Week of 9/28: Chapters 5 and 6
Week of 9/28: Chapters 5 and 6 CHEM-1070-40
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This 3 page Class Notes was uploaded by Dkrefft on Monday October 12, 2015. The Class Notes belongs to CHEM-1070-40 at Tulane University taught by Lopreore, Courtney in Fall 2015. Since its upload, it has received 45 views. For similar materials see General Chemistry I in Chemistry at Tulane University.
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Date Created: 10/12/15
pter 5 Notes 928 Cha Ions in aqueous solution conduct electricity so they re called electrolytes Strong electrolytes substance completely ionizes in aqueous solution breaks apart NaCl MgCl2 o MgClzaq gt Mgzaq 2610161 Weak electrolytes only partially ionized in solution 0 CH3C00H aq lt gt Haq CHgCOO aq o acetic acid acetate 0 If it s a weak electrolyte draw a double arrow to show that it s reversible Bracket notation 0 Symbol concentration In 00050M HCI HClaq gt Haq Cl aq Haq 00050 M 01 aq 00050M MgCizaqi 00050M MgClzaqf 00M Mg2aqf 00050M Cl aqf 00100M Strong reactions What are Al3 and 5042 in 00165M ofAl2 5003 Write equation for disassociation o 14125043 gt 2Al3 35042 Stoichiometry O O O O O O 3 00165molAlzSO43 2molAl3 3 0 Al 1L 1molAlzSO43 00330 MAI aq 2 00165 molAlzSO43 3mol042 2 o 504 L 1molAlzSO43 00495 M 504 aq precipitation reactions cations and anions combine to form a solid that is insoluble Net ionic equations 0 AgN03aq Na aq gt Ag 3 NaN03 aqso AgI precipitant strong electrolytes Ag N 03A g1 are represented by their separate ions 0 Agaq4 93 aq Nai aq I aq gtAg1s Nai aq Negleaqa get rid of the spectator ions ions that appear on both sides 0 Net ionic equation Agaq I aq gt AgIs Solubility rules 0 Type them here Suppose Precipitant or no reaction A gN03 aq KBraq o Agaq N03 aq K aq Br aq o Agaq W Br aq gt AgBrs l Ki N93 o Agaq Br gt AgBrs Indicate whether a precipitant forms K2504aq FeBr3 aq 0 2K 002 Fe 331 gt FeSO42 KBrso no reaction acid base reactions 0 know the strong acids and bases p 161 0 weak acid has a weak tendency to produce H in a solution 0 Acids produce H in a solution proton donors 930 9999p Bases produce OH in solution proton recievers CH3C00H aq lt gt Haq CH3C00 aq weak HCl gt Haq Cl aq HCi H20 gt 1130 Cl NH3 H20 lt gt NH4 0H H20 1 ltgt Haq 0H aq H OH 10e 7 at 25 degrees Celsius H gt 10e 7 means it s acidic OOOOOOOO Neutralization Reaction 0 Acid base water salt 0 HCl aq NaOH aq gt NaCL aq H20 l 0 Strong They come apart completely in solution and have the ability to cancel H aq eieeiq9 Nai 0H39aq gt Naieeiqaar epeeq H20 1 o H OH gt H20 l to recognize acids v bases ionizable H atoms mean it s an acid HCZH302 or CH3600H Redox reductionoxidation 0 When substances gain 0 atoms oxidation 0 Lose 0 atoms Reduction 0 They always come in pairs When no oxygen is involved 0 The oxidation state of an element increases as electrons are lost 0 Reduction is if the oxidation state of an element decreases Mg02s 4H ZCl gt Mn2 H20 l Cl2g Mn is reduced and Cl is oxidized Zns Cu2aq gt Zn2aq Cu 3 oxidation Zns gt Zn2aq 2e reduction Cu2aq 2e gt Cus Make sure the electrons are equal and on opposite sides so that they cancel out Steps to balancing a redox reaction in an acidic solution Write half reactions For each reaction balance all atoms except for H and O by inspection Balance oxygen by adding H20 Balance the hydrogen by adding H Balance the charges by using electrons Equalize the number of electrons in the redox half reactions by multiplying one or both half reactions by the appropriate integer to make sure the electrons cancel out Add the half reactions and cancel the species common to both sides Check the number of atoms and that the charges balance Balance this in acetic solution 503239aq Mn04aq gt 504239aq Mn2aq Oxi 5503239aq H20 gt 502aq 2H 2e39 Red 2Mn04aq 8H Se gt Mn2aq 4H20 5503239aq 5H20 gt 55042aq 10H 10e 2Mn04aq 16H 10e gt 2Mn2aq 8H20 O O O O O PwNe o remember that the goal is to get the electrons to cancel 0 55032 441429 2Mn04 6H 0e gt 55042 1 9Hi 1 0e 2Mn2 3H2 O we reduced the H and the water and cancelled the electrons 0 55032 6H gt 55042 2Mn2 3H20 0 make sure the atoms and charges balance Fe2aq Mn04aq gt Fe3aq Mn2aq 0 oxidation 5Fe2aq gt Fe3aq e 0 reduction Mn04aq 8H Se gt Mn2aq 4H20 o 5Fe2aq Mn04aq 8H le gt 5Fe3aq Mn2aqe 4H20 o 5Fe2aq Mn04aq 8H gt 5Fe3aq Mn2aq 4H20 Balancing in a basic solution Use the method of balancing for the acidic but don t add yet Add the number of OH ions equal to the number of H ions to both sides of the overall equation On the side containing both H and OH combine them to form H20 Cancel common H20 molecules that appear on both sides In a basic solution you ll never have H in your answer Balance Mn04aq CN gt Mn02s OCN 0 red 2Mn04aq 4HJr 3e gt Mn02s 2H20 o oxi 3CN H20 gt OCN 2H 2e 0 aerate BCN 2H 2Mn04 aq 20H 6e gt 3 OCN 6Hi 6e 2Mn04 1H20 20H39reducing the H and water and cancelling electrons o SCN H20 2Mn0439aq gt 3 OCN 2Mn0439 1 H 29 20H 0 BCN H20 2Mn0439aq gt 3 OCN 2Mn0439 20H Acidic has H at the end Basic has H
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