Calculate the concentration of HF in aqueous solution of NH4F and NH4HF2 mixture

Question

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

Answer 1

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

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

Let:

$A$ 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:

$$C_{A_1}=C_{A_o}-x$$

$$C_{B_1}=C_{B_o}+x$$

$$C_{C_1}=x$$

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

$$K_3=K_1\;K_2=\frac{C_{B_1}\;C_{C_1}}{C_{A_1}}=\frac{(C_{B_o}+x)\;x}{C_{A_o}-x}$$

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

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

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

$$C_{C_2}=C_{C_1}-y=x-y$$

$$C_{D_2}=y$$

$$C_{E_2}=y$$

Using relationship between equilibrium concentrations and $K_a$:

$$K_{a}=\frac{C_{D_2}\;C_{E_2}}{C_{C_2}}=\frac{y^2}{x-y}$$

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

$$K_1=\left(\frac{S_A}{M_A}\right)^2=\left(\frac{\pu{630g/L}}{\pu{57g/mol}}\right)^2=122.2$$

$$K_2=\left(\frac{S_B}{M_B}\right)^{-2}=\left(\frac{\pu{835g/L}}{\pu{37g/mol}}\right)^{-2}=\pu{1.97e-3}$$

Then, we calculate $K_3$:

$$K_3=K_1\;K_2=(122.2)\;(\pu{1.97e-3})=0.241$$

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

$$\rho_o=X_{Ao}\;\rho_A+X_{Bo}\;\rho_B+X_{Wo}\;\rho_W=(0.08)(\pu{1500g/L})+(0.35)(\pu{1010g/L})+(0.57)(\pu{1000g/L})$$

$$\rho_o=\pu{1043.5g/L}$$

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

$$C_{Ao}=x_{Ao}\;\frac{\rho_o}{M_A}=0.08\;\frac{\pu{1043.5g/L}}{\pu{57g/mol}}=\pu{1.46mol/L}$$

$$C_{Bo}=x_{Bo}\;\frac{\rho_o}{M_B}=0.35\;\frac{\pu{1043.5g/L}}{\pu{37g/mol}}=\pu{9.87mol/L}$$

Solving for $x$:

$$\frac{(9.87+x)\;x}{1.46-x}=0.241$$

$$x=\pu{0.0347mol/L}$$

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

$$K_a=\pu{6.76e-4}$$

So we can solve for $y$ now:

$$\frac{y^2}{0.0347-y}=\pu{6.76e-4}$$

$$y=\pu{0.00452mol/L}$$

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

$$C_{C_2}=x-y=\pu{(0.0347-0.00452)mol/L}=\pu{0.0302mol/L}$$

Answer 2

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.

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

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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