Respuesta :
NH4^+ + OH^- ==> NH3 + H2O
Set up an ICE chart.
initial:
NH4^+ = mL x M = 100 x 0.1 = 10 millimoles
NH3 = mL x M = 80 x 0.20 = 16 millimoles.
change:
we add 0.2 g NaOH which is 0.2/40 = 0.005 moles or 5 millimoles.
NH3 = +5 millimoles
NH4^+ = -5 millimoles
equilibrium:
NH3 = 16 + 5 = 21 millimoles.
NH4^+ = 10 - 5 = 5 millimoles.
You may substitute millimoles in place of concn (since millimoles/mL = molarity and the mL (180 mL) appears in both numerator and denominator) OR you can divide millimoles/180 mL to arrive at concn for both base and acid and substitute those numbers. Plug those into the HH equation and solve for pH.
For part d, just set up the HH equation and pH = pKa + log (base/acid). The question is asking you to calculate pH if base and acid were equal. So plug in the same number (any number you choose) for base and acid and calculate. Note that the log of 1 = 0
Set up an ICE chart.
initial:
NH4^+ = mL x M = 100 x 0.1 = 10 millimoles
NH3 = mL x M = 80 x 0.20 = 16 millimoles.
change:
we add 0.2 g NaOH which is 0.2/40 = 0.005 moles or 5 millimoles.
NH3 = +5 millimoles
NH4^+ = -5 millimoles
equilibrium:
NH3 = 16 + 5 = 21 millimoles.
NH4^+ = 10 - 5 = 5 millimoles.
You may substitute millimoles in place of concn (since millimoles/mL = molarity and the mL (180 mL) appears in both numerator and denominator) OR you can divide millimoles/180 mL to arrive at concn for both base and acid and substitute those numbers. Plug those into the HH equation and solve for pH.
For part d, just set up the HH equation and pH = pKa + log (base/acid). The question is asking you to calculate pH if base and acid were equal. So plug in the same number (any number you choose) for base and acid and calculate. Note that the log of 1 = 0
