If you want to prepare a solution that has #"pH" = 8.51# at #25^@ "C"#, what initial concentration of #"NaF"# would you need to achieve?

#sf(F^-)# is the co - base of #sf(HF)#:

#sf(F^(-)+H_2OrightleftharpoonsHF+OH^-)#

You need to look up #sf(pK_b)# for the #sf(F^-)# ion which = 10.82.

From the ICE table you get this expression:

#sf(pOH=1/2(pK_a-log[F^-])#`

#sf(pOH=14-pH=14-8.51=5.49)#

#:.##sf(5.49=1/2(10.82-log[F^-])#

#sf(1/2log[F^-]=-0.08)#

#sf(log[F^-]=-0.16)#

#:.##sf([F^-]=0.69color(white)(x)"mol/l")#

I am also getting approximately #"0.70 M"#. Specifically I get #"0.708 M"#. You might want to try doing the problem backwards and work from the solved concentration to check that the #"pH"# still turns out to be close to #8.51#.

As I mentioned earlier, you were given the pH which allows you to find the equilibrium concentration of #"H"^(+)#. Note the main equilibrium in this problem is:

#"F"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HF"(aq) + "OH"^(-)(aq)#

#"I"" "c" M"" "" "" "-" "" "" ""0 M"" "" ""0 M"#
#"C"" "-x " ""M"" "-" "" "+x " M"" "+x" M"#
#"E"" "(c - x) "M"" "-" "" "" "x" M"" "" "x" M"#

where we let #c# be the initial concentration of #"F"^(-)#, which is equal to that of #"NaF"# by their 1:1 stoichiometry.

You should have the #K_a# in your book, but I shall use the #"pKa"# of #3.17# for #"HF"#, which corresponds to a #"pKb"# of #10.83# for #"F"^(-)#, since #"pKw" = "pKa" + "pKb" = 14# at #25^@ "C"#.

Therefore, #K_b = 10^(-"pKb") = 10^(-10.83) = 1.479xx10^(-11)#. So, the #K_a# in your book should be around #6.671xx10^(-4)#. Now, using the definition of #K_b#:

#K_b = (["HF"]["OH"^(-)])/(["F"^(-)]) = x^2/(c - x)#

Since we were given that #"pH" = 8.51#, we know that #"pOH" = 14 - "pH" = 5.49# at #25^@ "C"#. Therefore:

#["OH"^(-)]# at equilibrium is #10^(-"pOH") = 3.236xx10^(-6)# #"M" = x#.

This means we already solved for #x#. So, we can plug it back into #K_b#:

#K_b = (3.236xx10^(-6) "M")^2/(c - 3.236xx10^(-6) "M") = 1.479xx10^(-11)#

By the small #K_b# approximation, since #K_b "<<" 10^(-5)#, we can say that the true answer allows us to neglect the #-3.236xx10^(-6) "M"# term that is being subtracted in the denominator with minimal error.

As a result, our evaluation simplifies to (where #K_b# implicitly has units of #"M"# since it had units of #"M"^2/"M"# from its definition):

#(1.479xx10^(-11) "M")c = (3.326xx10^(-6) "M")^2#

#=> color(blue)(c = ["F"^(-)] = ["NaF"] = "0.708 M")#