The exact experiement is the same one listed in this statistics report for a classes results:
https://www.researchgate.net/publication/252787534_Determination_of_Ksp_DG_DH_and_DS_for_the_Dissolution_of_Calcium_Hydroxide_in_Water_a_General_Chemistry_Experimentbut basically we were given some numbers and asked to calculate the enthalpy and entropy of Ca(OH)
2 The given information is:
diluted HCl concentration = 0.01469 M HCl
Room Temp = 23 °C = 296.15 K
High Temp = 36 °C = 309.15 K
Average volume of titrant delivered (room temp) = 24.33 mL = 0.02433 L
Average volume of titrant delivered (high temp) = 21.20 mL = 0.02120 L
Note: This was supposed to be a lab we performed ourselves but due to covid-19 we were not able to do this lab and were just given that information to complete the rest of the calculations with. I have asked my instructor for help but they seem to be very busy and I am not confident I will hear back from them at all so I am trying to get some help from the internet community.
I had this all done but then I realized that I shouldn't have halved the moles of OH
- because even though
Ca(OH)
2 Ca
2+(aq) + 2OH
-(aq)the moles of HCl added should be equal to the moles of OH
- already in solution because the actual reaction we are studying is:
H
3O
+(aq)+OH
-(aq) 2H
2O
(l)Which means that that total moles of H
3O
+= total moles of OH
-So the total moles of H
3O
+ is
[tex] \frac {24.33 mL}{1} * \frac {1 L}{1000mL} * 0.01469 \frac {mol}{L} = 3.574x10^{-4} mol H^{+} = 3.574x10^{-4}mol H_{3}O^{+}[/tex]
The samples that were titrated had a total volume of 10.00 mL which is equal to 0.010 L so the molarity (concentration) of OH
- is
[itex] \frac {3.574x10^{-4}mol}{0.010L} = 0.03574 \frac {mol}{L}=0.03574 M[/itex]OH
-So the real question is am I on the right track with this or is the [OH
-] ≠ [H
3O
+]?
The next thing to do is solve for K
eq=[itex] \frac {1}{2} [OH^{-}]^{3}= \frac {1}{2} (0.03574)^{3} = 2.2826x10^{-5} [/itex]
This K expression comes from K
sp=[Ca
2+][OH
-]
2And this is a saturated solution so the stoichiometry gives the relationship
[tex][Ca^{2+}]= \frac {1}{2}[OH^{-}][/tex]
so K
sp=[itex] \frac {1}{2}[OH^{-}][OH^{-}]^{2}= \frac {1}{2} [OH^{-}]^{3}[/itex]
That is how I get to the K
eq above
which is fine but the paper that I linked above gives the tabulated results as 2.15x10
-6 so this is off by a factor of 10, but if my procedure is good then that is fine, I just want to make sure that I am going about this correctly. The rest of the paper is to find Gibbs Free Energy and using the given info from both solutions to solve a linear system of equations for ΔH° and ΔS° which is something I am comfortable with, I just wanted some feedback on the first part of the work.
Thanks to anyone in advance for looking at this.
EDIT: ok so I ran through the equations with a 1:1 ratio this time and they seem to make even less sense, if anyone can tell me what is going on here it would be appreciated, I will write down all of my calculations below.
Room Temp [OH
-] = 0.03574 M (from the LaTeX above)
Now the "high temp" calculations:
[itex] \frac {21.20 mL}{1} * \frac {1 L}{1000mL} * 0.01469 \frac {mol}{L}=3.114x10^{-4}mol[/itex] OH
-the volume is still 0.010 L (should this be the TOTAL volume, as in the initial 10.00mL + 21.20mL, or just the original 10.00 mL? That could be why the numbers are so off but you want to know the initial [OH
-] before adding any acid so it makes more sense as using 10.00mL but please tell me if I am wrong)
so the concentration is [itex] \frac {3.114x10^{-4}}{0.010L}=0.03114 M[/itex] OH
- at "high temperature" (36°C)
now using the K
sp=[itex] \frac {1}{2}[OH^{-}]^{3}= \frac {1}{2} (0.03114)^{3}=1.5098x10^{-5}[/itex]
So K
sp(room temp) = [itex]2.2826x10^{-5}[/itex]
and
K
sp(hot) = [itex]1.5098x10^{-5}[/itex]
Then I need to find ΔG°
c and ΔG°
h, where c is for 'cool' and h is for 'hot'
in general ΔG°=[itex]-RTln(K)[/itex]
so
ΔG°
c=[itex]-(8.314 \frac {J}{mol*K})*296.15K*ln(2.2826x10^{-5})=26.3149 \frac {kJ}{mol}[/itex]
ΔG°
h=[itex]-(8.314 \frac {J}{mol*K})*296.15K*ln(1.5098x10^{-5})=27.3327 \frac {kJ}{mol}[/itex]
Now to solve for enthalpy and entropy I can use the two equations:
1. ΔG°
c=ΔH°-T
cΔS°
2. ΔG°
h=ΔH°-T
hΔS°
for equation 1:
[tex]26.3149 \frac {kJ}{mol}=\Delta H°-(296.15K)*\Delta S°[/tex]
and equation 2:
[tex]27.3327 \frac {kJ}{mol}=\Delta H°-(309.15K)*\Delta S°[/tex]
solving for ΔH° in equation 2:
[tex]\Delta H°=27.3327 \frac {kJ}{mol}+(309.15K)\Delta S°[/tex]
Using this ΔH° for equation 1:
[tex]26.3149 \frac {kJ}{mol}=(27.3327 \frac {kJ}{mol}+(309.15K)\Delta S°)-296.15K*\Delta S°[/tex]
simplifying:
[tex]26.3149 \frac {kJ}{mol}=27.3327 \frac {kJ}{mol}+13K*\Delta S°[/tex]
[tex]-1.0178 \frac {kJ}{mol}=13K*\Delta S°[/tex]
or
[tex]13K*\Delta S°=-1.0178 \frac {kJ}{mol}[/tex]
so ΔS°=[itex] \frac {-1.0178 \frac {kJ}{mol}}{13.00K}=-0.078292 \frac {kJ}{mol*K}= -78.2923 \frac {J}{mol*K}[/itex]
then using the [itex] \frac {kJ}{mol*K}[/itex] value of ΔS° in equation 2 for ΔH°:
[tex] \Delta H°=27.3327 \frac {kJ}{mol} +(309.15K) \Delta S° =27.3327 \frac {kJ}{mol} +(309.15K)*(-0.078292 \frac {kJ}{mol*K}) [/tex]
[tex] \Delta H°=3.12873 \frac {kJ}{mol}[/tex]
this is a very different result from what I got before, when I used [OH
-]=0.5[H
3O
+] when ΔH°<0 and ΔS°<0 (which line up with published results but doesn't make sense as the moles should be equal) but now ΔH°>0 and ΔS°<0 and the numbers are way different from the published ones in the paper above. They give ΔH°=-16.86 kJ/mol and ΔS°=-150.7 J/(mol*K) which was much more in line with the numbers I got from the 2:1 assumption.
Using the 2:1 assumption I get ΔH°=-24.21 KJ/mol and ΔS°=-187.6 J/(mol*K) which is basically in line with the published results.
please explain if you can