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Material Type: Notes; Professor: Laude; Class: PRINCIPLES OF CHEMISTRY II; Subject: Chemistry; University: University of Texas - Austin; Term: Unknown 1989;
Typology: Study notes
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Lecture 1 4 : Wrapping up pH calculations with a surefire stepwise process for working problems If we make the right assumptions about K values (far apart) and concentrations (large) then we can apply approximations so that there are only six equations needed to solve acid base problems. Even better, there are only five variables to consider in using these equations: Ka, Kb, [H+], [OH-], C acid, C base
Strong acid [H+] = Ca Strong base [OH-] = Cb Weak acid [H+^ ] = (Ka Ca ) 1/ Weak base [OH -^ ] = (Kb Cb ) 1/ Acid buffer [H+^ ] = Ka Ca /Cb Basic buffer [OH -^ ] = Kb Cb /Ca
So there isn't a lot of complexity at the bottom of this. The hard part is figuring out which equation to use and what each of the variables is. To accomplish this task, we use the following procedure: 1) strip away all the extraneous information (spectator ions), 2) identify strong acids and bases, 3) identify weak acids and bases, 4) determine if you should neutralize, 5) perform neutralization calculation, 6) decide whether to work the problem as an acid or a base. Once these steps are done, the problem is greatly simplified to the point that you can use the table above to work a calculation. The back of this page shows every possible type of starting conditions and how they reduce to one of the problems above.
Getting rid of spectator ions. Always eliminate the ions that do nothing: all alkali metals and alkali earths (Na+^ , K +^ , Ca++^ ) and all conjugate bases of strong acids (Cl-^ , NO 3 -^ , ClO 4 - , I -^ , Br -^ ). Thus NH 4 Cl is NH 4 +^ NaOH is just OH-^ KCOOH is just COOH-
Identify strong acids and bases. Strong acids are HCl, HNO 3 , H 2 SO 4 , HClO 4 , HBr, Hl. Strong bases are NaOH, KOH, Mg(OH) 2 , Ba(OH) 2 and other alkali metal or earth hydroxides. Notice what happens when you get rid of spectator ions for strong acids and bases.
HCl become H +^ HNO 3 becomes H +^ NaOH becomes OH-^ Mg(OH) 2 becomes 2OH -
In other words, all strong acids are H +^. All strong bases are OH-^.
And how do you represent a weak acid? HA (instead of HCH 3 CH 2 COO which only serves to confuse you). And how do you represent a weak base: A-^ (instead of NaCH 3 CH 2 COO which only serves to confuse you).
By the time you are through with step 3, you will have identified the presence of all acids and bases. You should have only six possible symbols representing them: H+^ or OH-^ for strong acids and bases HA or BH+^ for weak acids B or A-^ for weak bases
Any other terminology is a waste of time on a test without much time.
for example:
Starting Materials Materials after neutralization
Equation to use Sample problem Calculate pH
Strong acid alone H +^ pH = -log [H+] 0.2 M HNO 3
Strong acid and weak acid H+^ and HA (ignore HA)
pH =-log [H +^ ] 0.2 M HNO 3 and 0.4 M acetic acid Strong acid and weak base H+^ and HA (ignore HA)
pH = -log [H +^ ] 0.2 M HNO 3 and 0.1 M sodium acetate
Strong base alone OH -^ alone pOH = -log [OH-^ ] 0.1 M Ba(OH) 2
Strong base and weak base
OH -^ and A- (ignore A-)
pOH = -log [OH -^ ] 0.1 M Ba (OH) 2 and 0.1M sodium acetate Strong base and weak acid OH-^ and A- (ignore A -^ )
pOH =-log [OH -^ ] 0.4 M Ba(OH) 2 and 0.1M ammonium chloride
Weak acid HA or BH+^ [H +] = (K (^) aC (^) a)1/2^ 0.3 M acetic acid
Equivalent strong acid and weak base
ammonia
Weak base A-^ or B [OH - ] = (K (^) bC (^) b) 1/2^ 0.2 M NH (^3)
Equivalent strong base and weak acid
A -^ or B [OH -^ ] = (K (^) bCb)1/2^ 0.1M NaOH and 0.1M acetic acid
Weak acid and conjugate weak base
HA and A [H +^ ] = K (^) a Ca /Cb 0.2 M acetic acid and 0.1M sodium acetate Strong acid and weak base HA and A [H+^ ] =K (^) a Ca /Cb 0.2 M HCl and 0.4 M sodium acetate
Weak base and conjugate weak acid
ammonium chloride
ammonium chloride