Posted by Jessica on Saturday, November 26, 2011 at 8:14pm.
You need to do ICE charts for both of these. If we call aspirin, HA, it makes it easier to write.
millimoles HA = 20 x 0.07222 = 1.44
mmoles NaOH = 10 x 0.2 = 2.0
...........HA + NaOH ==> NaA + H2O
initial...1.44..................
add.............2.0............
change...-1.44.-1.44......1.44...1.44
equil.....0....0.556.......1.44..1.44
This tells you that the solution is one of NaOH with its salt but the pH will be determined by the strong base, NaOH.
M NaOH = mmoles/mL = 0.556/30 = ?
Convert to pH.
For b part, RNH2.
mmoles RNH2 = 40 x 0.10 = 4.0
mmoles HCl @0 mL = 0
mmoles HCl @ 20 = 20 x 0.1 = 2
mmoles HCl @ 40 = 40 x 0.1 = 4
etc.
.........RNH2 + HCl ==> RNH3^+ + Cl^-
initial...0.......................
added...........0
added..........2.0
added..........4.0
etc
change.....................
At zero mL, you have just the free weak base.
.........RNH2 + HOH ==> RNH3^+ + OH^-
initial...0.1.............0......0
change....-x.............x.......x
equil....0.1-x...........x.......x
Kb = (RNH3^+)(OH^-)/(RNH2)
Substitute from the ICE chart and solve for x = OH^-, then convert to pH.
For all points between zero and the equivalence point you have a buffered solution consisting of the salt, RNH3^+ and the free base, RNH2.
Use the Henderson-Hasselbalch equation for all of these.
The pH at the equivalence point is determined by the hydrolysis of the salt. Write the equation and set up an ICE chart for that.
Everything after the equivalence point is determined by excess HCl and pH = -log(H^+).
Post your work if you need further assistance.
For problem 1, if it were changed to 20 mL of the .07222M aspirin solution and 5mL of .2M NaOH, how would solving it differ? I tried
...........HA + NaOH ==> NaA + H2O
initial..0014444..................
add..............001............
change...-.001.-.001.....+.001...+.001
equil...0004444 0......+.001..+.001
And I got the answer as pH=1.75, when the answer is pH=4.88.
Furthermore for the original problem 1, I got 12.27 when the answer is 12.31. Is the .04 difference in pH negligible?
Sorry, one more question.
For
.........RNH2 + HCl ==> RNH3^+ + Cl^-
initial...0.......................
added...........0
added..........2.0
added..........4.0
etc
change.....................
the initial RNH2 is 0? Shouldn't it be the initial moles in the problem (.004moles of RNH2) and the initial HCl would be the moles depending on how much you added (like .002moles)
I have two questions. The first is are you using aspirin as the acid or as the sodium salt of the acid? Second question, if acid, then what are you using for Ka?
If you using it as the Na salt, I've probably worked #1 wrong.
I don't know why I wrote zero for initial RNH2? I guess all of those other zeros and I had zero on my mind. The initial concn of RNH2 is 0.1M isn't it? Yes, it's 0.004 moles and that divided by 40 mL or 0.004/0.040 = 0.1M (but the 0.1 M is given in the problem, too).
Don't get confused with the moles in the ICE charts and the Henderson-Hasselbalch equation. I use millimoles there and almost NEVER (almost never really means almost never) convert to concn in moles/L BECAUSE in buffered solutions the volumes ALWAYS cancel. Therefore, I convert everything to moles or millimoles and use them directly without converting back to concn. Technically, its the concn that must go there and not moles or millimoles. To be honest about it, I always docked my students for doing that, now I'm doing it. :-). It's a very convenient shortcut; however, if we use it we must remember that when a RATIO is not involved, one MUST convert to concn first before substituting into Ka or something like that.
Regarding your first post, I believe I used aspirin as the weak acid. In a previous problem which was related to these problems, it stated "aspirin is a weak acid with a Ka of 3.0x10^-5"
For part b of problem 2, I used
..RNH2 + HCl--> RNH3 + Cl
I.+.001..+.002...0......0
C.-x......-x.....+x....+x
E+.001-x..+.002-x..+x...+x
Is this correct? My issue with these problems is that I don't know when to use the variables (like +x and -x for the C in ICE) or whether to use numbers like .001-.001 and .002-.001 for the C row in ICE.
I tried it the way I wrote above but I keep getting 9.55 when the answer is 10.81
Thank you!
Oops, sorry I meant .004 under initial moles of RNH2 not .001. Regardless I get an answer of 9.84 instead of 10.81
It's getting tough to follow the thread but here is how I do the repost of 5mL of 0.2 M NaOH.
Without drawing the ICE chart, that gives us 1.44 mmoles acid to start and we add 5 x 0.2 = 1 mmole base.
That forms 1 mmole of the conjugate base (A^-) and leaves 1.44-1.00 = 0.44 mmole of the original acid.
pH = pKa + log(base)/(acid)
pH = 4.522 + log(1/0.44)
pH = 4.522 + log (2.27)
pH = 4.522 + 0.3565 = 4.8785 which I would round to 4.88.
(Note: Technically, as a follow up to my previous post, (base)/(acid) means concn base/concn acid and that is mmoles/mL (or moles/L). Therefore I SHOULD have said (base) = 1.0 mmole/25 mL = ? and (acid) = (1.44 mmoles/25 mL) = ? BUT since we use
log (base)/(acid) that becomes (1.0/25) and acid is (0.44/25) and the 25 cancels. It will ALWAYS cancel because the volume of the two are always the same solution so (mmoles/V)/(mmoles/V) is just mmoles/mmoles. :-).
..........RNH2 + HOH ==> RNH3^+ + OH^-
initial...0.1..............0........0
change.....-x..............x........x
equil.....0.1-x.............x........x
Kb = (RNH3^+)(OH^-)/(RNH2)
6.4E-4 = (x)(x)/(0.1-x).
x^2 = 6.4E-5 if we make the assumption that 0.1-x = 0.1. Then
x = sqrt 6.4E-5 = 8.00E-3 = (OH^-)
So pOH = 2.097 which converted to pH = 11.90. We probably should solve the quadratic and not make the assumption that 0.1-x = 0.1. If you do that the pH = 11.88 for pH at zero mL.