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777888

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equilibrium test questions
« on: January 07, 2005, 10:29:01 AM »
I have these questions on the test...I am concerned that I did them wrong...Can somebody see and teach me? Thank you :)
1. When Al3(PO4) dissolves, what is the ratio of Al3+ to PO4(3-)? [Is it 3:1?]

2. If Pb(Cl)2 is dissolving in water, and [Pb2+]=9.2x10^-2, what is the value of Ksp? [Can I use mol ratio for equilibrium reaction? Would the answer be Ksp=[Pb2+][Cl-]^2=(9.2x10^-2)(2x9.2x10^-2)^2=3.1x^10-3?]

3.If 50mL of water is added to 250mL and 0.1mol/L citric acid, and the pH at equilibrium is 4.6, find Ka.
[n=cV=0.1x0.25=0.025
c(new)=0.025/0.3=0.0833mol/L

[H+]10^-4.6=2.51x10^-5mol/L
    [HA]               [H+]               [A-]
I   0.0833           0                   0
C  -2.51x10^-5    +2.51x10^-5    +2.51x10^-5
E   0.08327     2.51x10^-5           2.51x10^-5
In the bolded part of the ICE table(initial concentratoin), should I use 0.08327(after adding water) or 0.1(before adding water?]

4.Find the molar solubility of Zn(OH)2 in a 0.1mol/L solution of Ca(OH)2, Ksp Zn(OH)2=7.7x10^-17.
[I calculated as follows:
Ca(OH)2 -> Ca2+ + 2OH-
0.1M                         0.2M
Zn(OH)2  <-> Zn2+ + 2OH-
   [Zn2+]   [OH-]
I      0         0.2
C     +x        +2x
E       x         0.2+2x
And molar solubility [Zn(OH)2]=x
My problem is that I found that Ca(OH)2 is insoluble...would that affect the Ca(OH)2 dissociation rxn?(I calculation [OH-]=0.1x2=0.2, I have assumed 100% dissolving?!]

5.Is it possible for titration of NaOH with a very weak acid like HCN(Ka=6.2x10^-10) to have a initial pH=5.3 and midpoint(halfway to equivalence point) pH=9.2...that would give a equivalence point with pH higher than 10...and the vertical part of the curve will be very short...is that possible?



6.Which best describes a concentrated weak acid?
a)has a low pH
b)has a high pH
c)most of the acid is in dissolved state while small amounts exist as molecules
d)most of the acid exist as molecule while small amounts exist as ions
e)none of the above
[I chose d, but would a also right?]
« Last Edit: January 08, 2005, 02:07:28 AM by 777888 »
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Offline Donaldson Tan

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Re:equilibrium test questions
« Reply #1 on: January 07, 2005, 05:15:13 PM »
(1) & (2) is right. (3) looks dodgy.

HA, H+, A- are reflected in terms of number of moles, not concentration, in the tables below:

assuming citric acid is monobasic,

in 250ml solution at equilibrium,
HA    H+    A-
eqbm(0.025 - X)    X    X
Ka = X2/[0.250*(0.025-X)]

in 300ml solution at equilibrium (upon addition of 50ml),
HA    H+    A-
eqbm(0.025 - X - Y)    X + Y    X + Y
Ka = (X + Y)2/[0.300*(0.025-X-Y)]

let X be amount of citric acid dissociated in 250ml solution
let Y be amount of citric acid dissociated upon addition of 50ml

given eqbm pH in 300ml solution is 4.6,
(X+Y)/(300x10-3) = 10-4.6
equation(1): X + Y = 3x10-5.6

assuming both 250ml & 300ml solution are at same temperature, then both solutions share the same Ka value, hence:

equation(2): X2/[0.250*(0.025-X)] = (X + Y)2/[0.300*(0.025-X-Y)]

the value of the RHS of equation (2) can be determined by substituting (1) into it. The numerical value corresponds to the required Ka

Ka = 7.54 x 10-9
« Last Edit: January 08, 2005, 08:33:26 AM by geodome »
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Offline Donaldson Tan

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Re:equilibrium test questions
« Reply #2 on: January 07, 2005, 05:27:48 PM »
regarding question(4),
Zn(OH)2 <-> Zn2+ + 2OH-
Ksp of Zn(OH)2 = 7.7x10-17
Ca(OH)2 <-> Ca2+ + 2OH-
Ksp of Ca(OH)2 = 5.5x10-6

the Ksp of Calcium Hydroxide is about 7x107 times of that of Zinc Hydroxide. In another words, calcium hydroxide is so much more soluble in water than zinc hydroxide, so addition of zinc hydroxide has minimal effect on the dissolution equilibrium of calcium hydroxide,
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"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Demotivator

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Re:equilibrium test questions
« Reply #3 on: January 07, 2005, 07:53:33 PM »
Q3
1) This is a bad question because citric acid is triprotic and it is confusing what Ka they are talking about. But let's assume they really intend a monoprotic acid.

