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### Topic: Zumdhal Chapter 8 Marathon Problem Help (Applications of Aqueous Equilibria)  (Read 8661 times)

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#### rgrocks

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• Mole Snacks: +0/-0 ##### Zumdhal Chapter 8 Marathon Problem Help (Applications of Aqueous Equilibria)
« on: September 21, 2007, 08:11:45 AM »
Let me type the problem and my methods
139)
A 225-mg sample of a diprotic acid is dissolved in enough water to make 250.0mL of solution. The pH of this solution is 2.06. A saturated solution of calcium hydroxide (Ksp- 1.3e-6) is prepared by adding excess calcium hydroxide to pure water and then rmeoving the undissolved solid by filltration. Enough of the calcium hydroxide solution is added to the solution of acid to reach the second equivalence point. The pH at the second equivalence point (as determined by pH meter) is 7.96. The first dissociation constant for acid (Ka1) is 5.90e-2. Assume that the volumes of solutions are additive, all solutions are at 25degC, and that Ka1 is at least 1000 times greater than Ka2.
a) calculate the molar mass of the acid
b. calculate the second dissociation constant for the acid Ka2

my method,
first to calculate molar mass of the acid
to calculate molar mass i set up a table (concentrations)
H2A    +    H2O   <---------->   H3O+    +     HA-
I    mm/225      -                          0               0
C    -x             -                        +x              +x
E   mm/225-x    -                          x               x

b/c pH =2.06 we know [H] = .0087096=x
Ka1= [HA][H]/[H2A] =  x2/[(mm-225)-x]   mm= 2.24,

please check this... this step is crucial to the rest of my method... right now im iffy if the concentration of H2A really is mm/225... can someone verify or prove differently my proof is

molarity = mmol/mL

mmol= (molar mass)1000

molarity = molarmass(1000)/mL

L=1000mL

molarity = molarmass/L       MM/.225 or 1000MM/mL   but using this method i get a molar mass < 1 which cant be right... any help?
« Last Edit: September 21, 2007, 09:16:08 AM by rgrocks »

#### Borek ##### Re: Zumdhal Chapter 8 Marathon Problem Help (Applications of Aqueous Equilibria)
« Reply #1 on: September 21, 2007, 08:29:34 AM »
b - Henderson-Hasselbalch equation.
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#### rgrocks

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• Mole Snacks: +0/-0 ##### Re: Zumdhal Chapter 8 Marathon Problem Help (Applications of Aqueous Equilibria)
« Reply #2 on: September 21, 2007, 08:33:24 AM »
continuing my train of thought...

Ca(OH)2  <-------->  Ca2+        +      2OH1-   Ksp- =1.3e-6
final      -                        x                     2x

Ksp = [OH]2[Ca]  = 4x3
x= .0068753443M  [OH] = .0137506887M

#### rgrocks

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• Mole Snacks: +0/-0 ##### Re: Zumdhal Chapter 8 Marathon Problem Help (Applications of Aqueous Equilibria)
« Reply #3 on: September 21, 2007, 08:53:04 AM »
actually i think i figured out how to to b already but it seems my "a" method is wrong, i used dimentional analysis on it and it doesnt come out right... how can i find molar mass?

#### Borek ##### Re: Zumdhal Chapter 8 Marathon Problem Help (Applications of Aqueous Equilibria)
« Reply #4 on: September 21, 2007, 09:08:23 AM »
You have to find molar concentration first. You don't need ICE - just solve Ka1 for [H2A].
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#### rgrocks

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• Mole Snacks: +0/-0 ##### Re: Zumdhal Chapter 8 Marathon Problem Help (Applications of Aqueous Equilibria)
« Reply #5 on: September 21, 2007, 09:17:56 AM »
You have to find molar concentration first. You don't need ICE - just solve Ka1 for [H2A].

i dont understand what you mean... i thought in order to solve Ka1 i need the ICE... could you show me what you mean?

#### Borek ##### Re: Zumdhal Chapter 8 Marathon Problem Help (Applications of Aqueous Equilibria)
« Reply #6 on: September 21, 2007, 09:48:26 AM »
Ka1 = [H+][HA-]/[H2A] - that's definition. Solve for [H2A] - you know all other variables as they were either given in the question, or they are known thanks to the dissociation stoichiometry.
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#### rgrocks

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• Mole Snacks: +0/-0 ##### Re: Zumdhal Chapter 8 Marathon Problem Help (Applications of Aqueous Equilibria)
« Reply #7 on: September 21, 2007, 10:42:43 AM »
Ka1 = [H+][HA-]/[H2A] - that's definition. Solve for [H2A] - you know all other variables as they were either given in the question, or they are known thanks to the dissociation stoichiometry.

I see what your saying... b/c pH=2.06  [H+] = .0087096 = [HA-]
so i get

0.00870962/5.90e-2=[H2A]  = 0.0012857

to get the number of mmols
[H2A] x 250mL = 0.3214285mmol = 225mg   225/0.321485 = 700.00 g/mol   can someone double check please

#### AWK

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« Reply #8 on: September 24, 2007, 04:37:23 AM »
In this case you should solve a more precise equation:

Ka1 = [H+][HA-]/[H2A]=x2/(c-x)
AWK

#### Borek ##### Re: Zumdhal Chapter 8 Marathon Problem Help (Applications of Aqueous Equilibria)
« Reply #9 on: September 24, 2007, 06:48:28 AM »
What you have found is [H2A], but you are interested in total concentration - that is [H2A] + [HA-] (you already you already know [HA-]). That's why AWK suggests to use full equation - which will give exactly the same result once solved for c; doesn't matter which way you will follow.

However, there is something wrong - are you sure your sample mass was 0.225mg and not 0.225g?
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