What is the magnitude change in pH when 1.8g of NaOH is added 1L of a solution that is 0.100M in NH3 and 0.100 in NH4Cl? Kb for NH3 1.8 * 10^-5

Answer 1

pH of the original solution before NaOH is added is
pH = pKa + log (base)/(acid)
pH = 9.26 + log (0.1/0.1) = 9.26

millimols NH3 = 0.1 x 1000 = 100
mmols NH4Cl = 0.1 x 1000 = 100
mols NaOH added = 1.8/40 = 0.045 mols or 45 millimols.

.............NH4^+ + OH^- = NH3 + H2O
I............100......0......100
add..................45.............
C............-45....-45.......+45
E.............55......0........45

Plug the E line into the HH equation and solve for pH of the solution, then you want the difference between that and the original pH.

Answer 2

thank you