Quote from: geodome on January 07, 2005, 05:15:13 PM
given eqbm pH in 300ml solution is 4.6,
(X+Y)/(300x10-3) = 10-4.6
equation(1): X + Y = 3x10-5.6

assuming both 250ml & 300ml solution are at same temperature, then both solutions share the same Ka value, hence:

equation(2): X2/[0.250*(0.025-X)] = (X + Y)2/[0.300*(0.025-X-Y)]

the value of the RHS of equation (2) can be determined by substituting (1) into it. The numerical value corresponds to the required Ka

Ka = 7.54 x 10-6

2) I don't see how you obtain  7.54 x 10-6
RHS,
  ( 3x10^-5.6)^2/(.3)(.025-0 approx)  =  1.2x10^-8.2 ?  

3) Solving for a system of two equations for Ka is unnecessary.
 You only need to calculate the final state because one can start with citric as completely molecular and let it dissociate into its final volume (beginning with .025 mole/.3 L = .0833M as 777888 calculated):
Using concentrations,
 .0833 - x  ->  x    x
 x = 2.51 x 10^-5  (from PH)

Ka = x^2/( .0833 - x) ~ 6.3x10^-10/.0833 = 7.56x10^-9
« Last Edit: January 07, 2005, 07:59:00 PM by Demotivator »
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777888

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Re:equilibrium test questions
« Reply #4 on: January 07, 2005, 08:49:39 PM »
Quote from: Demotivator on January 07, 2005, 07:53:33 PM
Q3
1) This is a bad question because citric acid is triprotic and it is confusing what Ka they are talking about. But let's assume they really intend a monoprotic acid.2) I don't see how you obtain  7.54 x 10-6
RHS,
  ( 3x10^-5.6)^2/(.3)(.025-0 approx)  =  1.2x10^-8.2 ?  

3) Solving for a system of two equations for Ka is unnecessary.
 You only need to calculate the final state because one can start with citric as completely molecular and let it dissociate into its final volume (beginning with .025 mole/.3 L = .0833M as 777888 calculated):
Using concentrations,
 .0833 - x  ->  x    x
 x = 2.51 x 10^-5  (from PH)

Ka = x^2/( .0833 - x) ~ 6.3x10^-10/.0833 = 7.56x10^-9
oh no, for Q3, I typed in the wrong number, I should type 0.0833 in the [HA]initial! ;D
So am I right for using this ICE table? (I just find the new concentration of HA after adding water, but I am not quite sure if I should use the original 0.1mol/L or this one...
    [HA]              [H+]              [A-]
I  0.0833          0                  0
C  -2.51x10^-5    +2.51x10^-5    +2.51x10^-5
E  0.08327    2.51x10^-5          2.51x10^-5

Would 0.1mol/L work? (instead of 0.0833mol/L)
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777888

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Re:equilibrium test questions
« Reply #5 on: January 07, 2005, 08:56:37 PM »
Quote from: geodome on January 07, 2005, 05:27:48 PM
regarding question(4),
Zn(OH)2 <-> Zn2+ + 2OH-
Ksp of Zn(OH)2 = 7.7x10-17
Ca(OH)2 <-> Ca2+ + 2OH-
Ksp of Ca(OH)2 = 5.5x10-6

the Ksp of Calcium Hydroxide is about 7x107 times of that of Zinc Hydroxide. In another words, calcium hydroxide is so much more soluble in water than zinc hydroxide, so addition of zinc hydroxide has minimal effect on the dissolution equilibrium of calcium hydroxide,

oic...so I guess it's valid to put 0.2mol/L for [OH-]initial !? (Can I assume Ca(OH)2 100% dissociates into ions)
  [Zn2+]  [OH-]
I      0        0.2
C    +x        +2x
E      x        0.2+2x=0.2(assumption)
Ksp=7.7x10^-17=x(0.2)^2
x=1.925x10^-15
[Zn(OH)2] at equilibrium(molar sobility=1.9x10^-15
« Last Edit: January 07, 2005, 11:09:52 PM by 777888 »
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Demotivator

