# 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.

Apr 9, 2015

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

$H C l {O}_{4}$ 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 $H C l {O}_{4}$ is about -10, while the pKa of ${F}^{-}$ is somewhere higher than 3.17, the pKa of $H F$. The pKas are not close enough that they are both considered weak acids and bases relative to each other; their ${K}_{a} s$ are over 13 orders of magnitude apart. In fact, it's more likely that $K F$ will exist in solution as ${F}^{-}$, while $H C l {O}_{4}$ donates a proton to form $H F$.

In the previous sense, $H F$ 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. $H C l {O}_{4}$ protonates the ionized ${F}^{-}$ from $K F$ and the buffer is formed then. It's kind of a tricky question.

Apr 9, 2015

Yes, this is a buffer with pH = 3.04.

$\text{KF}$ is the salt of a strong base and the weak acid $\text{HF}$.

The ${\text{F}}^{-}$ ion is basic and will react completely with a strong acid like ${\text{HClO}}_{4}$.

The $\textcolor{red}{\text{molecular equation}}$ is

$\text{KF(aq)" + "HClO"_4"(aq)" → "HF(aq)" + "KClO"_4"(aq)}$

The color(red)("ionic equation" is

$\text{K"^+"(aq)" + "F"^(-)"(aq)" + "H"_3"O"^+"(aq)" + "ClO"_4^(-)"(aq)" → "HF(aq)" + "H"_2"O(l)" + "K"^+"(aq)" + "ClO"_4^(-)"(aq)}$

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

$\text{F"^(-)"(aq)" + "H"_3"O"^+"(aq)" → "HF(aq)" + "H"_2"O(l)}$

The first problem is to figure out how much ${\text{F}}^{-}$ reacts and how much is unreacted.

$\text{Moles of F"^(-) = 0.045 cancel("L") × "0.60 mol"/(1 cancel("L")) = "0.027 mol}$

$\text{Moles of H"_3"O"^+ = 0.025 cancel("L") × "0.60 mol"/(1 cancel("L")) = "0.0015 mol}$ We have a buffer, because we have a solution of a weak acid $\text{HF}$ and its conjugate base ${\text{F}}^{-}$.

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

${\text{HF" + "H"_2"O" → "H"_3"O"⁺ + "F}}^{-}$; $\text{p"K_"a} = 3.14$

"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.