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NEET-XII-Chemistry

Previous Year Paper year:2017

with Solutions - page 4
Qstn# 62 Prvs-QstnNext-Qstn
  • #62
    pH of a saturated solution of ``Ca(OH)_2``
    is 9. The
    solubility product ``(K_{
    sp}) `` of ``Ca(OH)_2`` is:
    (1)`` 0.5 × 10^{-15 }
    (2) ``0.25 × 10^{-10}``
    (3)`` 0.125 × 10^{-15}``
    (4)`` 0.5 × 10^{-10}``
    digAnsr:   1
    Ans : ( 1 )
    Sol. + +⇀↽ 22Ca(OH) Ca 2OH
    pH = 9 Hence pOH = 14 - 9 = 5
    [OH-] = 10-5 M
    Hence [Ca2+] =
    5
    10
    2
    Thus K
    sp
    = [Ca2+][OH-]2
    =
    
     
     
     
    5
    5 210 (10 )
    2
    = 0.5 × 10-15