Question

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?

Answer 1

`sf([HF]=0.69color(white)(x)"mol/l")`

Explanation:

`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")`

Answer 2

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`.

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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")`