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Make up 2-solutions - Version 074 Make up 2 laude(51635...

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Version 074 – Make up 2 – laude – (51635) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 6.0 points A 40 mL sample of 0 . 25 M NaCHOO is titrated with 0 . 2 M HCl. What is the pH of the solution after 140 mL of HCl has been added? 1. -1.0 2. 3.0 3. 0 4. 1.0 correct 5. 2.0 6. 2.1 Explanation: After the CHOOH has been neutralized, 0 . 018 mol of HCl remain in a solution with a total volume of 180 mL . [H + ] = 0 . 018 mol 180 mL = 0 . 1 M pH = 1 002 6.0 points For a weak diprotic acid system (H 2 A) which of the following statements are true? I) H 2 A and A 2 - form a bu ff er. II) K a1 > K a2 III) [A 2 - ] > [HA - ] in acidic solution IV) HA - is amphiprotic. 1. II only 2. II and IV only correct 3. I and III only 4. I and IV only Explanation: H 2 A and HA - or HA - and A 2 - will form bu ff ers, but H 2 A and A 2 - will not form a bu ff er. For polyprotic acids, protons become harder to remove (dissociate) as the ion be- comes more charged. Therefore K a1 > K a2 for this system. In acidic solution (access pro- tons) all ions will be protonated. HA - is amphiprotic because it can accept a proton or it can donate a proton. 003 6.0 points Which of the following equations does not require any assumptions in order to provide an accurate solution? 1. all require assumptions correct 2. [OH - ] = ( K b · C b ) 1 / 2 3. [H + ] = ( K a · C a ) 1 / 2 4. [H + ] = C a 5. [OH - ] = K b · C b C a Explanation: All of these equation assume, at the very least, that autoprotolysis is negligible. 004 6.0 points Which K a value(s) would you use when calcu- lating the pH of a system involving NaH 2 PO 4 ? 1. K a 3 2. K a 2 , K a 3 3. K a 1 , K a 2 correct 4. K a 2 5. K a 1 Explanation: If the only form present is H 2 A - , then it is the amphiprotic case, and you use K a 1 and K a 2 .
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Version 074 – Make up 2 – laude – (51635) 2 005 6.0 points If a solution of HOCl is titrated by a strong base such as NaOH, what is the pH at the half equivalence point? The ionization constant is 3 . 5 × 10 - 8 for HOCl. 1. pH = 7.5 correct 2. pH = 4.5 3. pH = 3.5 4. pH = 5.5 5. pH = 6.5 Explanation: K a = 3 . 5 × 10 - 8 At half the equivalence point, this solution is a bu ff er in which [HOCl] = [OCl - ]. pH = p K a + log [OCl - ] [HOCl] = p K a = - log(3 . 5 × 10 - 8 ) = 7 . 45593 006 6.0 points Estimate the pH of 0.10 M Na 2 HPO 4 (aq) given p K a1 = 2 . 12, p K a2 = 7 . 21, and p K a3 = 12 . 68 for phosphoric acid.
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