## Multi-Equilibrium Calculations for HF Solution and HF-F- Solution

Zhang Ying, Quan Xinjun,

Abstract

In HF solution and HF-F- solution, there are not only the dissociations of hydrofluoric acid but also the reactions between hydrogen fluoride and fluoride ion, so the equilibrium concentration of each component is the final product of two equilibria. Theoretically, the correct result can be obtained only if the two equilibria are considered simultaneously. However, much attention has been paid to the dissociation equilibrium of HF than to the HF-F- association in teaching, as a result students are used to take the wrong calculation method when confronting such problems. In order to develop students' comprehensive abilities to deal with the calculations of complex systems, the simple and accurate formulas for calculating the ion concentrations in HF solution and HF-F- solution are derived based on multi-equilibrium. By comparison, it is found that both equilibria must be considered in all cases, except that the concentration of HF solution is less than 1.91×10-2 mol·L-1 which can be approximately calculated as monobasic weak acid.

Keywords： Hydrofluoric acid ; Dissociation equilibrium ; Multi-equilibrium ; Complex system ; Calculation

Zhang Ying. Multi-Equilibrium Calculations for HF Solution and HF-F- Solution. University Chemistry[J], 2021, 36(8): 2009043-0 doi:10.3866/PKU.DXHX202009043

## 1 问题的由来

0.25 mol·L−1的HF溶液中[H+]为：

$K_{\mathrm{a}}^{\ominus}=\frac{\left[\mathrm{H}^{+}\right]^{2}}{c-\left[\mathrm{H}^{+}\right]}$

$\left[\mathrm{H}^{+}\right]=\sqrt{K_{\mathrm{a}}^{\ominus} \cdot c}$

## 2 HF水溶液氢离子浓度的计算

$\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]+\left[\mathrm{HF}_{2}^{-}\right]$

$c=\left[\mathrm{H}^{+}\right]+[\mathrm{HF}]+\left[\mathrm{HF}_{2}^{-}\right]$

$c=\left[\mathrm{F}^{-}\right]+[\mathrm{HF}]+2\left[\mathrm{HF}_{2}^{-}\right]$

\begin{aligned}&\text { 反应 }(\mathrm{III}): \mathrm{H}^{+}+\mathrm{F}^{-} \leftrightarrows \mathrm{HF} \\&\beta_{1}=\frac{[\mathrm{HF}]}{\left[\mathrm{H}^{+}\right]\left[\mathrm{F}^{-}\right]}=\frac{1}{K_{\mathrm{a}}^{\ominus}}\end{aligned}

\begin{aligned}&\text { 反应 }(\mathrm{IV}): \mathrm{H}^{+}+2 \mathrm{~F}^{-} \leftrightarrows \mathrm{HF}_{2}^{-} \\&\beta_{2}=\frac{\left[\mathrm{HF}_{2}^{-}\right]}{\left[\mathrm{H}^{+}\right]\left[\mathrm{F}^{-}\right]^{2}}=\frac{K^{\ominus^{\prime}}}{K_{\mathrm{a}}^{\ominus}}\end{aligned}

\begin{aligned}c &=\left[\mathrm{H}^{+}\right]+[\mathrm{HF}]+\left[\mathrm{HF}_{2}^{-}\right] \\&=\left[\mathrm{H}^{+}\right]+\beta_{1}\left[\mathrm{H}^{+}\right]\left[\mathrm{F}^{-}\right]+\beta_{2}\left[\mathrm{H}^{+}\right]\left[\mathrm{F}^{-}\right]^{2} \\&=\left[\mathrm{H}^{+}\right]\left(1+\beta_{1}\left[\mathrm{~F}^{-}\right]+\beta_{2}\left[\mathrm{~F}^{-}\right]^{2}\right)\end{aligned}

\begin{aligned}\left[\mathrm{H}^{+}\right] &=\left[\mathrm{F}^{-}\right]+\left[\mathrm{HF}_{2}^{-}\right] \\&=\left[\mathrm{F}^{-}\right]+\beta_{2}\left[\mathrm{H}^{+}\right]\left[\mathrm{F}^{-}\right]^{2}\end{aligned}

