If you add 0.42 g NH4Cl (MW = 53.5 g/mol) to 200 mL of 0.34 M NH3 solution,?

Question

For NH3, Kb = 1.8 x 10-5

a. 3.81
b. 4.74
c. 7.92
d. 9.25
e. 10.19

Answer 1

e. 10.19

pH = pKa + log([NH3]/[NH4+])
pH = -log 5.55 x 10^-10 + log((0.068)/(0.00785))
pH = 10.19

Ka = Kw/Kb = 1 x 10^-14/1.8 x 10-5