What is the molarity of an aqueous ammonia solution that has an OH- concentration of 0.0011 M? Kb = 1.8x10-5
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.........NH3 + HOH ==> NH4^+ + OH^-
initial...x.............0.......0
change.-1.1E-3.........1.1E-3..1.1E-3
equil..x-1.1E-3........1.1E-3..1.1E-3
Kb = (NH3)(OH^-)/(NH3)
Substitute into the Kb expression and solve for x = (NH3)