HOCN from this result. HOCN + H 2 O H 3 O + + OCN - 0.0100M HCN pH 2.77 = 10 -2.77 = 1.70 x 10 -3 [H 3 O + ] HOCN + H 2 O H 3 O + + OCN - Initial 0.0100 Change Equilibrium 0.00170 HOCN + H 2 O H 3 O + + OCN - Initial 0.0100 0 0 Change -X +X +X Equilibrium 0.0100 - X 0.00170 +X HOCN + H 2 O H 3 O + + OCN - Initial 0.0100 0 0 Change -0.00170 +0.00170 +0.00170 Equilibrium 0.0083 0.00170 0.00170 K a = ([H 3 O + ][OCN - ])/[HOCN] K a = ((0.00170)(0.00170))/0.0083 = 3.48 x 10 -4 Chem 162-2007 exam II review session 46
ET: Given initial concentration of and equilibrium concentration of weak base, find Kb. 91 (mod.) Codeine (C 18 H 21 NO 3 ) is a derivative of morphine, a weak base, that is used as an analgesic, narcotic, or antitussive. It was once commonly used in cough syrups but is now available only by prescription because of its addictive properties. If the pH of a 1.7 x 10 -3 M solution of codeine is 9.59, calculate K b . RNH 2 + H 2 O RNH 3 + + OH - 1.7 x 10 -3 M codeine = RNH 2 pH = 9.59 pOH = 14.00 - 9.59 = 4.41 10 -pOH = [OH - ] 10 -4.41 = 3.89 x 10 -5 = [OH - ] RNH 2 + H 2 O RNH 3 + + OH - Initial 1.7 x 10 -3 M 0 0 Change Equilibrium +3.89 x 10 -5 RNH 2 + H 2 O RNH 3 + + OH - Initial 1.7 x 10 -3 M 0 0 Change -X +X +X Equilibrium 1.7x10 -3 -X +X +3.89 x 10 -5 RNH 2 + H 2 O RNH 3 + + OH - Initial 1.7 x 10 -3 M 0 0 Change -3.89 x 10 -5 +3.89 x 10 -5 +3.89 x 10 -5 Equilibrium 1.66 x 10 -3 +3.89 x 10 -5 +3.89 x 10 -5 K b = ([RNH 3 + ][OH - ])/[RNH 2 ] ((+3.89 x 10 -5 ) x (+3.89x10 -5 ))/(1.66 x 10 -3 ) = 9.11 x 10 -7 = K b Chem 162-2007 exam II review session 47
SKIP ET: Given equilibrium concentration and Ka, find initial concentration 24. A saturated solution of benzoic acid, C 6 H 5 COOH, has pH = 2.93. What amount of benzoic acid is dissolved* in a total of 1.00 L of this solution? K a = 6.3x10 -5 for C 6 H 5 COOH *What is the initial concentration of benzoic acid in solution? A . 0.023 mol B. 0.16 mol C. 0.30 mol D. 0.080 mol E. 0.016 mol pH 2.93 [H + ] = 10 -2.93 = 0.001175 K a = 6.3 x 10 -5 HB + H 2 O B - + H 3 O + HB(aq) + H 2 O(l) B - (aq) + H 3 O + (aq) Initial Y Change Equilibrium +0.001175 HB(aq) + H 2 O(l) B - (aq) + H 3 O + (aq) Initial Y 0 0 Change -X +X +X Equilibrium Y - X +X +0.001175 HB(aq) + H 2 O(l) B - (aq) + H 3 O + (aq) Initial Y 0 0 Change -0.001175 +0.001175 +0.001175 Equilibrium Y - 0.001175 +0.001175 +0.001175 K a = ([B - ][H 3 O + ])/[HB] = 6.3 x 10 -5 ([0.001175][0.001175])/[Y - 0.001175] = 6.3 x 10 -5 Y = 2.2 x 10 -2 Chem 162-2007 exam II review session 48
POLYPROTIC ACIDS Chem 162-2004 Exam II Acids and Bases - Chapter 14 Solutions of acids and bases, K a , K b , pH, pOH, % dissoc. ET: Define polyprotic acid H 2 CO 3 is a diprotic acid with K a1 = 4.3 x 10 -7 ; K a2 = 5.6 x 10 -11 Calculate the pH of 0.025 M H 2 CO 3 . A. 6.27 B . 3.98 C. 4.49 D. 5.02 E. 5.63 H 2 CO 3 + H 2 O HCO 3 - + H 3 O + K a1 = 4.3 x 10 -7 HCO 3 - + H 2 O CO 3 2- + H 3 O + K a2 = 5.6 x 10 -11 H 2 CO 3 H 2 O HCO 3 - H 3 O + Initial 0.025 0 0 Change -X +X +X Equilibrium 0.025-X +X +X ([HCO 3 - ][H 3 O + ])/[H 2 CO 3 ] = 4.3 x 10 -7 ([X][X])/[0.025-X] = 4.3 x 10 -7 X = 1.035 x 10 -4 = [H 3 O + ] = [HCO 3 - ] Considering only this first step, pH = -log(1.035 x 10 -4 ) = 3.99 Based on the extremely small K a2 , the HCO 3 - should dissociate very little to provide additional [H 3 O + ], so we can probably just use the first ICE table to get the [H 3 O + ].

#### You've reached the end of your free preview.

Want to read all 84 pages?

• Spring '08
• siegal
• pH