ml

Total V
(mL and L)

MBVB (mols base)

MAVA (mols acid)

[NH4+] = (mols acid)/ V
0 90
0.090
(0.090)(0.50) = 0.0450 --- ---
2 92
0.092
0.0450 - 0.0018 = 0.0432 (0.9)(0.002) = 0.0018 0.0018/0.092
10 100
0.100
0.0450 - 0.009 = 0.036 (0.9)(0.010) = 0.009 0.009/0.100
25 115
0.115
0.0450 - 0.0225 = 0.0225 (0.9)(0.025) = 0.0225 0.0225/0.115
40 130
0.130
0.0450 - 0.0324 = 0.0126 (0.9)(0.036) = 0.0324 0.0324/0.130
49 139
0.139
0.0450 - 0.0441 = 0.0009 (0.9)(0.049) = 0.0441 0.0441/0.139
50 140
0.140
0.0450 - 0.0450 = 0 (0.9)(0.05) = 0.0450 0.0450/0.140
51 141
0.141
0.0450 (0.9)(0.051) - 0.0450 = 0.0009 0.0009/0.141
60 150
0.150
0.0450 (0.9)(0.06) - 0.0450 = 0.009 0.009/0.150
80 170
0.170
0.450 (0.9)(0.08) - 0.0450 = 0.027 0.027/0.170
100 190
0.190
0.450 (0.9)(0.10) - 0.0450 = 0.045 0.045/0.190
 So you've unionized, er, organized yourselves. Now you can moveforth and prosper with some equilibrium fun. But it's really nobig deal. The key is to move as deliberately as time allows,making sure your math is tighter than Carmen Electra's dress for the Oscars. It's easy to lose a lot of points really quickly if you're at all careless. Anyway, in case youstart wondering if we are freaks or something (which we are) thefollowing math solutions were not done in our heads but ratherwith the solver function on a TI-83 (your best friend). All ofthese equilibrium problems originate from this derived equation: (the assumption is that the reaction will shift to the right,producing X M OH¯). NH3 + H2O --> NH4+ + OH¯ Y = initial [NH4+], Z = initial [NH3] Z - X Y+X X Kb = [NH4+][OH¯] / [NH3] --> Kb = [Y + X][X] / [Z - X]Here's another fun chart (rember pOH + pH = 14):

ml

Math Junk

pOH

pH
0 0= (X)(X) / (0.5 - X) - 1.8 E-5
X = [OH¯] = 0.0030
-log(0.0030) = 2.5 11.5
2 0 = (0.0018/0.092 + X)(X) / (0.0432/0.092 - X) - 1.8 E-5
X = 4.2 E-4
-log(4.2 E-4) = 3.4 10.6
10 pOH = pKb - log [ (0.036/0.10) / (0.009/0.10) ] 4.1 9.9

*** Max Buffer Point! pOH = pKb ***
25 pOH = pKb - log [ ( 0.0225/0.115) / (0.0225/0.115) ] = pKb 4.7 9.3
40 pOH = pKb - log [ (0.009/0.13) / (0.036/0.13) ] 5.3 8.7
49 0 = (0.0441/0.139 + X)(X) / (0.0009/0.139 - X) - 1.8 E-5
X = 3.7 E-7
-log(3.7 E-7) = 6.4 7.6
50 0 = (X)(X) / (0.045/0.14 - X) - 5.6 E-10
X = 1.3 E-5
--- -log(1.3 E-5) = 4.9

(Note: at 50 mL, we have reached the equivalence point. Thus, all of the NH3 has been titrated at this point and we must switch to reaction 2, using the same type of equilibrium equation. Now, however, we are solving for [H3O+].)
 NH4+ + H2O --> NH3 + H3O+ Y = initial [NH4+], Z = initial [NH3] Y - X Z + X X
51 0 = (X + 0.0009/0.141)(X) / (0.045/0.141 - X) - 5.6 E-10
X = 2.8 E-8 (but we already have (0.0009/0.141) M H3O+)
--- -log(0.0009/0.141) = 2.2

Okay, here's what's going on here. The concentration of H3O+ in solution is the sum of the amount produced by the reaction of NH4+ with H2O and the amount produced by the dissociation of the leftover HCl that hasn't reacted with the base. In comparison with the amount produced by the dissociation, X (from the reaction of NH4+) is so small at this point as to be inconsequential in the equation, and we're left with [H3O+] = [HCl]. Now wouldn't it be nice if we had the concentration of HCl handy? Oh, wait! I knew that first chart would come in handy! Just divide the MAVA value by the volume of the solution and you're money.
60 [ H3O+ ] = 0.009/0.15 = 0.060 --- -log(0.060) = 1.2
80 [ H3O+ ] = 0.027/0.170 = 0.159 --- -log(0.159) = 0.799
100 [ H3O+ ] = 0.045/0.190 = 0.236 --- -log(0.236) = 0.627
 So there you have it. Good job! You've made it this far. Now, get set for the nastiest type of chemistry problem you have ever encountered: Polyprotic Titrations.