There is a similar question on this site, but here i have a mixture.

I have a mixture of NH4F (35% weight) and NH4HF2 (8% weight) in water. How do I calculate the weight % expressed in HF? or the molar concentration of HF?

I arrived to calculate the mol/L of each component of the solution, being 10.49M of NH4F and 1.64M of NH4HF2. Now how do I calculate the HF conc?

Thank you

  • 2
    $\begingroup$ As usually in all but standard scenarios with ready to use expressions: Formulate all equilibrium equations, amount inventory equations and charge balance equation. Use substitution method to solve the equation set and to express desired concentration as the function of the known. Use simplifying approximations based on strong inequalities wherever applicable. $\endgroup$
    – Poutnik
    Nov 8, 2022 at 11:17
  • 3
    $\begingroup$ Not that at such high concentrations, all calculations are off by large extent. And I do mean large. $\endgroup$
    – Poutnik
    Nov 8, 2022 at 11:23
  • $\begingroup$ No problem about the accuracy of the calculation, i just need an idea of the weith or volume ratio expressend in NH4F:HF:H2O $\endgroup$
    – Ale
    Nov 8, 2022 at 11:29
  • $\begingroup$ @Ale Are the concentrations given initial or final? $\endgroup$
    – Sam202
    Nov 8, 2022 at 14:23
  • $\begingroup$ Concentrations given are initial ones $\endgroup$
    – Ale
    Nov 9, 2022 at 9:37

2 Answers 2


The initial equilibrium systems that lead to formation of $\ce{HF}$ as an intermediate are:

$$ \require{cancel} \begin{align} \ce{NH4HF2 (aq)&<=>\cancel{\ce{NH4+(aq)}}+\cancel{\ce{F- (aq)}}+HF (aq)} \tag{$K_1$} \\ \ce{\cancel{\ce{NH4+(aq)}}+\cancel{\ce{F- (aq)}} &<=> NH4F(aq)} \tag{$K_2$} \\ \hline \ce{NH4HF2(aq) &<=> NH4F(aq) + HF(aq)} \tag{$K_3$} \end{align} $$


$A$ represent $\ce{NH4HF2}$

$B$ represent $\ce{NH4F}$

$C$ represent $\ce{HF}$

$D$ represent $\ce{H+}$

$E$ represent $\ce{F-}$

$W$ represent $\ce{H2O}$

At equilibrium, the concentrations of $A,B,C$ are:




Using the relationship between the overall equilibrium constant $K_3$ and the individual ones, $K_1$ and $K_2$:


$\ce{HF}$ dissociates partially in aqueous solution:

$$\ce{HF (aq)<=>H+(aq) + F-(aq)}$$

The concentrations of $C,D,E$ after dissociation are:




Using relationship between equilibrium concentrations and $K_a$:


We can calculate $K_1$ and $K_2$ using the solubilities of $A$ and $B$ at 25°C:



Then, we calculate $K_3$:


Then, we calculate the initial density of the mixture (i.e. when only $A,B,W$ are present):



Next, we calculate $C_{Ao}$ and $C_{Bo}$:



Solving for $x$:



The $K_a$ of $\ce{HF}$ at 25°C is approximately:


So we can solve for $y$ now:



Finally, we can calculate the concentration of $\ce{HF}$ in the final solution:


  • $\begingroup$ @Poutnik $K_{sp}$ (in this case, $K_1$ and $K_2$) can be estimated using solubility of $\ce{NH4HF2}$ and $\ce{NH4F}$ in water. $\endgroup$
    – Sam202
    Nov 8, 2022 at 18:27
  • $\begingroup$ $\ce{NH4HF2(aq)}$ means $\ce{NH4+(aq) + HF2-(aq)}$ // $\ce{NH4F(aq)}$ means $\ce{NH4+(aq) + F-(aq)}$ $\endgroup$
    – Poutnik
    Nov 8, 2022 at 23:42
  • $\begingroup$ Some sources show $\ce{NH4HF2}$ as $\ce{NH4^+ + HF + F-}$. Check ionic structure. $\endgroup$
    – Sam202
    Nov 8, 2022 at 23:46
  • $\begingroup$ If there is not enough initial F- to deny dissociation of HF2-, according to the equilibrium. But the solution has plenty of F-. Reaction F- + HF <=> HF2- has K near 4. // Note that very concentrated HF very raises in acidity and pure liquid HF has acidity comparable to sulphuric acid, due 3 HF <=> H2F+ + HF2-. ( + forming $\ce{H_{n-1}F_{n}-}$ ) $\endgroup$
    – Poutnik
    Nov 8, 2022 at 23:54
  • $\begingroup$ I misunderstood one of the moderators. I thought that syntax was for units only. Thanks for the correction. $\endgroup$
    – Sam202
    Nov 9, 2022 at 0:02

There are these possible equilibrii, but the first 2 do not apply, as there are no solid salts. (5) can be neglected due low enough $\mathrm{pH}$.

\begin{align} \ce{NH4HF2(s) &<=> NH4+(aq) + HF2-(aq)} \tag{1}\\ \ce{NH4F(s) &<=> NH4+(aq) + F-(aq)}\tag{2}\\ \\ \ce{F-(aq) + HF(aq) &<=> HF2-(aq)}\tag{3}\\ \ce{HF(aq) &<=> H+(aq) + F-(aq)}\tag{4}\\ \\ \ce{NH4+(aq) &<=> NH3(aq) + H+(aq)}\tag{5} \end{align}

For the reaction (3), $\log{K} = 0.6$, i.e. $K \approx 3.98=\frac{[\ce{HF2-}]}{[\ce{HF}][\ce{F-}]}$ so very rough estimation of [HA] can be done.

$$[\ce{HF}]=\frac{[\ce{HF2-}]}{K[\ce{F-}]}=\pu{\frac{1.64}{3.98 \times 10.49} mol L-1 HF} \approx \pu{0.04 mol L-1 HF}$$

For the reaction (4), $\mathrm{p}K_\mathrm{a} = 3.17$

$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log{\frac{\ce{[F-]}}{\ce{[HF]}}}=3.17 + \log{\frac{10.49}{0.04}} \approx 5.6$

For the molar percentage of $\ce{HF}$, there is needed the solution density. You know it, or can measure it. Then it is trivial calculation, reversed to the ones you used to calculate molar concentrations of the salts.

$$w\%=\frac{c[\pu{mol L-1}]M[\pu{g mol-1}]}{\rho[\pu{g L-1}]} \cdot 100$$

An alternative approach, using Excel solver with Least Squares approach to solve the set of nonlinear equations:

Conc of variables values units equations
F- x 10.98 mol/L Ka = xt/z
HF2- y 1.32 mol/L K=y/xz
HF z 0.030 mol/L x+2y+z=a+2b
H+ t 0.0000018 mol/L x+y=a+b
5.7 pH
NH4F a 10.49 mol/L
NH4HF2 b 1.64 mol/L
Ka 0.000676083
K 3.98
  • $\begingroup$ OK, thank you. What about the weight ratio of HF, NH4F and H2O? $\endgroup$
    – Ale
    Nov 9, 2022 at 9:44

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