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Re:equilibrium test questions
« Reply #6 on: January 07, 2005, 09:03:45 PM »
Q3
No, 0.1mol/L would not  work.
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Demotivator

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Re:equilibrium test questions
« Reply #7 on: January 07, 2005, 09:15:23 PM »
Q3
And if they really meant triprotic acid, it would be:
H3A ->  3H+  +  A-
and a different Ka.
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777888

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Re:equilibrium test questions
« Reply #8 on: January 07, 2005, 09:33:28 PM »
Quote from: Demotivator on January 07, 2005, 09:15:23 PM
Q3
And if they really meant triprotic acid, it would be:
H3A ->  3H+  +  A-
and a different Ka.
What is the formula for critic acid? Is that organic? (If so, I would be safe for assuming monprotic acid since we haven't learn any organic chemistry!) But for triprotic acid, how could there be a Ka? As I know, it will have 3 Ka values...
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Demotivator

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Re:equilibrium test questions
« Reply #9 on: January 07, 2005, 09:51:44 PM »
Yes organic with a complex formula.
There's Ka1, Ka2, Ka3  for the three steps of dissociation
When the reactions are combined,

Ka1xKa2xKa3 = Ka = [H]^3[A]/[H3A]
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777888

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Re:equilibrium test questions
« Reply #10 on: January 07, 2005, 10:35:51 PM »
Quote from: Demotivator on January 07, 2005, 09:51:44 PM
Yes organic with a complex formula.
There's Ka1, Ka2, Ka3  for the three steps of dissociation
When the reactions are combined,

Ka1xKa2xKa3 = Ka = [H]^3[A]/[H3A]
I have a question...how do you know that citric acid is triprotic?
By the way, for Q6, will concentrated weak acid have a low pH(choice A)? What's the difference between strong acid and concentrated weak acid?
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Demotivator

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Re:equilibrium test questions
« Reply #11 on: January 07, 2005, 10:50:12 PM »
Only if one has seen the structure,  or knows the IUPAC  (formal) name, or has been told.

Q6  d is best because,
( even if conc weak acid can have low PH (how low is low? too fuzzy. there is a realistic limit though)),
     a can apply to strong acids as well.
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777888

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Re:equilibrium test questions
« Reply #12 on: January 07, 2005, 11:05:22 PM »
Quote from: Demotivator on January 07, 2005, 10:50:12 PM
Only if one has seen the structure,  or knows the IUPAC  (formal) name, or has been told.

Q6  d is best because,
( even if conc weak acid can have low PH (how low is low? too fuzzy. there is a realistic limit though)),
     a can apply to strong acids as well.

Q6: I guess low means low as a strong acid...
Will a concentrated weak acid have a pH low as a strong acid!
I think d is the best answer but this answer applies to ALL weak acids...dilute or concentrated.
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Demotivator

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Re:equilibrium test questions
« Reply #13 on: January 07, 2005, 11:19:28 PM »
If the strong acid and weak acid have similar strong concentrations, the strong acid will have a lower ph. If conc of strong acid is low enough, the weak acid will have lower ph.
There is more of a limit to  how low a ph a weak acid can go to than there is for a strong acid.
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777888

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Re:equilibrium test questions
« Reply #14 on: January 07, 2005, 11:27:11 PM »
Quote from: Demotivator on January 07, 2005, 11:19:28 PM
If the strong acid and weak acid have similar strong concentrations, the strong acid will have a lower ph. If conc of strong acid is low enough, the weak acid will have lower ph.
There is more of a limit to  how low a ph a weak acid can go to than there is for a strong acid.
so by any mean (d) would be the best answer...

geodome is not here yet...so I will just ask you something about question 4 if you have time :)

4.Find the molar solubility of Zn(OH)2 in a 0.1mol/L solution of Ca(OH)2, Ksp Zn(OH)2=7.7x10^-17.
Question4 :
(Can I assume Ca(OH)2 100% dissociates into ions) and put [OH-]initial=0.2mol/L in the ICE table?
  [Zn2+]  [OH-]
I      0        0.2
C    +x        +2x
E      x        0.2+2x=0.2(assumption)
« Last Edit: January 07, 2005, 11:28:09 PM by 777888 »
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