$\beta_{2} c\left[\mathrm{~F}^{-}\right]^{2}+\alpha\left[\mathrm{F}^{-}\right]-c=0$

α表达式代入式(10)，得：

$\frac{K^{\ominus^{\prime}}}{K_{\mathrm{a}}^{\ominus}}\left[\mathrm{F}^{-}\right]^{3}+\left(\frac{1}{K_{\mathrm{a}}^{\ominus}}+\frac{K^{\ominus^{\prime}}}{K_{\mathrm{a}}^{\ominus}} c\right)\left[\mathrm{F}^{-}\right]^{2}+\left[\mathrm{F}^{-}\right]-c=0$

$K^{\ominus \prime}\left[\mathrm{F}^{-}\right]^{3}+\left(1+K^{\ominus \prime} c\right)\left[\mathrm{F}^{-}\right]^{2}+K_{\mathrm{a}}^{\ominus}\left[\mathrm{F}^{-}\right]-K_{\mathrm{a}}^{\ominus} c=0$

 c/(mol·L−1) [F−]/(mol·L−1) 相对误差 [H+]/(mol·L−1) 相对误差 式(1) 式(2) 本法 式(1) 式(2) 本法 1.00 × 10−3 5.39 × 10−4 – 5.38 × 10−4 0.19% 5.39 × 10−4 – 5.39 × 10−4 0 5.00 × 10−3 1.49 × 10−3 – 1.47 × 10−3 1.36% 1.49 × 10−3 – 1.50 × 10−3 −0.67% 1.00 × 10−2 2.21 × 10−3 – 2.16 × 10−3 2.31% 2.21 × 10−3 – 2.25 × 10−3 −1.78% 1.91 × 10−2 3.17 × 10−3 – 3.02 × 10−3 4.97% 3.15 × 10−3 – 3.27 × 10−3 −3.67% 3.19 × 10−2 4.18 × 10−3 – 3.86 × 10−3 8.29% 4.19 × 10−3 – 4.41 × 10−3 −4.99% 5.00 × 10−2 5.31 × 10−3 – 4.71 × 10−3 12.7% 5.31 × 10−3 – 5.77 × 10−3 −7.97% 7.50 × 10−2 6.57 × 10−2 – 5.55 × 10−3 18.4% 6.57 × 10−2 – 7.44 × 10−3 −11.7% 0.100 7.63 × 10−3 – 6.17 × 10−3 23.7% 7.63 × 10−3 – 9.00 × 10−3 −15.2% 0.250 – 1.25 × 10−2 8.07 × 10−3 54.9% – 1.25 × 10−2 1.75 × 10−2 −28.6% 0.500 – 1.77 × 10−2 9.21 × 10−3 92.2% – 1.77 × 10−2 3.07 × 10−2 −42.3% 0.750 – 2.17 × 10−2 9.71 × 10−3 123% – 2.17 × 10−2 4.38 × 10−2 −50.5% 1.00 – 2.51 × 10−2 1.00 × 10−2 151% – 2.51 × 10−2 5.73 × 10−2 −56.2%

(1) 对于HF水溶液，无论是[F]还是[H+]，仅考虑HF的解离平衡与同时考虑HF溶液中的两个平衡相比计算结果总是存在误差(c =1.00 × 10−3 mol∙L−1例外是由于有效数字限制的缘故)，且HF溶液浓度越大，相对误差越大。

(2) [F]的误差为正误差，[H+]的误差为负误差。

(3) 当HF溶液浓度一定时，[F]误差的程度大于[H+]误差的程度。

(4) c =1.91 × 10−2 mol∙L−1时，[F]的相对误差为5%，刚好达到无机化学计算允许的最大误差，表明当c < 1.91 × 10−2 mol∙L−1时，可以近似地按一元弱酸同时计算F和H+离子浓度，而当c ≥ 1.91 × 10−2 mol∙L−1时，必须同时考虑HF溶液的两个平衡计算F和H+离子浓度。

(5) c ≥ 2.50 × 10−1 mol∙L−1时，c/Kaϴ ≥ 400，按照一元弱酸离子浓度最简式计算的[F]的相对误差大于等于54.9%，[H+]的相对误差亦大于等于−28.6%，远远超出无机化学计算允许误差要求，故最简式完全不适用于HF水溶液离子浓度计算。

