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%%
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CONJUGATE ACID BASE PAIRS -&K"%*$854*I4*23"*6%"."8,"*$%*/I."8,"*$<*/*6%$2$8*GDF):*
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HI%(=J&'% 12*G%55K% g%
HNO2(aq) + H2O(!) ! NO2!(aq) + H3O+(aq)
acid base conjugate conjugate base acid
0*G=8*%SW%
NH3(aq) + H2O(!) ! NH4+(aq) + OH! (aq)
base acid conjugate conjugate acid base
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Kw = [H+][OH!] = 1.0 " 10!14 at 25°C
pH = ! log10[H+] low pH = acidic high pH = basic pOH= ! log10[OH!] pKw = ! log10 Kw = 14 pH + pOH = ! log10[H+] + ! log10[OH!] = 14
HI%(=J&'% 12*G%55K% 5b%
What is an electrolyte? A substance that conducts electricity when dissolved in solution
MJ Bojan Chem 112
Which Bulbs Light Up? Distilled H2O
Tap H2O
NaCl(aq)
0.1M HCl(aq)
0.1M CH3COOH
sugar(aq)
CH3OH
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Bulb Wattage 2.5 7.5 25 40
19
STRONG ACIDS
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HI%(=J&'% 12*G%55K% Kh%
When dissolved in water, weak acids only partially dissociate to form H+ and the corresponding BrJnsted conjugate base.
WEAK ACIDS
Weak acids are weak electrolytes: PARTIAL dissociation into ions +Y&G3.*)L%%% %%
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General Approach to Equilibrium Constant Problems
1. Write the balanced reaction.
2. Write the expression for Keq.
3. Set up a data table: (may need algebraic unknowns) I initial conditions C changes in concentrations E equilibrium concentrations
4. Substitute equilibrium concentrations into the expression for Keq and solve.
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HI%(=J&'% 12*G%55K% K:%
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CHA = 0.10 M acetic acid: CH3COOH What is the pH? What is the % dissociation? %%%Compare this to the [H+] of 0.10 M HI. %%%%HI%(=J&'% 12*G%55K% KF%
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Examples: H2CO3(aq) ⇆ H+ (aq) + HCO3
!(aq) Ka1 = 4.3 ! 10"7 HCO3
!(aq) ⇆ H+(aq) + CO32!(aq) Ka2 = 5.6 ! 10"11
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q&%V=0%)=G*%3=.>30=B"%&"#$)%+,&-* \$%#75/* X/(* X/;* X/>*1&0/='#"% SK1R@% 4.3 " 10!7 5.6 " 10!11 AAA%32=)32=0#"% S@jR:% 7.5 " 10!3 6.2 " 10!8 4.2 " 10!13 D-.V-0=-)% SKDR@% 1.7 " 10!2 6.4 " 10!8 AAA%D-.V-0#"% SKDR:% .&0E*% 1.2 " 10!2 AAA%RY&.#"% SK1KR:% 5.9 " 10!2 6.4 " 10!5 AAA%!)"=0/#"% SK16S6R6% 8.0 " 10!3 2.6 " 10!12 AAA%1#40#"% S@16SFRC% 7.4 " 10!4 1.7 " 10!5 4.0 " 10!7
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H3PO4 (aq) ⇆ H+(aq) + H2 PO4!(aq) Ka1 = 7.5 ! 10"3
H2PO4!(aq) ⇆ H+(aq) + HPO4
2!(aq) Ka2 = 6.2 ! 10"8
HPO42!(aq) ⇆ H+(aq) + PO4
3!(aq) Ka3 = 4.2 ! 10"13
WEbaWhE@_L*+L_]V*
HI%(=J&'% 12*G%55K% @h%
P#)40#/-B='%"-08*%V=0%32=)32=0#"%&"#$%NaOH is added to a solution of H3PO4 until the pH
reaches 12. Which species are present in the solution?
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HI%(=J&'% 12*G%55K% @K%
D40='E%(&)*)%^%+Y&G3.*)%• What is the pH of a 0.028 M solution of NaOH?
• What is the [H+] concentration of a solution of Ca(OH)2 with pH 11.68?
HI%(=J&'% 12*G%55K% @@%
M+!q%(!D+D%&0*%9*&N%*.*"40=.>4*)%M2*'%$#))=.8*$%#'%9&4*0c%9*&N%/&)*)%='.>%3&0B&..>%0*&"4%4=%
V=0G%RST%&'$%42*%"=00*)3='$#'E%(0[')4*$%&"#$;%%%%]2*>%7*&"4%7+x+7DO(_`%%9#42%9&4*0%4=%V=0G%RST;%% This is an example of a hydrolysis reaction.
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Hydrolysis: when a cation or anion reacts with H2O to form H3O+(aq) or OH"(aq)
M+!q%(!D+D%Weak bases can be neutral molecules Example: NH3T*amines NH3(aq) + H2O(!) ⇆ NH4
+(aq) + OH!(aq)
Or Anions (any ion derived from a weak acid) Example: F", NO2
", CH3COO"
F!(aq) + H2O(!) ⇆ HF(aq) + OH!(aq) %%HI%(=J&'% 12*G%55K% @F%
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HI%(=J&'% 12*G%55K% @C%
]2#'E)%4=%q\RM%Acid Dissociation HA + H2O(l) ⇆ H3O+(aq) + A!(aq)
Equilibrium constant = Ka
3q&%r%T.=E%q&%
Ka = [H+][A!][HA]
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+,-#.#/0#-G%"=')4&'4%r%q/%3q/%r%T.=E%q/%
Kb = [HB+][OH!][B]
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