1. Dismiss Notice
  2. Download free Tapatalk for iPhone or Tapatalk for Android for your phone and follow the SDN forums with push notifications.
    Dismiss Notice
Dismiss Notice
Visit Interview Feedback to view and submit interview information.

Freezing Point Dep/Boiling Point Elev

Discussion in 'MCAT Study Question Q&A' started by rockyDoctora, Jul 31, 2011.

  1. rockyDoctora

    2+ Year Member

    Joined:
    Nov 10, 2009
    Messages:
    140
    Likes Received:
    0
    Status:
    Pre-Medical
    Hey guys, so I did my Kaplan FL #5 yesterday and just been going over the PS section and the very first question is related to the thread title.

    The passage talks about how acids dissociate more in ammonia than in water since ammonia acts as a base.

    Then they say that the student prepares these two solutions:

    Solution X: 2 mols of CH3COOH dissolved in 2000g of ammonia (liq)
    Solution y: 2 mols of CH3COOH dissolved in 2000g of water
    [Kf_amm = -0.957 C/m ; Kf_water = -1.86 C/m]

    The question then asks whether the relative freezing point depression of X or Y is larger.





    The answer to the question was Solution X. Their explanation was that the acid dissociates more readily in ammonia than in water. They talk about how the dissociation constant of the acid is about 10^-5 in water (wasnt given in passage but roughly) while in ammonia, it would be full dissociation. They say that this factor is much more significant than the difference in kf's.

    I don't understand this explanation.... We know that the magnitude of kf of water is double that of ammonia. However, I don't understand how the dissociating factor is so much more significant. Lets say that none of the acid dissociated in water so the value of i in [deltaT = iKm] is 1 (the acid is still dissolved in water right?), and in full dissociation in ammonia, i = 2.

    So if we do the math here:

    kf_water = 2 x kf_ammonia
    2 x i_water = i_ammonia
    m_water = m_ammonia

    So the 2's both cancel out and the depression should be around equal? Please tell me if I am wrong... This is frustrating me.
     
    #1 rockyDoctora, Jul 31, 2011
    Last edited: Jul 31, 2011
  2. Note: SDN Members do not see this ad.

  3. plzNOCarribbean

    Joined:
    May 21, 2010
    Messages:
    496
    Likes Received:
    4
    Status:
    Pre-Medical
    I follow where your logic is coming from and I think you are correct.

    I think if you assume full dissociation of ammonia and say its Twice as much as water, its essential like multiplying the Kf of ammonia by 2.

    2(-0.957) > -1.867

    This is a rather obscure and difficult question. Did the passage specifically state that ammonia caused complete acid dissociation and water didnt, or did they mention that it was twice as much as water. If you think about it, ammonia is a rather weak base, so this is quite the leap they are making.

    Ammonia is a weak acid (Pka ~ 38) and a weak base, since it is not charge. Its conjugate base, the amide is a very strong base but ammonia itself is not, which is why I find it rather questionable that they would assume complete dissociation in ammonia and not water, since both are uncharged species.
     
  4. rockyDoctora

    2+ Year Member

    Joined:
    Nov 10, 2009
    Messages:
    140
    Likes Received:
    0
    Status:
    Pre-Medical
    The passage mentioned that the acid dissociated more readily in ammonia than water because ammonia is a base.

    The more crucial question on my mind is why this dissociation is so significant.

    I am doing it right when I say that i_ammonia = 2 (at most)? Because the most the acid can dissociate to is by forming the solutes H+ and RCOO- while if it doesn't dissociate in water, the only solute is RCOOH.

    If this is wrong, then I would like someone to explain for me. If I'm right, then this really is a bad question.
     

Share This Page