PH = = PKa+LG[c(HB)/c(NaB)]= = PKa = = 5

(2)HB & lt； = = & gtH+ + B-

[H+] == 10^-5mol/L

C (HB) = = 0.1* 50/150 =1/30 mol/L.

Dissociation degree =10-5/(1/30) *100% = = 0.03%.