## 3 HF-F−溶液氢离子浓度的计算

HF-F溶液氢离子浓度的计算也存在容易出错的情况。以HF-NaF溶液为例，若HF初始浓度为c1，NaF初始浓度为c2，以往在遇到这种情况时考虑到HF的酸性较强会根据式(13)计算[H+]：

$\left[\mathrm{H}^{+}\right]=K_{\mathrm{a}}^{\ominus} \frac{[\mathrm{HF}]}{\left[\mathrm{F}^{-}\right]}=K_{\mathrm{a}}^{\ominus} \frac{c_{1}-\left[\mathrm{H}^{+}\right]}{c_{2}+\left[\mathrm{H}^{+}\right]}$

$\left[\mathrm{H}^{+}\right]+\left[\mathrm{Na}^{+}\right]=\left[\mathrm{H}^{+}\right]+c_{2}=\left[\mathrm{F}^{-}\right]+\left[\mathrm{HF}_{2}^{-}\right]$

$c_{1}=\left[\mathrm{H}^{+}\right]+[\mathrm{HF}]+\left[\mathrm{HF}_{2}^{-}\right]\\$

$c_{1}+c_{2}=\left[\mathrm{F}^{-}\right]+[\mathrm{HF}]+2\left[\mathrm{HF}_{2}^{-}\right]$

\begin{aligned}c_{1} &=\left[\mathrm{H}^{+}\right]+[\mathrm{HF}]+\left[\mathrm{HF}_{2}^-\right] \\&=\left[\mathrm{H}^{+}\right]+\beta_{1}\left[\mathrm{H}^{+}\right]\left[\mathrm{F}^{-}\right]+\beta_{2}\left[\mathrm{H}^{+}\right]\left[\mathrm{F}^{-}\right]^{2} \\&=\left[\mathrm{H}^{+}\right]\left(1+\beta_{1}\left[\mathrm{~F}^{-}\right]+\beta_{2}\left[\mathrm{~F}^{-}\right]^{2}\right)\end{aligned}

\begin{aligned}\left[\mathrm{H}^{+}\right]+c_{2} &=\left[\mathrm{F}^{-}\right]+\left[\mathrm{HF}_{2}^{-}\right] \\&=\left[\mathrm{F}^{-}\right]+\beta_{2}\left[\mathrm{H}^{+}\right]\left[\mathrm{F}^{-}\right]^{2}\end{aligned}

$\beta_{2} c_{1}\left[\mathrm{~F}^{-}\right]^{2}+\alpha\left[\mathrm{F}^{-}\right]-\alpha c_{2}-c_{1}=0$

α表达式代入式(19)，得：

$\frac{K^{\ominus \prime}}{K_{\mathrm{a}}^{\ominus}}\left[\mathrm{F}^{-}\right]^{3}+\left(\frac{1}{K_{\mathrm{a}}^{\ominus}}+\frac{K^{\ominus^{\prime}}}{K_{\mathrm{a}}^{\ominus}} c_{1}-\frac{K^{\ominus^{\prime}}}{K_{\mathrm{a}}^{\ominus}} c_{2}\right)\left[\mathrm{F}^{-}\right]^{2}+\left(1-\frac{c_{2}}{K_{\mathrm{a}}^{\ominus}}\right)\left[\mathrm{F}^{-}\right]-c_{1}-c_{2}=0$

$K^{\ominus \prime}\left[\mathrm{F}^{-}\right]^{3}+\left(1+K^{\ominus \prime} c_{1}-K^{\ominus \prime} c_{2}\right)\left[\mathrm{F}^{-}\right]^{2}+\left(K_{\mathrm{a}}^{\ominus}-c_{2}\right)\left[\mathrm{F}^{-}\right]-K_{\mathrm{a}}^{\ominus}\left(c_{1}+c_{2}\right)=0$

