Answer:
a) pH = 8.823
b) pH = 11.653
c) pH = 5.033
Step-by-step explanation:
a) HBr is strong acid:
HBr + H2O ↔ H3O+ + Br-
∴ [ HBr ] ≅ [ H3O+ ] = 1.5E-9 M
⇒pH = -log (1.5E-9) = 8.823
b) KOH is a strong base:
KOH + H2O ↔ K+ + OH-
⇒pOH = -log [ OH- ] = -log ( 0.0045 ) = 2.346
∴ 14 = pH + pOH
⇒ pH = 14 - 2.346 = 11.653
c) conjugated base:
NH4CL + H2O ↔ HCl + NH4OH
Kb = 1.75E-5
Kw = 1E-14
⇒ Kh = Kw/Kb = 5.714E-10
∴ Kh = [ H3O+ ]² / ( 0.15 - [ H3O+ ] )......from mass balanced and load balanced
⇒ [ H3O+ ]² = (5.714E-10) * ( 0.15 - [ H3O+ ] )..... quadratic formula
⇒ [ H3O+ ] = 9.2577E-6 M
⇒ pH = -log (9.2577E-6) = 5.033