# 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.

• Is this a homework or book question? If so we'll need to see the exact question to help. Also we have a "homework" policy that requires you to show some work. // If this is some sort of other inquiry, please explain the context better so we can figure out how to help.
– MaxW
Mar 23, 2020 at 18:29
• It's not homework. It's for my work. A reviewer asked me to calculate this for a paper I want to publish. But I'm so lost, I haven't done these calculations since the first year of Uni. Like I have some ideas on how to calculate this, but I'm not entirely sure. I prepared the solutions of NH4F and NH4HF2 in water and I need to calculate the amount of HF in these solutions. The concentrations of NH4F is 5 M and NH4HF2 is 0,18 M. Mar 23, 2020 at 18:35
• Ok, for a paper. In that case you'll want "reasonable" calculations not just some tripe 'theoretical" calculations. In other words, your solutions are so concentrated that you'll need activities, not the raw concentrations to do the calculations. Not sure where you'd find that data.
– MaxW
Mar 23, 2020 at 18:47
• I think theoretical are enough, the reviewer asked to estimate the amount of HF. Mar 23, 2020 at 18:49
• $$\mathrm{K_2}=\dfrac{\ce{[HF2^-]}}{\ce{[HF][F^-]}}=4.7$$ Mar 24, 2020 at 16:30

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 $$\ce{[HF] + [H+] \ll [H2F-]}$$

• The NH4F and NH4HF2 are two separate solutions, c(NH4F)=5M, and c(NH4HF2)=0.18 M. For each of these solutions, that are not mixed I need to determine the concentration of HF. In the expression K2 for NH4F solution I only know the concentration of F- which is 5, but for conc. of HF2- and HF are unknown. This is where I get lost, because I don't know if I can assume this concentrations are maybe the same, or one is the same as F- Mar 23, 2020 at 21:51

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 $$[\ce{HF}] = \pu{0.116 M}$$

• Don't think that you can neglect $\ce{NH4+}$ as a weak acid.
– MaxW
Mar 24, 2020 at 20:27