Problem: 180 mL of 0.300 M HNO3 are slowly titrated with 0.900 M KOH. What is the pH of the solution when 0, 3, 10, 25, 40, 58, 60, 64,75, 90, and 100 mL of KOH have been added?
	Solution: Confucius says that we have a strong-strong titration.Why? Check the list, braniac. In case you were curious (and youdon't have to be), the following reactions occur in this problem.
	HNO3 + H2O --> NO3¯ + H30+ (this reaction occurs when mol HNO3 > mol KOH) KOH + H2O --> K+ + OH¯ (this reaction occurs when mol HNO3 < mol KOH) KOH + HNO3 --> KNO3 + H2O (this is the titration reaction) You don't have to worry about rxn. #3, because it doesn't change the [H+] or [OH¯] in the solution.
	Moving right along, you can start plugging away now. First,though, you probably want to get organized.
	Probably the best way to attack this bear is to set up a neato chart. I don't know, maybe a chart like this one (check it out now, an explanation follows):

ml (Base added)	Total V (mL and L)	MAVA (mols acid)	MBVB (mols base)
0	180 mL 0.180 L	(0.300)(0.180) = 0.0540	0
3	183 mL 0.183 L	0.0540 - 0.0027 = 0.0513	(0.9)(0.003) = 0.0027
10	190 mL 0.190 L	0.0540 - 0.009 = 0.0450	(0.9)(0.010) = 0.009
25	205 mL 0.205 L	0.0540 - 0.0225 = 0.0315	(0.9)(0.025) = 0.0225
40	220 mL 0.220 L	0.0540 - 0.0360 = 0.0180	(0.9)(0.040) = 0.0360
58	238 mL 0.238 L	0.0540 - 0. 0522 = 0.0018	(0.9)(0.058) = 0522
60	240 mL 00.240 L	0.0540 - 0. 0540 = 0	(0.9)(0.060) = 0.0540 (The equivalence point!!!)
64	244 mL 0.244 L	0	(0.9)(0.064) = 0.0576 0.0576 - 0.0540 = 0.0036
75	255 mL 0.255 L	0	(0.9)(0.075) = 0.0675 0.0675 - 0.0540 = 0.0135
90	270 mL 0.270 L	0	(0.9)(0.090) = 0.081 0.081 - 0.0540 = 0.027
100	280 mL 0.280 L	0	(0.9)(0.10) = 0.09 0.09 - 0.0540 = 0.036

That was a roller coaster ride of fun, wasn't it? In case you're lost (doubt it), a chart like this is key for staying organized. It keeps track of how many moles of acid you have as you titrate with base (you might care because strong acids/bases dissociate completely--so concentrated strong acid = [H30+] and concentrated strong base = [OH¯] ).

Anyway, if you take another look at the chart, up until the equivalence point (when mols acid = mols base) the MAVA column lists how many moles of H30+ are left in solution after the acid reacts with the base (rxns. 1 & 3 are happening). After the equivalence point, all the acid has been used up and the base is by its lonely self. Now the MBVB column tells you how much OH¯ is in solution (because rxns. 2 & 3 are happening). Got that? Now, you lucky stud, you can find pH values. Watch and learn.

ml (Base added)	[H3O+]	[OH¯]	pH = -log ([H+]) = 14 + log ([OH¯])
0	0.300	-----	-log(0.300) = 0.523
3	0.0153/0.183 = 0.280	-----	-log(0.280) = 0.553
10	0.045/0.190 = 0.237	-----	-log(0.237) = 0.625
25	0.0315/0.205 = 0.157	-----	-log(0.157) = 0.804
40	0.0180/0.220 = 0.0818	-----	-log(0.0818) = 1.09
58	0.0018/0.238 = 0.00756	-----	-log(0.00756) = 2.12
60	-----	-----	7
64	-----	0.0036/0.244 = 0.0148	14 + log(0.0148) = 12.2
75	-----	0.0135/0.255 = 0.016	14 + log(0.0529) = 12.7
90	-----	0.0270/0.270 = 0.100	14 + log(0.100) = 13.0
100	-----	0.0360/0.280 = 0.129	14 + log(0.129) = 13.1

So there you have it. That wasn't so bad, was it? All right, then, since you probably think you're pretty smart now, why don't you move on to Weak/Strong Titrations?