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I'm redoing a quiz from my Chem II course. I'm having trouble with one specific part of the problem.
Part 1) Find pH of a sol'n that is 0.50M HA (Ka = 1.2 x 10^-7) and 0.40 M NaA.
I calculated the pH using Handerson-H***elbalch:
pH = pKa + log (A-/HA)
pH = -log(1.2 x 10^-7) + log (0.40/0.50)
pH = 6.8
Part 2) Find pH at the equivalence point when 10mL of the above sol'n is titrated with 1.00M of NaOH (strong base).
I have no idea how to do this. I think the equivalence point is when [HA] = [A-], but I have no idea where to go from there. Help?
Part 3) What's the best indicator (given a chart) for the above titration?
This one will be no problem, since I'm just looking for a colorband that goes through the given pH approximately in the middle of the band.
====================================
Update: Here's what I eventually found:
I have (initially) 5 mmol of HA and 4 mmol of A- (via the NaA). Meaning that when I titrate, the sol'n with NaOH, the table looks like:
Given that I insert 5 mmol of NaOH from a 1.0M NaOH sol'n, that means I added 5mL of NaOH sol'n, increasing the total volume to 15mL.
This means the [A-] is 9mmol / 15 mL, or 0.60M.
New ICE, this time with concentrations:
Using this equation form means I need Kb. Kb = Kw/Ka (and assuming Standard Temperature + Pressure); Kb = (10^-14) / (1.2 x 10^-7) = (8.3333 x 10^-8).
Kb = [HA][OH-] / [A-]
8.3333 x 10^-8 = (x^2) / (0.6-x)
**Assuming x is small → x = sqrt(0.6 * 8.3333 * 10^-8) = 2.236 x 10^-4 (x is sufficiently small).
x = [OH-]. -log[x] = -log[OH-] = pOH = 3.65
At 25C (standard temperature+pressure is 25C @ 1atm), pH = 14 - pOH.
pH = 10.35
Part 1) Find pH of a sol'n that is 0.50M HA (Ka = 1.2 x 10^-7) and 0.40 M NaA.
I calculated the pH using Handerson-H***elbalch:
pH = pKa + log (A-/HA)
pH = -log(1.2 x 10^-7) + log (0.40/0.50)
pH = 6.8
Part 2) Find pH at the equivalence point when 10mL of the above sol'n is titrated with 1.00M of NaOH (strong base).
I have no idea how to do this. I think the equivalence point is when [HA] = [A-], but I have no idea where to go from there. Help?
Part 3) What's the best indicator (given a chart) for the above titration?
This one will be no problem, since I'm just looking for a colorband that goes through the given pH approximately in the middle of the band.
====================================
Update: Here's what I eventually found:
I have (initially) 5 mmol of HA and 4 mmol of A- (via the NaA). Meaning that when I titrate, the sol'n with NaOH, the table looks like:
|
HA
|
+
|
NaOH
|
←→
|
NaA
|
+
|
HOH
|
| I |
5 mmol
|
5 mmol
|
4 mmol
|
--
|
| C |
-5 mmol
|
-5 mmol
|
+5 mmol
|
--
|
| E |
0 mmol
|
0 mmol
|
9 mmol
|
--
|
Given that I insert 5 mmol of NaOH from a 1.0M NaOH sol'n, that means I added 5mL of NaOH sol'n, increasing the total volume to 15mL.
This means the [A-] is 9mmol / 15 mL, or 0.60M.
New ICE, this time with concentrations:
|
A-
|
+
|
HOH
|
←→
|
HA
|
+
|
OH-
|
| I |
0.6
|
--
|
0
|
0
|
| C |
-x
|
--
|
+x
|
+x
|
| E |
0.6-x
|
--
|
x
|
x
|
Using this equation form means I need Kb. Kb = Kw/Ka (and assuming Standard Temperature + Pressure); Kb = (10^-14) / (1.2 x 10^-7) = (8.3333 x 10^-8).
Kb = [HA][OH-] / [A-]
8.3333 x 10^-8 = (x^2) / (0.6-x)
**Assuming x is small → x = sqrt(0.6 * 8.3333 * 10^-8) = 2.236 x 10^-4 (x is sufficiently small).
x = [OH-]. -log[x] = -log[OH-] = pOH = 3.65
At 25C (standard temperature+pressure is 25C @ 1atm), pH = 14 - pOH.
pH = 10.35
even then it prohably would be wrong because i am terrible at chemistry :D