Calculate the amount of NH42SO4 in grams which must be added to 500 ml of 0.2 M NH3 to yield a solution of pH = 9, Kb   for  NH3 = 2 × 10-5

Calculate the amount of NH42SO4 in grams which must be added to 500 ml of 0.2 M NH3 to yield a solution of pH = 9, Kb   for  NH3 = 2 × 10-5

  1. A

    32.48 g

  2. B

    42.48 g

  3. C

    13.20 g

  4. D

    62.48 g

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    Solution:

    POH= -log Kb + log NH4+NH4OH

    Let ‘a’ millimoles of NH4+ is added to a solution  

    millimoles of NH4OH = 500 × 0.2 = 100

        NH4+ = salt = a

    Given PH= 9 ;  POH= 5 

       5 = - log 2 × 10-5  + log a100

       a = 200  millimoles = 0.2 moles

    Moles of NH42SO4 added = a2=0.1   mol

       W NH42SO4 = 0.1 × 132 = 13.2g

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