ChemistryCalculate pH of 0.002 N   NH4OH having  2%  dissociation (log 4 = 0.6)

Calculate pH of 0.002 N   NH4OH having  2%  dissociation (log 4 = 0.6)

  1. A

    7.6

  2. B

    8.6

  3. C

    9.6

  4. D

    10.6

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    Solution:

    NH4OH is a weak base and partially dissociated

                                      NH4OH    NH4+ + OH-

    Concentration              1                    0            0

    before dissociation

    Concentration            1-α                α          α

    after dissociation

    C=2×10-3  ; α=2100       OH- = Cα = 2 × 10-3 × 2100 = 4 × 10-5 M

    pOH = -log OH-

            = -log 4 × 10-5 = 4.4 pH=14-4.4 = 9.6

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