Asked by Natalie
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)
Answers
Answered by
DrBob222
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.
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.