CH302 Spring 2011 Practice Exam 2

CH302 Spring 2011 Practice Exam 2 - CH302 Spring 2011...

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CH302 Spring 201 1 Practice Exam 2 - calculator free 1. What will be the pH in a titration in which 50.0 mL of 0.120 M HNO 3 (aq) is added to 25.0 mL of 0.240 M KOH(aq)? A. 0.74 B. 7 C. 13.33 D. 13.38 Answer: B 2. What are the acid dissociation constants that would be used to solve the buffer equation for a solution made with NaLiHA and Na2LiA that derive from a triprotic weak acid? A. first B. second C. third D. first and second E. second and third Answer: C 3. What is the pH of a 1 M Lithium maleiate solution? The pKas for this salt of a diprotic acid are 2 and 6. Answer Answer: Removing the spectator ions for this diprotic acid, what remains is a base, A = , which because the pK values are far enough apart, can be treated as a simple weak base calculation. The rest of the problem can be done without calculator. If the pK a2 is 6 then the pK b1 is 8. The OH - concentration from the simple weak base calculation is then 10 -4 and the pH is 10, which makes sense for a weak base calculation. 4. What can you say about the pH of 10 -9 M solutions of HCl, citric acid, acetic acid, ammonia and lithium hydroxide? Answer Answer: The pH of a solution depends on how many protons are in solution. When acids and bases are added to water they alter the neutral pH due to the number of protons and hydroxides being equal. However the number of protons or hydroxides added, whatever the original source (strong acid or base, weak, polyprotic. ...) must be significantly greater than the number of protons coming from the water if the pH of solution is going to change appreciably. A 10 -9 M solution of any compound, no matter how strong an acid or base it is, will not alter the solution pH which will remain right around pH 7 (depending on temperature of course.) 5. Which of the following 1 M weak base solutions will have an error of about 1% in calculating the pH if the Kw is not considered? A. A weak base with pKb = 10 B. A weak base with pKb = 14 C. A weak base with pKb = 12 D. A weak base with pKb = 2 Answer: A When 40.0 mL of 1 ! 10 " 4 M Ba(OH) 2 are mixed with 60.0 mL of 2 ! 10 " 5 M La 2 (SO 4 ) 3 ( K sp for La(OH) 3 is 1.0 ! 10 " 19 and K sp for BaSO 4 is 1.1 ! 10 " 10 )
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6. A. a precipitate of BaSO 4 is obtained. B. a precipitate of La(OH) 3 is obtained. C. two precipitates are obtained. D. no precipitate is obtained. Answer: B 7. In performing a buffer neutralization calculation which has a weak acid, a weak base and a strong acid or base present initially, if you follow through all the steps in the neutralization if can take a lot of work to find the equilibrium pH. Why is it that this problem can often be done quite efficiently in seconds without a calculator? Answer
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CH302 Spring 2011 Practice Exam 2 - CH302 Spring 2011...

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