 c(HF)/(mol·L−1) c(NaF)/(mol·L−1) c(HF)/c(NaF) [H+]/(mol·L−1) 相对误差 本法* 式(13) 1.00 × 10−2 0.100 0.100 4.33 × 10−5 6.26 × 10−5 44.6% 2.50 × 10−2 0.100 0.250 1.14 × 10−4 1.56 × 10−4 36.8% 5.00 × 10−2 0.100 0.500 2.56 × 10−4 3.12 × 10−4 21.9% 7.50 × 10−2 0.100 0.750 4.22 × 10−4 4.67 × 10−4 10.7% 0.100 0.100 1.00 6.19 × 10−4 6.22 × 10−4 0.485% 0.150 0.100 1.50 1.10 × 10−3 9.30 × 10−4 −15.5% 0.250 0.100 2.50 2.44 × 10−3 1.54 × 10−3 −36.9% 0.500 0.100 5.00 7.79 × 10−3 3.04 × 10−3 −61.0% 0.750 0.100 7.50 1.53 × 10−2 4.49 × 10−3 −70.7% 1.00 0.100 10.0 2.44 × 10−2 5.91 × 10−3 −75.8% 0.100 1.00 × 10−2 10.0 5.38 × 10−3 4.24 × 10−3 −21.2% 0.100 1.33 × 10−2 7.50 4.63 × 10−3 3.59 × 10−3 −22.5% 0.100 2.00 × 10−2 5.00 3.50 × 10−3 2.70 × 10−3 −22.9% 0.100 4.00 × 10−2 2.50 1.85 × 10−3 1.50 × 10−3 −18.9% 0.100 8.00 × 10−2 1.25 8.30 × 10−4 7.74 × 10−4 −6.75% 0.100 0.100 1.00 6.19 × 10−4 6.22 × 10−4 0.485% 0.100 0.133 0.750 4.16 × 10−4 4.70 × 10−4 13.0% 0.100 0.200 0.500 2.20 × 10−4 3.14 × 10−4 42.7% 0.100 0.400 0.250 6.88 × 10−5 1.57 × 10−4 128% 0.100 1.00 0.100 1.20 × 10−5 6.30 × 10−4 425%

*为了保证计算结果具有较高的准确度，在第一步计算[F]过程中，当[F]与c(NaF)数值接近时，适当增加了[F]计算结果有效数字的选取位数

(1) 对于HF-F溶液，在所讨论的浓度和比例范围内，采用式(13)与同时考虑HF溶液中的两个平衡相比计算结果总是存在误差，且相对误差的大小既与缓冲比c(HF)/c(NaF)有关，也与c(HF)和c(NaF)的大小有关。其中[H+]的相对误差以缓冲比等于1时为最小，当缓冲比偏离于1时相对误差都增大。

(2) 在我们所计算的20个溶液中，除了c(HF)/c(NaF) = 1时外，其他溶液仅考虑HF的解离平衡，与同时考虑HF溶液中的两个平衡相比[H+]计算结果相对误差都大于无机化学计算的允许误差，说明式(13)对于HF-F溶液来说是不适用的。

## 4 结语

(1) 氢氟酸非一般意义上的一元弱酸，在HF溶液和HF-F溶液中存在着HF的解离和HF与F化合两个反应，溶液中各物种的平衡浓度都是两个平衡共同作用的产物，只有将两个平衡同时考虑才能得到正确的结果。

(2) 在多重平衡计算中选取有关平衡中相同离子的浓度做未知数有利于方程的消元，继而方便利用在线计算器进行计算。

(3) 对于HF溶液，只有当c < 1.91 × 10−2 mol∙L−1时，才可以近似地按一元弱酸同时计算[H+]和[F]，而最简式[H+] = (Kac)1/2完全不适合氢氟酸溶液中氢离子浓度的计算。

(4) 对于计算HF-F溶液[H+]来说，除c(HF)/c(NaF) = 1时外，仅考虑HF解离平衡得到的计算公式是不适用的。

(5) 在离子平衡的计算中，要关注次反应对主反应的影响。除了本文所讨论的HF溶液和HF-F溶液外，诸如计算AgCl在NaCl溶液中的溶解度、HgS在NaS溶液中的溶解度等也需要分别考虑[AgCl2]、[HgS2]2−对AgCl、HgS沉淀溶解平衡的影响。这些过程既涉及比较复杂的计算，又涉及元素化合物的性质，对于培养学生全面分析问题的能力很有帮助。

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