C142A Ch 8 2011 lecture notes

# C142A Ch 8 2011 lecture notes - Class Average 64 100 = 1 90...

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Exam #2 Statistics Class Average: 64 100 = 1% 90 - 99 = 10% 80 - 89 = 16% 70 - 79 = 19% 60 - 69 = 15% 50 - 59 = 13% 40 - 49 = 11% 30 - 39 = 7% <30 = 7% Average grade of top 10% = 94 The average test score for students that did NOT get the clicker score they could have gotten from just being there and guessing (i.e., 60%) is 15 points below the average for the rest of the class. This 15 points translates into a GPA difference dropping from a bit above class average (2.9) to well below class average (1.6).

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Which salt will form a basic solution when placed into water? KBr NH4Cl Na C l Li N O 3 K3PO 4 Non e o f th a b v 0% 0% 0% 0% 0% 0% 1. KBr 2. NH 4 Cl 3. NaCl 4. LiNO 3 5. K 3 PO 4 6. None of the above
The Stepwise Dissociation of Phosphoric Acid Calculating ions concentrations in a certain molarity of phosphoric acid H 3 PO 4 (aq) + H 2 O (l) ! H 2 PO 4 - (aq) + H 3 O + (aq) K a = 7.5x10 -3 H 2 PO 4 - (aq) + H 2 O (l) ! HPO 4 2- (aq) + H 3 O + (aq) K a = 6.2x10 -8 HPO 4 2- (aq) + H 2 O (l) ! PO 4 3- (aq) + H 3 O + (aq) K a = 4.8x10 -13 Net: H 3 PO 4 (aq) + 3 H 2 O (l) ! PO 4 3- (aq) + 3 H 3 O + (aq) Order from biggest K a to smallest: Just like in the case with several different acids, each K a is so much smaller then the one above it that its contribution can be neglected wrt changing any concentrations of previous products!! Phosphoric acid is a weak acid, and normally only looses one proton in solution, but it will lose all three when titrated with a strong base.

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Example 7.15 (P268): More complex salts Predict whether an aqueous solution of each of the following salts will be acidic, basic, or neutral. a) NH 4 C 2 H 3 O 2 b) NH 4 CN c) Al 2 (SO 4 ) 3 Solution: get needed K from K a = K w /K b , or K b = K w /K a a) The ions are the ammonium and acetate ions, K a for NH 4 + is 5.6 x 10 -10 , and K b for C 2 H 3 O 2 - is 5.6 x 10 -10 . Since they are equal, the solution will be neutral and the pH close to 7. b) The solution will contain the ammonium and cyanide ions, the K a value for NH 4 + is 5.6 x 10 -10 , and Since K b for CN - is much larger than K a for NH 4 + , this solution will be basic. c) This solution contains the hydrated Aluminum ion and the sulfate ion. K a for Al(H 2 O) 6 3+ = 1.4 x 10 -5 , for sulfate, K b = 8.3 x 10 -13 ; therefore this solution will be acidic. K b (for CN - ) = = 1.6 x 10 -5 K w K a (for HCN)
Ch. 7 (problem #96 in the back of Ch. 7) QUESTION: Calculate the pH of a 0.416 M solution of C 5 H 5 NHF. Solution: This salt dissociates into C 5 H 5 NH + and F - : C 5 H 5 NH + is the conjugate acid of the weak base C 5 H 5 N (pyridine), which has K b = 1.7x10 -9 , so we know its K a = K w / K b F - is the conjugate base of the weak acid HF, which has K a,HF = 7.2x10 -4 , so we know its K b,F- = K a,HF / K w . How do we decide which one will win in determining the pH of the solution? Larger! How do we calculate [H + ]? Best way: Couple the 2 acid equilibria , with one in reverse, so that H

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## This note was uploaded on 01/18/2012 for the course CHEM 142A taught by Professor Campbell during the Fall '11 term at University of Washington.

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C142A Ch 8 2011 lecture notes - Class Average 64 100 = 1 90...

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