calculate the ph of 0.08M solution of hypochlorous acid,HOCL.The ionization constant of the acid is 2.5x 10 -5 .determine the % dissociation of HOCL?
Answers
Answered by
7
The dissociation equation of HOCI
Initial concentration (M)0.08M 0 0
Change to reach
Equilibrium concentration(M)-X+X+X
equilibrium concentration(M)0.08-X X X
Ka=(H3O^+)(CIO-)÷(HOCL)
Ka=X²/(0.08-x)
As x<<0.08 then 0.08-x=0.08
x^-/0.08=2.5×10^-5
x²=2.0×10^-6
x=1.41×10^-3
Hence (HOCI)dissociated =1.41×10^-3M &(H+)=1.41×10^-3M
Therefore % dissociation=(HOCI)dissociate/(HOCI)INITIAL)×100
=1.41×10^-3/0.08
pH=-log(H+)
pH=-log(1.41×10^-3)
pH=2.85
Initial concentration (M)0.08M 0 0
Change to reach
Equilibrium concentration(M)-X+X+X
equilibrium concentration(M)0.08-X X X
Ka=(H3O^+)(CIO-)÷(HOCL)
Ka=X²/(0.08-x)
As x<<0.08 then 0.08-x=0.08
x^-/0.08=2.5×10^-5
x²=2.0×10^-6
x=1.41×10^-3
Hence (HOCI)dissociated =1.41×10^-3M &(H+)=1.41×10^-3M
Therefore % dissociation=(HOCI)dissociate/(HOCI)INITIAL)×100
=1.41×10^-3/0.08
pH=-log(H+)
pH=-log(1.41×10^-3)
pH=2.85
Similar questions