Unidentified Solutions Form

Unidentified Solutions Form - Solutions Part 1 1 Solution:...

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Solutions Part 1 1 Solution: We have not done a problem like this before. However, the solution is easy to figure out. Usually when we do a weak acid problem we are solving for the hydrogen ion concentration. Let’s write the solution for that problem and see if it helps us to figure out how to solve this one. K a ± x 2 0.058 " x This time we want the K a so it is clear that what we are looking for is the value of x .So ask yourself the question: “how can I find x ?” Oh yes, I see x can be found from the pH. x ± H ² ± 10 " pH Also the value of A " must be the same as H ² because every HA that dissociates makes one H ² and one A " . HA v H ² ² A " Knowing this I can find the K a . K a ± x 2 " x K a ± ± 10 " pH   2 " 10 " pH ± 9. 217 ± 10 " 7 2 Solution: All of these choices are salts and they dissociate to ions in water. Only the ion O 2 " is basic 3 Solution: H 2 SO 4 ² NaOH v HSO 4 " ² Na ² ² H 2 O, HSO 4 " is a weak acid so the resulting mixture will be acidic. All of the other acids produce neutral products (materials that aren’t acids or bases) resulting in pH ± 7.0 4 Solution: The titration reaction is NH 4 ² ² NaOH v NH 3 ² H 2 O The weak base NH 3 makes the equivalence point basic. 5 Solution: Here are the half cell reactions:
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Mn(s) v Mn 2 ± (aq) ± 2 e " oxidation Cu 2 ± ( aq ) ± 2 e " v Cu( s )r e d u c t i o n The oxidizing agent is the species that was reduced in doing the oxidizing. In other words, if one reactant is oxidized the other reactant must be the oxidizing agent. 6 Solution: Percent ionization is %ioniz. ² H ± H 2 CO 3 initially ± 100 0.2461 % ² H ± 0.071 M ± 100 ² 10 " 3.758 0.071M ± 100 ² 0.2461 So we can find the H ± concentration. . H ± ² ± 0.071 M /100 ² 1. 747 ± 10 " 4 The equilibrium constant is K a ² H ± HCO 3 " H 2 CO 3 ² ¡ 1. 747 ± 10 " 4 ¢¡ 1. 747 ± 10 " 4 ¢ c " ¡ 1. 747 ± 10 " 4 ¢ ² 4. 311 ± 10 " 7 7 Solution: 8 Solution:
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Blue arrows show the donation of electron pairs. a. Step 1: Two electrons in the OH bond of water transfer to the O atom leaving behind H ± . b. Step 2: Two electrons on the carbon in CN " are donated to the H ± to create HCN c. CN " is a Lewis base because it donates an electron pair .CN " is also a Lowry-Bronsted acid because it accepts a proton. H ± is a Lewis acid because it accepts an electron pair from CN " . d. Note that triple bonds often act as Lewis acids. But in this case the triple bond isnt involved in the reaction. The triple bond doesn’t change in this reaction. 9 Solution: This is the equilibrium reaction of the weak acid: HNO 2 ± aq   ± H 2 O ± e   v H 3 O ± ± aq   ± NO 2 " ± aq   ± adding HC e increases the H 3 O ± concentration. LeChateliers principle says the HNO 2 equilibrium shifts back towards reactants. This will decrease the percent of reaction (% ionization) of HNO 2 ± adding KNO 2 increases the NO 2 " concentration. LeChateliers principle says the HNO 2 equilibrium shifts back towards reactants. This will decrease the percent of
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This note was uploaded on 05/12/2011 for the course BIOL 1202 taught by Professor Gregg during the Spring '08 term at LSU.

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Unidentified Solutions Form - Solutions Part 1 1 Solution:...

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