To produce a conjugate base it must be acting as an acid Therefore it is a

# To produce a conjugate base it must be acting as an

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To produce a conjugate base, it must be acting as an acid. Therefore it is a proton donator forming CO 2 3 . 016 10.0 points The buildup of lactic acid in muscles causes pain during extreme physical exertion. The K a for lactic acid (C 2 H 5 OCOOH) is 8 . 4 × 10 4 . Calculate the pH of a 0 . 06 M solution of lactic acid. 1. 2.149 2. 11.851 3. 2.174 correct 4. 2.297 5. 2.043 6. 6.348 7. 3.533 8. 1.926 Explanation: K a = 8 . 4 × 10 4 [C 2 H 5 OCOOH] = 0 . 06 You MUST use the quadratic formula to solve this one. If you use the approximation, your answer will be more than 5% off. C 2 H 5 OCOOH + H 2 O Initial 0.06 Final 0 . 06 - x C 2 H 5 OCOO + H 3 O + Initial - - Final x x K a = bracketleftbig C 2 H 5 OCOO bracketrightbig bracketleftbig H 3 O + bracketrightbig [C 2 H 5 OCOOH] = x 2 0 . 06 - x = 8 . 4 × 10 4 x 2 + 8 . 4 × 10 4 x - 5 . 04 × 10 5 = 0 To solve the quadratic equation, x = - b ± b 2 - 4 ac 2 a Since b 2 - 4 ac = ( - 8 . 4 × 10 4 ) 2 - 4 (1) ( - 5 . 04 × 10 5 ) = 2 . 02 × 10 4 , then x = ( - 8 . 4 × 10 4 ) ± 2 . 02 × 10 4 2 (1) = (0 . 00669 , - 0 . 00753) , x = bracketleftbig H 3 O + bracketrightbig = 0 . 00669, so pH = - log bracketleftbig H 3 O + bracketrightbig = - log(0 . 00669) = 2 . 174 017 10.0 points
kamalska (mk23835) – HW05 – Quigley – (104001) 6 The pH of 0 . 1 M HClO 2 (chlorous acid) aqueous solution was measured to be 1 . 2. What is the value of p K a for chlorous acid? 1. 0.96 correct 2. 2.57 3. 1.20 4. 0.11 5. 3.91 6. 1.40 Explanation: M = 0 . 1 M pH = 1 . 2 Analyzing the reaction with molarities, HClO 2 + H 2 O H 3 O + + ClO 2 0 . 1 - 0 0 - x - x x 0 . 1 - x - x x [H 3 O + ] = [ClO 2 ] = 10 pH = 10 1 . 2 = 0 . 0630957 mol / L The K a is K a = [H 3 O + ][ClO 2 ] [HClO 2 ] = (0 . 0630957) 2 0 . 1 - 0 . 0630957 = 0 . 107876 and the p K a is p K a = - log(0 . 107876) = 0 . 967077 . 018 10.0 points What is the pH of a 0 . 21 M solution of potas- sium generate (KR-COO)? K a for the generic acid (R-COOH) is 2 . 7 × 10 8 . 1. 10.445 correct 2. 3.425 3. 10.295 4. 6.431 5. 10.575 6. 7.569 7. 10.815 8. 7.000 9. 10.205 10. 3.555 Explanation: M KR COO = 0 . 21 M K a = 2 . 7 × 10 8 It’s a salt of a weak generic acid (KA). Get it? Generic acid makes generic ions. Ha! This means you need a K b for the weak base A . Use K b = K w K a and you’ll get the K b = 3 . 7037 × 10 7 . You CAN use the ap- proximation for the equilibrium which means that [OH ] = radicalbig K b · C A - = 0 . 000278887 M pH = 14 - pOH = 14 + log(0 . 000278887) = 10 . 4454 019 10.0 points Adding KBr to pure water would result in what kind of solution? 1. acidic 2. neutral correct 3. basic Explanation: 020 10.0 points Consider a solution that is 0.10 M in a weak
kamalska (mk23835) – HW05 – Quigley – (104001) 7 triprotic acid which is represented by the gen- eral formula H 3 A with the following ionization constants: K 1 = 1 . 0 × 10 3 , K 2 = 1 . 0 × 10 8 , K 3 = 1 . 0 × 10 12 . Which species is present in the lowest concentration in 0.10 M H 3 A solution? 1. HA 2 2. A 3 correct 3. H 3 A 4. H 3 O + 5. OH Explanation: 021 10.0 points What is the pH of 0 . 3 M ammonia if the K a for its conjugate acid, ammonium chloride, is 5 . 5 × 10 10 . 1. 11.4 correct 2. 4.3 3. 0.002 4. 8.7 5. 2.6 Explanation: [NH 3 ] = 0 . 3 M K a = 5 . 5 × 10 10 K a · K b = K w K b = K w K a = 1 × 10 14 5 . 5 × 10 10 = 1 . 81818 × 10 5 Analyzing the reaction using molarities, NH 3 + H 2 O NH + 4 + OH ini 0 . 3 0 0 0 Δ - x 0 + x + x fin 0 . 3 - x 0 x x K b = [NH + 4 ][OH ] [NH 3 ] = x · x 0 . 3 - x = 1 . 81818 × 10 5 Assuming 0 . 3 - x 0 . 3, x 2 0 . 3 = 1 . 81818 × 10 5 x = radicalBig (0 . 3)(1 . 81818 × 10 5 ) = 0 . 0023355 = [OH ] K w = [H + ][OH ] [H + ] = K w [OH ] = 1 × 10 14 0 . 0023355 = 4 . 28174 × 10 12 Thus pH = - log([H] + ) = - log(4 . 28174 × 10 12 ) = 11 . 3684 022 10.0 points What is [H 3 O + ] in a solution made by dissolving 0.100 mole of sodium acetate (NaCH 3

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• Fall '08
• QUIGLEY
• pH, greater extent