The acid dissociation reaction for #HF# is the following:
#HF->H^(+)+F^-#
The #K_a# expression is: #K_a=([H^(+)][F^-])/([HF])=6.6xx10^(-4)#
The pH of this solution could be found by: #pH=-log[H^(+)]#.
We will need to find the #[H^(+)]# using #ICE# table:
#" " " " " " " " "HF" " " " " " "->" " " " H^(+) +" " " "F^-#
#"Initial" " "6.2xx10^(-3)M" " " " " " " "0M" " " " " "0M#
#"Change" " "-xM" " " " " " " " " "+xM" " " " "+xM#
#"Equilibrium" ""(6.2xx10^(-3)-x)M" " "xM" " " " " " "xM#
now, we can replace the concentrations by their values in the expression of #K_a#:
#K_a=([H^(+)][F^-])/([HF])=6.6xx10^(-4)=(x xx x)/(6.2xx10^(-3)-x)#
Solve for #x=1.7xx10^(-3)M#
#=>pH=-log[H^(+)]=-log(1.7xx10^(-3))=2.77#
Acids & Bases | pH of a Weak Acid.
