Example Problem


El Problema: You got 90.0 ml of 0.500 M NH3 and your friend the madscientist decides to slowly add 0.9 M HCl to it, ruining perfectlygood ammonia. What is the pH reading on the CBL (your nerdyfriend is using a pH probe) when 0, 2, 10, 25, 40, 49, 50, 51, 60, 80, and 100mL of HCl have been added to solution? (this is not a trickquestion, although it may be hard to believe that a CBL couldactually work) Kb of NH3 = 1.8 E-5 (1.8 * 10-5)


Solution: As you look at this problem, terror may initially gripyour heart when you realize, while scanning your huge list ofstrong acids and bases, that NH3 is a weak base. Luckily, aftercarefully running down the list, you find that HCl lives among thewild strong acid clan (your head would probably explode if youhad to deal with a weak acid and a weak base at the same time). Anyway, remembering the wise words you read in the theorysection, it dawns on you that NH3 does not ionize completely. Translation: Get ready for equilibrium fun! Fortunately,things are still pretty damn simple, you lucky dawg (ominous notefor things to come again should be taken). To get organized, youprobably would want to start by writing out both of the reallycomplicated reactions that you will need during the problem. Hey, and if you're feeling really adventurous and/or you reallywant to impress Mr. McAfoos (not an easy task), you can evenwrite out the reaction between NH3 and HCl, seemingly a completewaste of space to an amateur, but trust us, it's the most easilyoverlooked part of this problem and nailing its involvementseparates los chicos de los hombres (spanish dictionarytranslation required).


  1. NH3 + H2O --> NH4+ + OH¯
    Kb = 1.8 E-5
  2. NH4+ + H2O --> NH3 + H3O+
    Ka = Kw/Kb = 1.0 E-14 / 1.8 E-5 = 5.6 E-10
  3. NH3 + HCl --> NH4+ + Cl¯


Sorting through the purposes of these equations, you'll need thefirst until all of the NH3 has been converted to NH4+, the secondafter this conversion has completely taken place, and the thirdbecause one of its products, NH4+, is involved in the equilibriumequations for the first two reactions. The far reaching impactof this is an additional column on the dreaded chart o' doom thataccounts for the NH4+ produced from this reaction (the amount of which is equalto the mols of NH3 that reacts with HCl--you'll see). Movingright along, you'll save yourself mucho tiempo if you remember yourgood friends Harry Henderson and Dick Hasselbalch and theirequation involving pOH:


pOH = pKb - log ( [B] / [HB+] )

Specific to this problem, the equation is written as:
pOH = pKb - log ( [NH3] / [NH4+] )


Remember though, this equation works only for added volumes ofacid between 10% and 90% of the equivalence point. Speakin' o yeolde equivalence point (insert heavy Irish accent), lets find it:


MBVB = MAVA ---> (0.5M)(90mL) = (0.9)(VA) ---> VA = 50 mL


So ye olde equivacating point is 50 mL. Using partial fractionsand Euler's method, we see that H/H can be used between 5 mL and45 mL. Well, partner, there's nothing to do now but to go towork. In my right hand, straight from the folks of Omaha,Nebraska, I hold the all powerful chart. We'll catch up with youlater. Be prepared for a pop quiz.


ml
(Acid added)

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
(Acid added)

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.


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