Hydrogen Fluoride Buffer Solutions
The subject of Dynamics And Equilibria is more complex than this, and a complete treatment of that subject will not be attempted here.
A "complete" equilibrium constant which honestly describes the state of an HF buffer solution, derives from the chemical reaction:
[math]\displaystyle{ HF\rightleftharpoons F^{1-}+H^{+} }[/math]
And takes the form:
[math]\displaystyle{ K_{c}=\frac{\left[F^{1-}\right]\left[H^{+}\right]}{\left[HF\right]} }[/math],
Where the square brackets denote the concentration of the particles enclosed. But, a quantity which is more frequently stated is, the ratio:
[math]\displaystyle{ K_{a}=\frac{\left[F^{1-}\right]}{\left[HF\right]} }[/math],
Which is exact when the solution is neutral, and when the pH of water is (7). Thus, [math]\displaystyle{ \left(K_{a}\right) }[/math] is exact when [math]\displaystyle{ \left[H^{+}\right]=10^{-7} }[/math]. Some sources have stated [math]\displaystyle{ \left(K_{a}\right)=6.6\cdot10^{-4} }[/math], which means that its potency logarithm, [math]\displaystyle{ \left(pK_{a}\right)=3.18 }[/math]. But what some sources may state is, that it's difficult to measure [math]\displaystyle{ \left(K_{a}\right) }[/math] directly, because as soon as one has added some amount of [math]\displaystyle{ \left(HF\right) }[/math] to the solution, the solution becomes acidic, and [math]\displaystyle{ \left[H^{+}\right]\gt 10^{-7} }[/math]. If we were to assume that [math]\displaystyle{ \left(K_{a}\right)==6.6\cdot10^{-4} }[/math] when [math]\displaystyle{ \left[H^{+}\right]=10^{-7} }[/math], then it will also follow that...
[math]\displaystyle{ K_{c}=K_{a}\cdot10^{-7}=6.6\cdot10^{-11} }[/math].
[math]\displaystyle{ \left(pK_{c}\right) }[/math] Or [math]\displaystyle{ \left(K_{c}\right) }[/math] can be used to describe the general behavior of the buffer solution. Thus, if it's desired that [math]\displaystyle{ \left[H^{+}\right]=10^{-4} }[/math], then this value can just be plugged in to the equation for [math]\displaystyle{ \left(K_{c}\right) }[/math], and the ratio between [math]\displaystyle{ \left[F^{1-}\right] }[/math] and [math]\displaystyle{ \left[HF\right] }[/math] defined. The problem which then remains in preparing the buffer solution is, that neither [math]\displaystyle{ \left[F^{1-}\right] }[/math] nor [math]\displaystyle{ \left[HF\right] }[/math] will equal the amount of the compounds, say, [math]\displaystyle{ \left(NaF\right) }[/math] and [math]\displaystyle{ \left(HF\right) }[/math] that were initially added. The increase of [math]\displaystyle{ \left[H^{+}\right] }[/math] above [math]\displaystyle{ \left(10^{-7}\right) }[/math] may consume some [math]\displaystyle{ \left(HF\right) }[/math] and add to the final concentration of [math]\displaystyle{ \left(F^{1-}\right) }[/math]. In that case, what the lab technicians may do is:
- Rely on the effectiveness of the buffer solution to compensate. Or,
- Assume that acid has been added by itself, restate [math]\displaystyle{ \left[HF\right] }[/math] as a more complex expression which depends on the known amount of that, minus [math]\displaystyle{ \left[H^{+}\right] }[/math], pretend that [math]\displaystyle{ \left[F^{1-}\right] }[/math] and [math]\displaystyle{ \left[H^{+}\right] }[/math] are equal as the numerator of [math]\displaystyle{ \left(K_{c}\right) }[/math], and solve the quadratic to determine [math]\displaystyle{ \left[H^{+}\right] }[/math]. Or,
- Subtract [math]\displaystyle{ \left[H^{+}\right] }[/math] from [math]\displaystyle{ \left[F^{1-}\right] }[/math] in order to determine how much conjugate base must be added to the solution. In this case, adding a negative amount of base isn't possible.
If [math]\displaystyle{ \left[H^{+}\right] }[/math] is close to its neutral value, their initial difference may be neglected as part of an approximation (but the buffer solution still offer increased resistance to fatigue as the acid gets used).
Also, it can be explained that if [math]\displaystyle{ \left[HF\right] }[/math] is to be maximized, say, in order to improve its glass-etching performance, then actually adding a strong acid will be helpful, as doing so increases [math]\displaystyle{ \left[H^{+}\right] }[/math], thereby elevating [math]\displaystyle{ \left[HF\right] }[/math] with respect to [math]\displaystyle{ \left[F^{1-}\right] }[/math].
The example below is a scratch sheet, for formulating a buffer solution, where it's allowable for [math]\displaystyle{ \left[HF\right] }[/math] actually to be [math]\displaystyle{ \left(10\;mol\:/L\right) }[/math], and where the target [math]\displaystyle{ \left[H^{+}\right] }[/math] is more realistically [math]\displaystyle{ \left(10^{-5}\right) }[/math]...
- [math]\displaystyle{ K_{c}=\frac{\left[F^{1-}\right]\left[H^{+}\right]}{\left[HF\right]}=6.6\cdot10^{-11} }[/math]
- [math]\displaystyle{ \left[H^{+}\right]=10^{-5} }[/math]
- [math]\displaystyle{ \left[HF\right]=1.0\cdot10^{+1}\;mol\:/L }[/math]
- [math]\displaystyle{ \frac{\left[F^{1-}\right]}{mol\:/L}=\frac{6.6\cdot10^{-11}}{10^{-5}}\cdot10^{+1} }[/math]
- [math]\displaystyle{ =6.6\cdot10^{-5} }[/math]
- [math]\displaystyle{ \frac{\left[F^{1-}\right]}{mol\:/L}-\left[H^{+}\right]=5.6\cdot10^{-5} }[/math]
A type of conclusion which can be drawn is, that specifically in the case of Hydrogen Fluoride, the buffer solution's main purpose is to reduce the acidity, and thus to reduce the rate at which the solution etches, below the rate at which an uncontrolled acid would etch, and on the basic side of neutrality for water. This has clear applications in, say, IC manufacture.
