Determine the ph of the following solutions: (a) 1.5 x 10-9 m hbr (aq) (think about this one carefully) (b) 0.0045 m koh (aq) (c) 0.15 m nh ci (aq)

a) pH = 8.823

b) pH = 11.653

c) pH = 5.033

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

pie chart because it is in percents ( % )

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying