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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: 777888 on January 07, 2005, 10:29:01 AM

Title: equilibrium test questions
Post by: 777888 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?]
Title: Re:equilibrium test questions
Post by: Donaldson Tan 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
Title: Re:equilibrium test questions
Post by: Donaldson Tan 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,
Title: Re:equilibrium test questions
Post by: 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.

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
Title: Re:equilibrium test questions
Post by: 777888 on January 07, 2005, 08:49:39 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)
Title: Re:equilibrium test questions
Post by: 777888 on January 07, 2005, 08:56:37 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
Title: Re:equilibrium test questions
Post by: Demotivator on January 07, 2005, 09:03:45 PM
Q3
No, 0.1mol/L would not  work.
Title: Re:equilibrium test questions
Post by: 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.
Title: Re:equilibrium test questions
Post by: 777888 on January 07, 2005, 09:33:28 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...
Title: Re:equilibrium test questions
Post by: 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]
Title: Re:equilibrium test questions
Post by: 777888 on January 07, 2005, 10:35:51 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?
Title: Re:equilibrium test questions
Post by: 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.
Title: Re:equilibrium test questions
Post by: 777888 on January 07, 2005, 11:05:22 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.
Title: Re:equilibrium test questions
Post by: 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.
Title: Re:equilibrium test questions
Post by: 777888 on January 07, 2005, 11:27:11 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)
Title: Re:equilibrium test questions
Post by: 777888 on January 08, 2005, 12:44:10 AM
7. For the rxn: A + B + 200kJ <-> C + D, predict the effects after drecreasing [D].
a)release energy
b)increase [C]
c)both a and b

[b is definitely right, but I'm not quite sure about a...the temperature would decrease, but the 200kJ is in the left side of the rxn...]
Title: Re:equilibrium test questions
Post by: Donaldson Tan on January 08, 2005, 08:51:05 AM
haha.. the tabulation method i taught u is now called ICE. Cute name..
Normally, exam-wise, if you arent given the Ksp of Ca(OH)2, you can safely assume 100% dissociation.

what is dH for question(7) ?
Title: Re:equilibrium test questions
Post by: 777888 on January 08, 2005, 11:19:17 AM
haha.. the tabulation method i taught u is now called ICE. Cute name..
Normally, exam-wise, if you arent given the Ksp of Ca(OH)2, you can safely assume 100% dissociation.

what is dH for question(7) ?
what is dH ???
Title: Re:equilibrium test questions
Post by: Donaldson Tan on January 08, 2005, 11:59:20 AM
7. For the rxn: A + B + 200kJ <-> C + D, predict the effects after drecreasing [D].
a)release energy
b)increase [C]
c)both a and b

what is deltaH for the reaction in question 7?
Title: Re:equilibrium test questions
Post by: 777888 on January 08, 2005, 12:20:58 PM
what is deltaH for the reaction in question 7?
+200kJ, but according to Le Chatelier's principle, the 200kJ will decrease...
Title: Re:equilibrium test questions
Post by: Donaldson Tan on January 08, 2005, 12:44:29 PM
deltaH of a reaction is different from applying Q of 200kJ to the reaction system.
Title: Re:equilibrium test questions
Post by: 777888 on January 08, 2005, 01:18:31 PM
deltaH of a reaction is different from applying Q of 200kJ to the reaction system.
But after the stress and the equilibrium shift, the temperature of the SYSTEM will decrease(because heat energy decreased), right? (releasing energy?!)
Title: Re:equilibrium test questions
Post by: 777888 on January 08, 2005, 01:45:21 PM
Let me make an example that is similar.
5.20mL of 0.20M HCN is titrated with 0.35M NaOH. Find
a)pH initially and
b)at midpoint(halfway to equivalence).

I calculated this way:
a)HCN+NaOH->H2O+NaCN
HCN+H2O<->H3O+ + CN- (hydrolysis)
Ka=6.2x10^-10=x^2 / 0.2 (assumption made)
pH=4.95

b)HCN +   NaOH->H2O+NaCN
n=0.004    n=0.002
V=0.02      c=0.35
V of NaOH required to reach midpoint=0.002/0.35=0.00571

After rxn:
HCN +   NaOH->H2O+NaCN
n=0.002  n=0            n=0.002
c=n/V(total)
c=0.1556                   c=n/V0.07779

HCN+H2O<->H3O+ +CN-(hydrolysis)
6.2x10^-10=0.07779x/0.1556 (assumption made)
pH=9.21
The results are surprising to me...the pH is so high at midpoint...all other weak acid/strong base quesitons I did before have midpoint at about 5 point something only. So did I calculated them right?
Title: Re:equilibrium test questions
Post by: 777888 on January 08, 2005, 02:20:08 PM
8.Which of the following is in a state of equilibrium
a)A container with 20mL of water
b)A container with an ice cube and water
c)both a and b
[would a container with 20mL of water be in a state of equilibrium]
Title: Re:equilibrium test questions
Post by: Donaldson Tan on January 08, 2005, 05:18:45 PM
(7) I'm not sure of your notation. I assume "..+200kJ" means u add 200kJ of heat, and it's not deltaH of the reaction. Adding heat to the system favours the endothermic reaction.

(8) for (a). H2O(l) <-> H2O (g), but for (b), there will be heat transfer from water to the ice that cause it to melt, hence no equilibrium.
Title: Re:equilibrium test questions
Post by: 777888 on January 08, 2005, 05:30:36 PM
(7) I'm not sure of your notation. I assume "..+200kJ" means u add 200kJ of heat, and it's not deltaH of the reaction. Adding heat to the system favours the endothermic reaction.

(8) for (a). H2O(l) <-> H2O (g), but for (b), there will be heat transfer from water to the ice that cause it to melt, hence no equilibrium.
7)Yes the reaction is endothermic:
A + B + heat <-> C + D
and then [D] is decreased, so shift right and [A], and heat all decreased. But heat decreased means temperature is lowed and is releasing energy, right? (heat energy is consumed...)

8b)H2O(l) <-> H2O(s) why not equilibrium? I thought that there will be equilibrium of melting and solidfying processes occuring at the same rate...
Title: Re:equilibrium test questions
Post by: Donaldson Tan on January 09, 2005, 09:27:16 AM
(8) Common sense lah, u put ice in water, what do u see? however, u find liquid water exist without any significant change in its mass..
Title: Re:equilibrium test questions
Post by: 777888 on January 09, 2005, 10:23:55 PM
Let me make an example that is similar.
5.20mL of 0.20M HCN is titrated with 0.35M NaOH. Find
a)pH initially and
b)at midpoint(halfway to equivalence).

I calculated this way:
a)HCN+NaOH->H2O+NaCN
HCN+H2O<->H3O+ + CN- (hydrolysis)
Ka=6.2x10^-10=x^2 / 0.2 (assumption made)
pH=4.95

b)HCN +   NaOH->H2O+NaCN
n=0.004    n=0.002
V=0.02      c=0.35
V of NaOH required to reach midpoint=0.002/0.35=0.00571

After rxn:
HCN +   NaOH->H2O+NaCN
n=0.002  n=0            n=0.002
c=n/V(total)
c=0.1556                   c=n/V0.07779

HCN+H2O<->H3O+ +CN-(hydrolysis)
6.2x10^-10=0.07779x/0.1556 (assumption made)
pH=9.21
The results are surprising to me...the pH is so high at midpoint...all other weak acid/strong base quesitons I did before have midpoint at about 5 point something only. So did I calculated them right?
OK! Thanks!
How about the titration calculations? Did I do it right?(espically the hydrolysis part...) :)