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A buffer solution is 1.00 M in NH3 and 1.10 M in NH4Cl. If 0.230 moles of NaOH are added to 1.00 L of the buffer, what is its pH? Assume the volume remains constant. Kb of NH3 = 1.8 multiplied by 10-5.

  • chemistry -

    Use the Henderson-Hasselbalch equation.
    NH3 + H^+ ==> NH4^+
    NH3 = 1.00 M from the problem.
    H^+ = whatever, but can be calculated if you want to know.
    NH4^+ = 1.10 M from the problem.
    Just for the fun of it I plugged in 1.00 for B and 1.10 for acid and found pH = 9.21.
    For the problem, if we add 0.230 moles NaOH, we must REMOVE 0.230 H^+, which shifts the above equilibrium to the left. NH4^+ is decreased by 0.230 and NH3 is increased by 0.230.
    Final moles NH3 = 1.00 + 0.230 = ??
    Final moles NH4^+ = 1.10 - 0.230 = ??
    Substitute into HH equation and solve. I get something close to 9.4 but check me out.

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