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What is pH of 0.1M solution of NH4Br, if the pKb of NH3 is 4.74?

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    .......NH4^+ + HOH ==> NH3 + H3O^+
    I......0.1..............0.....0
    C.......-x..............x.....x
    E.....0.1-x.............x......x

    Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(NH4^+)
    Substitute and solve for x = (H3O^+) and convert to pH.

    note: If pKb = 4.74 then Kb = 1.8E-5

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