What is the pH of a #6.2 * 10^-3# M solution of #HF#?

1 Answer
Jun 5, 2016

Answer:

#pH=2.77#

Explanation:

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.