Question

Calculate the pH of the following:

a) 0.25 M HClO (aq)
b) 0.015 M Sr (OH) (aq)
c) 0.015 M Hl (aq)
d) 0.15 M CH3NH2 (aq)

Answer 1

Ka of JClO= 4*10-8

ICE table   

               HClO                      H+        ClO-

initial                            0.25                         0             0

Change                       -x                             x              x

Ka= x²/(0.25-x)= 4*10^-8 whem solved using excel, [H+] =1*10-4, pH= 4

2. Sr(OH)2----> Sr+2 + 2OH-

Strontium hydroxixe is strong base

hecne 0.015M gives rise to 2*0.015=0.03M OH-

pOH= -log(0.03)=1.523, pH= 14-1.523 =12.477

3. HI is strong acid hence [H+] =0.015, pH= -log(0.015)= 1.823

4. Kb of CH3NH2= 4.4*10-4

CH3NH2+ H2O----->CH3NH3+ + OH-

Kb= [CH3NH3+] [OH-]/[CH3NH2] = x²/ (0.15-x)= 4.4*10^-4,   x= 0.0076 when solved using excel

pOH= -log(0.0076)= 2.12, pH= 14-2.12= 11.88