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All Message Boards : Equilibrium
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Reply
 Message 1 of 7 in Discussion 
From: MSN NicknameAlbert1145  (Original Message)Sent: 10/1/2008 5:51 AM
Hi Steve,

I'm doing some revision over the holidays, could you check some of my questions?

1. A salt MX3 has a Ksp value of 8.47E-15. What is the millimolar(mM) concentration of X- in a saturated solution of this salt?
I got 42.085mM

2. Co(g) + H2)(g) <==> CO2(g) + H2(g)
occurs in an ideal mixture of ideal gases. At 700K, Kp=5.10. At this temperature, calculate deltaG in kJ.
I got 9.48kJ

3. 8.9 of a high molecular weight biological compound is dissoved in water to make 500mL of solution. The resulting solution developed an osmotic pressure of 0.044atm at 25C. What is the molecular weight of the compound?
I got 2472.977


I'm don't know how to do this question:
The density of a 14.1M CH3OH solution is 0.858 g/mL. What is the molality of this solution? H2O is the solvent.

Thank you very much


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Reply
 Message 2 of 7 in Discussion 
From: MSN Nickname·Steve·Sent: 10/1/2008 9:10 PM
1. The equation for the solution process is
                     MX3 (s)    M3+ (aq)  +  3X�?/SUP> (aq)
Equilibrium:      �?nbsp;             x              3x
 
Therefore, Ksp  =  [M3+][X�?/SUP>]3,  8.47 X 10�?5  =  (x)(3x)3  =  27X4,  x  =  1.33 X 10�? M
 
So, the molar solubility is 1.33 X 10�? M, and the concentration of X�?/SUP> is 3x.
3(1.33 X 10�? M)  =  3.99 X 10�? M  or  0.399 mM.
 
 
2.  Looks OK, except it should be �?.48 kJ/mol (negative).
 
 
3.  I always do these molecular weight calculations from a colligative property in three steps.  In this case, the colligative property is osmotic pressure, Π.
 
a)  Calculate the concentration (molarity here).
     Π  =  MRT,  M  =  Π / RT  =  0.044 atm / (0.08206 L atm/mol K)(298.15 K)  =  0.0017984 M.
 
b)  From the concentration and volume, calculate the moles of solute.
     Molarity  =  moles / L,  moles  =  molarity X L  =  (0.0017984 mol/L)(0.500 L)  =  0.0008992 mol
 
c)  Divide the grams of solute by the moles of solute to get the molar mass in units of g/mol.
     8.9 g / 0.0008992 mol  =  9898 g/mol
 
 
4.  To convert molarity to molality, moles / kg of solvent, we can start by assuming that we have 1 liter of the solution.  This 1 liter of solution will then have 14.1 moles of CH3OH in it.  Convert 14.1 moles of CH3OH to grams.
 
Next, calculate the total mass of the 1 liter of solution:  mass  =  volume  X  density:
mass  =  1000 mL  X  0.859 g/mL  =  859 g
 
Now get the mass of the solvent alone by subtracting the mass of the CH3OH from the total mass of the solution:
mass of H2O  =  859 g  �?nbsp; grams of CH3OH
 
Convert grams of H2O to kg, and then plug the numbers into the molality formula:
molality  =  moles of CH3OH / kg of H2O
 
That should do it!

Reply
 Message 3 of 7 in Discussion 
From: MSN NicknameAlbert1145Sent: 10/2/2008 2:05 AM
Could you please guide me through question 2?

Reply
 Message 4 of 7 in Discussion 
From: MSN Nickname·Steve·Sent: 10/2/2008 6:02 AM
2.  I just plugged into the formula ΔG°  =  –RT lnKp :
     ΔG°  =  �?8.314 X 10�? kJ mol�? K�?)(700 K) ln(5.10)  =  �?.48 kJ/mol.
 
 
 
CO (g)  +  H2O (g)      CO2 (g)  +  H2 (g)
 
Keq  =   [CO2][H2]  
            [CO][H2O]
 
Since the concentrations of ideal gases are proportional to their pressures, we can say,
 
Kp  =    PCO2 PH2  
           PCO PH2O
 
which is used in the equation ΔG°  =  –RT ln Kp
 

Reply
 Message 5 of 7 in Discussion 
From: MSN Nickname·Steve·Sent: 10/2/2008 6:14 AM
Also in question 1, I noticed that I forgot to put in the equilibrium arrow.  Here is that part again:
 
                     MX3 (s)        M3+ (aq)    +    3 X�?/SUP> (aq)
Initially:            �?nbsp;                        0                       0
Equilibrium:      �?nbsp;                        x                       3x
 
Ksp  =  [M3+] [X�?/SUP>]3
 
8.47 X 10�?5   =   (x) (3x)3   =   27 X4
 

Reply
 Message 6 of 7 in Discussion 
From: MSN NicknameAlbert1145Sent: 10/3/2008 2:02 AM
Sorry Steve, I'm a bit confused with the equation equations.
I thought its ΔG° = –RT lnKc not ΔG° = –RT lnKp

Albert

Reply
 Message 7 of 7 in Discussion 
From: MSN Nickname·Steve·Sent: 10/3/2008 5:24 AM
When using the formula ΔG° = –RT ln K to calculate K, K will be Kc for reactions in solution and concentrations would be expressed in molarity units.  But if you have a gas phase reaction, K from this formula will be Kp and the concentrations of the substances are expressed in standard pressure units such as atmospheres.
 
The equilibrium constant obtained from ΔG° = –RT ln K is called the "thermodynamic" equilibrium constant.  In the thermodynamic equilibrium constant expression for a reaction involving gases, gases are always given as partial pressures; for solutions, always molarity.  If the equilibrium is heterogeneous and has some substances in solution and others in the gas phase, both units, molarity and atmospheres, are used.  It is confusing, but that is the convention! 

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