practice final

practice final - CHEM 6B PRACTICE FINAL EXAM(Prof Crowell...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chem 6B Practice Final 1 A Prof. Crowell CHEM 6B PRACTICE FINAL EXAM (Prof. Crowell) May 28th, 2010 FILL IN YOUR STUDENT ID NUMBER , NAME , and TEST FORM = A 1. For the following four salts, determine whether water solutions of each salt would be acidic, basic, or neutral? The salts are: 1. KNO 2 , 2. NH 4 Cl , 3. CsI , 4. CH 3 COONa a) 1. acidic, 2. basic, 3. neutral, 4. acidic b) 1. acidic, 2. basic, 3. basic, 4. acidic c) 1. neutral, 2. acidic, 3. basic, 4. neutral d) 1. neutral, 2. acidic, 3. neutral, 4. basic *e) 1. basic, 2. acidic, 3. neutral, 4. basic We can view these salts as a reaction between an acid and a base, and determine which is stronger. Alternatively, we can view the salt as composed of a cation and an anion, and discuss the relative acidic, basic, or neutral properties of each. I will give both views below. 222 2 2 22 2 2 KNO : ( ) ( ) ( ) ( ), ( ) ; ( ) overall basic or: (aq) + ( ) ( ) ( ). ( ) strong base + ( ) = weak a KNO s H O l K aq NO aq K aq neutral NO aq basic KOH HNO aq KNO s H O l KOH aq HNO aq     cid strong wins so overall basic 44 2 4 4 34 2 3 NH Cl: ( ) ( ) ( ) ( ( ) ( ) overall acidic or: (aq) + ( ) ( ) ( ). ()w e a k b a s e + () = NH Cl s H O l NH aq Cl aq NH aq weak acid Cl aq neutral NH HCl aq NH Cl s H O l NH aq HCl aq  strong acid strong wins so overall acidic 2 2 CsI: ( ) ( ) ( ) ( ; o v e r a l l n e u t r a l or: (aq) + ( ) ( ) ( ). ( ) strong base + ( ) = strong ac CsI s H O l Cs aq I aq Cs aq neutral I aq neutral CsOH HI aq CsI s H O l CsOH aq HI aq id overall neutral 33 2 3 3 2 3 CH COONa: ( ) ( ) ( ) ( ; or: (aq) + ( ) ( ) ( ). e a k CH COONa s H O l Na aq CH COO aq Na aq neutral CH COO aq weak base overall basic CH COOH NaOH aq CH COONa s H O l CH COOH aq acid + ( ) = strong base strong wins so overall basic NaOH aq 2. Consider the titration of 100.0 mL of 0.500 M NH 3 (K b = 1.8 x 10 -5 ) with 0.500 M HCl. At the stoichiometric point of this titration, the [H + ] is: a) 1.0 x 10 -7 M b) 1.8 x 10 -5 M *c) 1.2 x 10 -5 M d) 5.6 x 10 -10 M e) none of these. We first look at the stoichiometric reaction between the weak base 3 NH and the strong acid HCl . Since the concentrations are equal, we know that the volume added will need to be equal to have the same number of moles of each in order to reach the endpoint: 2 4 Stoich Rxn: ( ) ( ) ( ) ( 1/ 100ml 0.5M=50mmol 100 0.5M=50 0 -50 b NH aq H O aq H O l NH aq K K init change   -50 +50 0 0 50 mmol final 3 Alternatively, we could determine the volume needed to reach the endpoint using: [] 0.5 (100) 100 0 . 5 u Tu T NH M VV m l HCl M 
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Chem 6B Practice Final 2 A Prof. Crowell So we now have just the conjugate acid 4 NH in solution, and we need to consider its equilibrium reaction: 42 3 3 Equil Rxn: ( ) ( ) ( ) ( ), 50mmol/200ml=0.25M 0 0 -x a NH aq H O l NH aq H O aq K K init change  22 33 4 +x +x 0.25 x x [ ( )][ ( )] [ ( )] (0.25 ) (0 a final M x NH aq H O aq xx K NH aq x  14 5 3 5 55 .25) (0.25) (0.25)(1.0 10 ) (0.25) 1.18 10 [ ( )] 1.8 10 [ ( )] 1.2 10 log( ) log(1.18 10 ) 4.93 w a b K x xK x H O a q Kx Ha q x p H x x    3. For ammonia, NH 3 , K b is 1.8 x 10 - 5 . To make a buffered solution of pH = 10.0 at 298K, the ratio of NH 4 Cl to NH 3 must be: a) 1.8 : 1 b) 1 : 1.8 *c) 0.18 : 1 d) 1 : 0.18 e) 1.8 x 10 5 : 1 In this problem we are asked the ratio of NH 4 Cl to NH 3
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 11

practice final - CHEM 6B PRACTICE FINAL EXAM(Prof Crowell...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online