When #"0.182 M"# of #"NaF"# (whose conjugate acid, #"HF"#, has #K_a = 6.76 xx 10^(-4)#) associates in water, what is the #"pH"#?

#"pH" ~~ 8.22#

What is the percent association of #"NaF"# at this concentration?

What you should recognize is that #"NaF"# is a base (although it should be obvious from the question wording). Therefore, you do NOT use the #K_a#. You use the #K_b#.

From the autodissociation of water,

#2"H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "OH"^(-)(aq)#

#K_w = K_aK_b#

And thus, at #25^@ "C"# and #"1 atm"# like usual,

#K_b = (10^(-14))/(6.76 xx 10^(-4)) = 1.48 xx 10^(-11)#

This is for #"NaF"# acting as a base in WATER. The ICE table in molarities is:

#"NaF"(aq) + "H"_2"O"(l) rightleftharpoons "NaOH"(aq) + "HF"(aq)#

#"I"" "0.182" "" "" "-" "" "" "0" "" "" "" "0#
#"C"" "-x" "" "" "-" "" "+x" "" "" "+x#
#"E"" "0.182-x" "-" "" "" "x" "" "" "" "x#

Its mass action expression is:

#K_b = 1.48 xx 10^(-11) = x^2/(0.182 - x)#

Since #K_b# #"<<"# #10^(-5)#, we can readily invoke the small #bbx# approximation without any error to get:

#1.48 xx 10^(-11) ~~ x^2/0.182#

Thus,

#x = ["OH"^(-)] ~~ sqrt(1.48 xx 10^(-11) cdot 0.182) " M"#

#= ul(1.64 xx 10^(-6) "M")#

This first allows you to get the #"pOH"#:

#"pOH" = -log["OH"^(-)] = 5.78#

But since at #25^@ "C"# and #"1 atm"#, #"pH" + "pOH" = 14#, we have:

#color(blue)("pH") = 14 - 5.78 = color(blue)ul(8.22)#

Does this make logical sense for a BASE? Why?