[RobertLBio]

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_Experiment

but 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  Ca²⁺_(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:

H₃O⁺_(aq)+OH^-_(aq) 2H₂O_(l)

Which means that that total moles of H₃O⁺= total moles of OH^-

So the total moles of H₃O⁺ 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^-] ≠ [H₃O⁺]?

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=[Ca²⁺][OH^-]²

And 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[H₃O⁺] 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

[RobertLBio]

For some reason the modify button is longer available to me, I am not trying to move this to the top or anything like that, I just wanted to say that my second K_eq for the "hot" temp is correct, I just forgot to change the 269.15K to 309.15K on the post. I wonder why I can't modify this post anymore, if it's locked can the mods unlock it please? I won't make any more replies to this post unless someone replies to me but I just wanted to make sure that if someone decided to take the time to examine all of this that they didn't just say that I used the wrong temp twice.

Thank you again for anyone reading this.

EDIT: and I guess I should say that its not really [OH^-]=0.5[H₃O⁺] but that the moles of H₃O⁺ added should be equal to the moles of OH^- in the 10.00mL sample that was titrated so once you calculate the moles of HCl added to the equivalence point it should be equal to the amount of OH^- ions that were already in solution. But, when using that 1:1 assumption I get those very odd and off numbers, not just by factors but also by sign, so there must be some aspect of the fact that Ca(OH)₂ produces 2moles of OH^- per mole of Ca(OH)₂ dissolved in solution. I think that should clear things up in the original post.

[Borek]

I wonder why I can't modify this post anymore

Posts can be modified only for some time. Unfortunately, ability to modify posts is regularly abused by people removing or changing the content that was already answered/commented on, leaving the answers in vacuum and making it impossible for newcomers to understand the thread.

Your calculation of Ksp values looks OK to me.

[RobertLBio]

I wonder why I can't modify this post anymore

Posts can be modified only for some time. Unfortunately, ability to modify posts is regularly abused by people removing or changing the content that was already answered/commented on, leaving the answers in vacuum and making it impossible for newcomers to understand the thread.

Your calculation of Ksp values looks OK to me.

Ok so do you have any idea why my ΔH° is so 'high' and positive and why my ΔS° is only about half of the value reported in the linked document for the same basic experiment and the same temperature?

My instructor has told me that it is 1:1 the moles of H₃O⁺:moles of OH^- but she did tell me that she 'made up' the numbers and worked backwards to catch 'out and out cheating' so perhaps they are meant to be off from those published results?

[RobertLBio]

OK I talked to my instructor and gave her my numbers that 'dont add up' and they were correct, as I said before she said she 'made up the numbers' to make sure that there was no 'out and out cheating' by looking up other published values or other posts on forums.

Thanks to anyone that looked at this and to Borek for responding to me.

I will probably be back with more question this fall