How can I calculate the amount of HF in aquoues solution of NH4F and an aquoues solution of NH4HF2?

Dissociation of $\ce{NH4F}$ and $\ce{NH4HF2}$ in water is complete, I assume:
$$\ce{NH4F -> NH4+ + F-}$$
and
$$\ce{NH4HF2 -> NH4+ + HF2-}$$


Can I calculate the amount of $\ce{HF}$ in these solutions from this equilibrium?


$$\ce{HF + F- <=> HF2-}$$


$$\mathrm{K\_2}=\dfrac{\ce{[HF^{2-}]}}{\ce{[HF][F^-]}}=4,7$$


If I know the concentrations of NH4F in water and NH4HF2 in water.




**WRONG SOLUTION: OP clarified that there are two separate solutions, not a mixture of the salts.** 


Assuming:


(1) that concentrations can be used instead of activities (bad assumption...)  

(2) The concentration of the ions $\ce{H2F-}$ and $\ce{F-}$ is so large that the final concentrations will be the same as the initial concentrations.  

(3) Since HF is a relatively weak acid, little $\ce{HF}$ will dissociate into $\ce{H+}$ and $\ce{F-}$.


Using equilibrium:
$$\mathrm{K\_2}=\dfrac{\ce{[HF^{2-}]}}{\ce{[HF][F^-]}}=4.7$$


From 5 molar $\ce{NH4F}$, $\ce{[F-] = \pu{5 molar}}$


From 0.18 molar $\ce{NH4HF2}$, $\ce{HF\_2^- = \pu{0.18 molar}}$


So plugging the number in:


$$\mathrm{K\_2}=\dfrac{\ce{[HF^{2-}]}}{\ce{[HF][F^-]}} = \dfrac{0.18}{\ce{[HF]}\times 5}=4.7$$


$$\ce{[HF] = 0.008 molar}$$


For the final concentrations $\ce{[HF] \ll [H2F-]}$, so the assumption that the dissociation of $\ce{H2F-}$ can be ignored is valid. The $\ce{HF}$ will dissociate and acidify the solution, but still $\ce{[HF] + [H+] \ll [H2F-]}$


 2

Let us calculate first the $\ce{NH4F}$ problem. The ion $\ce{F-}$ reacts partially with water according to


$$\ce{F- + H\_2O <=> HF + OH-}$$


The equilibrium constant $K\_\mathrm{b}$ of this equilibrium is defined by


$$K\_\mathrm{b} = \frac{[\ce{HF}][\ce{OH-}]}{[\ce{F-}]}$$


The numerical value of $K\_\mathrm{b}$ is needed, but it is not tabulated. However it can be obtained from the dissociation constant $K\_\mathrm{a}$ of $\ce{HF}$ in water, which can be found in the tables : $K\_\mathrm{a} = 10^{-3.17}$. So $K\_\mathrm{b} = K\_\mathrm{w}/K\_\mathrm{a} = 10^{-14}/ 10^{-3.17} = 10^{-10.83}$. This numerical value is now introduced in the definition of $K\_\mathrm{b}$


We can simplify by writing : $[\ce{HF}]=[\ce{OH-}]= x$. If we admit that $x$ is small, $[\ce{F-}]$ = $\pu{5 M}$. Solving this equation gives : $x^2 = 5·10^{-{10.83}}$ and $x = 3.84 \cdot 10^{-6}$. This is the expected result : In $\ce{KF}$ $\pu{3 M}$, the concentration of $\ce{HF}$ is $\pu{3.84 \times 10^{-6} M}$.


Let's now calculate the problem of $\ce{NH4HF2}$ in a solution $\pu{0.18 M}$, where the following equilibrium is given :


$$\ce{F- + HF <=> HF2-}$$


In this solution, there was $\pu{0.18 M}$ $\ce{HF2-}$ in the beginning, from which $x$ is transformed into $\ce{HF}$ and $\ce{F-}$. So the final concentration of $\ce{HF2-}$ is $0.18 - x$. The equilibrium constant $K\_2$ of this equilibrium is


$$K\_2 = \frac{[\ce{HF2-}]}{[\ce{HF}][\ce{F-}]} = \frac{0.18-x}{x^2} = 4.7$$


This is an equation to solve, and the solution is $x = 0.116$. So the concentration of $\ce{HF}$ in $\pu{0.18 M}$ $\ce{NH4HF2-}$ is $[\ce{HF}] = \pu{0.116 M}$


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