Question

Buffer or no buffer? `"45 mL"` of `"0.60 M"` `"KF"` reacts with `"25 mL"` of `"0.60 M"` `"HClO"_4` to yield `"HF"` and `"KClO"_4`, all aqueous.

Answer 1

Yes. More info up above.

A buffer is a weak acid/base plus its salt. Like a mixture of acetic acid and sodium acetate.

`HClO_4` is a strong acid

KF is not a salt of this acid either

A buffer is a weak acid/base plus its soluble conjugate base/acid (which usually works well with an alkali metal as the cation).

The pKa of `HClO_4` is about -10, while the pKa of `F^-` is somewhere higher than 3.17, the pKa of `HF`. The pKas are not close enough that they are both considered weak acids and bases relative to each other; their `K_as` are over 13 orders of magnitude apart. In fact, it's more likely that `KF` will exist in solution as `F^-`, while `HClO_4` donates a proton to form `HF`.

In the previous sense, `HF` and `F^-` end up forming a buffer, which, as stated on the answer above, has a pH of 3.04, which reflects the pKa of HF of 3.17. `HClO_4` protonates the ionized `F^-` from `KF` and the buffer is formed then. It's kind of a tricky question.

Answer 2

Yes, this is a buffer with pH = 3.04.

`"KF"` is the salt of a strong base and the weak acid `"HF"`.

The `"F"^-` ion is basic and will react completely with a strong acid like `"HClO"_4`.

The `color(red)("molecular equation")` is

`"KF(aq)" + "HClO"_4"(aq)" → "HF(aq)" + "KClO"_4"(aq)"`

The `color(red)("ionic equation"` is

`"K"^+"(aq)" + "F"^(-)"(aq)" + "H"_3"O"^+"(aq)" + "ClO"_4^(-)"(aq)" → "HF(aq)" + "H"_2"O(l)" + "K"^+"(aq)" + "ClO"_4^(-)"(aq)"`

The `color(red)("net ionic equation"` is

`"F"^(-)"(aq)" + "H"_3"O"^+"(aq)" → "HF(aq)" + "H"_2"O(l)"`

The first problem is to figure out how much `"F"^-` reacts and how much is unreacted.

`"Moles of F"^(-) = 0.045 cancel("L") × "0.60 mol"/(1 cancel("L")) = "0.027 mol"`

`"Moles of H"_3"O"^+ = 0.025 cancel("L") × "0.60 mol"/(1 cancel("L")) = "0.0015 mol"`

We have a buffer, because we have a solution of a weak acid `"HF"` and its conjugate base `"F"^-`.

To calculate the pH, we use the Henderson-Hasselbalch Equation.

`"HF" + "H"_2"O" → "H"_3"O"⁺ + "F"^-`; `"p"K_"a" = 3.14`

`"pH" = "p"K_"a" + log(("[F"^(-)"]")/"[HF]") = 3.14 + log((0.012 cancel("mol"))/(0.015 cancel("mol"))) = 3.14 – 0.097 = 3.04`

Note that the ratio of the concentrations is the same as the ratio of the moles, because both species are in the same solution.