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.
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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