Shib Sagar posted an Question
August 31, 2020 • 00:46 am 30 points
  • IIT JAM
  • Chemistry (CY)

Please solve this problem

f ae group beca (b) KHC (d) CaCO, NaHCO, < KHCO, hjk

1 Answer(s) Answer Now
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  • Suman Kumar

    see this

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    Solubilities of carbonates and bicarbonates of alkali metals and alkaline earth metals increase down the group. 

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    Group-2 carbonates are sparingly soluble in water as their lattice energies are higher (it is due to increase in covalent nature).

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    so option D is correct Shib

  • comment-profile-img>
    Dinesh khalmaniya 1 Best Answer

    option D

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    sir please explain

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    wait..

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    video solution https://youtu.be/Kl6acJZCcZc

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    sir in this video they didn't told the reason

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    In the give metal bicarbonates, the anion (bicarbonate ion) is the same. Hence the solubility will depend on the size of cation. The size of Na+ ion is smaller than that of K+ ion. Hence sodium bicarbonate has high lattice energy and lower solubility than potassium bicarbonate.

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    CaCo3 is a strong ionic compound. Due to high electrostatic force of attraction the atoms are held so strongly together that you can not only break it easily but also cant dissole it. though CaCo3 decomposes in heat ,but it is insoluble in water in all conditions.

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    got it?

  • comment-profile-img>
    Priyanshu kumar

    Option D is correct answer shib

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    sir please explain

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    https://youtu.be/Kl6acJZCcZc check this shib

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    sir in this video they just told that which has highest and which has lowest solubility but didn't say the reason

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    Dear shib...solubility of bicarbonates is more than carbonates

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    And you know fajans rule....smaller the cation,larger the anion more the covalent character

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    and covalent character inversely proportional to solubility

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    between K+ and Na+...size of Na+ cation is smaller so it has more covalent charcter hence less solubility than khco3

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    so overall order becomes Khco3>Nahco3>caco3

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    now got this dear shib??

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