This is a follow up question again to this: http://www.enotes.com/homework-help/ok-this-follow-up-question-my-previous-one-http-463275 The answer I received there was really helpful and I just had...

This is a follow up question again to this: http://www.enotes.com/homework-help/ok-this-follow-up-question-my-previous-one-http-463275

The answer I received there was really helpful and I just had my lab where I need to use 1,90g KHT (KC4H5O6) in 100mL water. And then add 1,90g of KCl in the same solution to find out at which temperature precipitate will form.

and the concentration with just KHT in solution is: [K+]=[HT-] = 0,10M

K = 0,01 and InK = -4,60

Concentration with 1,90g KCl in solution:
[K+] = 1,254 and [HT-] = 0,10M

K=0,1254 and InK=-2,07

From a graph I got and calculated:
deltaH = 53988,6J/mol and deltaS = 131,21J/Kmol

I tried to use the same method to solve this, but the temperature I got from the calculations is weird. With just KHT present I got a T=580,81K and with KCl present I got T=473,7K 

These temperature seems so wrong cause we could observe precipitate in the lab at a lower temperature, also this doesn´t support the hypothesis that the temperature for which precipitate will form should be much higher with a common-ion in solution and this is an endoterm reaction too. 

So should I use a nother method to find at which precipitate will form or what went wrong? I tried to double check my calculations many times to, and it seems like the error wasn´t from there. 

Asked on by pham1403

1 Answer | Add Yours

llltkl's profile pic

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

Consider the process of dissolution of KHT only in 100 ml water,

Molar concentrations, `[K^+]=19.0/188.2=0.101`

`[HT^-]=0.101`

`K=[K^+][HT^-]=0.101*0.101=0.102`

`lnK=-4.5855`

Putting `DeltaH^o`  and `DeltaS^o` values from what you got from the graph,

`DeltaG^o=DeltaH^o-T*DeltaS^o` (where T is the temperature of saturation),

`=53988.6-T*131.21`

Also,

`DeltaG^o=-RTlnK`

`=-8.314*T*-4.5855`

`=38.1238T`

Equating the two expressions of `DeltaG^o,`

`53988.6-T*131.21=38.1238T`

`rArr T(131.21+38.1238)=53988.6`

`rArr T=53988.6/169.3338=318.84 K=45.7^oC`

-----------------------------------------------------------

Now, consider the process of dissolution of KHT in presence of 1.9g KCl in 100 ml water (provided that dissolution of all the salts is done at elevated temperature),

Molar concentrations, `[K^+]=19.0/188.2+19.0/74.6=0.35568`

`[HT^-]=0.101`

`K'=[K^+][HT^-]=0.35568*0.101=0.03592`

`lnK'=-3.32647`

`DeltaG^o=DeltaH^o-T'*DeltaS^o ` (where T' is the temperature of saturation now),

`=53988.6-T'*131.21`

Also,

`DeltaG^o=-RT'lnK'`

`=-8.314*T'*-3.32647`

`=27.6563T'`

Equating the two expressions of `DeltaG^o` ,

`53988.6-T'*131.21=27.6563T'`

`rArr T'(131.21+27.6563)=53988.6`

`rArr T'=53988.6/158.8663=339.85 K=66.7^oC`

This higher value of T' is commensurate to your idea of earlier precipitation of KHT (i.e. at a higher temperature) in presence of a common ion `(K^+)` . 

Sources:

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