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What is the original molarity of an aqueous solution of ammonia (NH3) whose pH is 11.10 at 25 degrees C? (Kb for NH3=1.8 x 10^-5)

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    NH3 + H2O ==> NH4^+ + OH^-

    Kb = (NH4^+)(OH^-)/(NH3) = 1.8E-5

    Convert pH 11.10 to pOH, then to OH, substitute into Kb expression for NH4^+ and OH^-, and solve for (NH3), the only unknown in the equation.

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