Given $\rm pK_a(HF) = 3.17$; the equilibrium constant for the reaction $\rm HF + F^- \rightleftharpoons HF_2^-$ is $\rm K_c ^\circ = 3.98$.

(1) To $\rm 100~mL~0.200~mol\cdot L^{-1}~HF$, solid $\rm KOH$ is added (volume change is negligible, same below) to obtain buffer solution 1 with $\rm pH = 3.50$. The mass of $\rm KOH$ required is $u~\rm g$.

(2) Given $\rm K_{sp}(LiF) = 1.84 \times 10^{-3}$, assuming the pH of the solution remains unchanged, and using the mass balance relationship of the solution, the solubility $\rm s$ ($\rm g\cdot L^{-1}$) of $\rm LiF$ in solution 1 can be calculated as $v~\rm g\cdot L^{-1}$.

(3) Assuming that solid $\rm NaF$ is added to buffer solution 1 instead, such that $[\rm Na^+]$ is equal to the $[\rm Li^+]$ calculated in the previous question, then the pH at equilibrium is $w$.

Then $vw/u$ is approximately (If the calculated result has an error greater than 5% from all options, choose A)
已知 $\rm pK_a(HF) = 3.17$；反应 $\rm HF + F^- \rightleftharpoons HF_2^-$ 的平衡常数$\rm K_c ^\circ = 3.98$。

（1）向$\rm 100~mL~0.200~mol\cdot L^{-1}~HF$中加入$\rm KOH$固体（忽略体积改变，下同）以得到$\rm pH = 3.50$的缓冲溶液1，所需$\rm KOH$的质量为$u~\rm g$。

（2）已知$\rm K_{sp}(LiF) = 1.84 \times 10^{-3}$，假定溶液pH不变，使用溶液的物料守恒关系，可算得$\rm LiF$在溶液1中的溶解度$\rm s$ ($\rm g\cdot L^{-1}$)为$v~\rm g\cdot L^{-1}$。

（3）假定改为向缓冲溶液1中加入$\rm NaF$固体使$[\rm Na^+]$与上问算得的$[\rm Li^+]$相等，则平衡时的pH为$w$。

则$vw/u$约为（若算得的结果与所有选项误差均大于5%则选A）
All other options are incorrect
1.47
2.08
0.26
1.92
4.06
3.88
0.67
3.12
2.50
4.44
5.61
6.06
1.71
3.54
7.89
其他选项均不正确
1.47
2.08
0.26
1.92
4.06
3.88
0.67
3.12
2.50
4.44
5.61
6.06
1.71
3.54
7.89
(1)

Let the final $\rm [F^-] = x~mol\cdot L^{-1}$

$\rm [HF] = (10^{-3.50}/10^{-3.17})x = 0.468x~mol\cdot L^{-1}$
$\rm [HF_2^-] = 3.98 \cdot 0.468x \cdot x = 1.86x^2~mol\cdot L^{-1}$

Material balance: $\rm [F^-] + [HF] + 2[HF_2^-] = (1 + 0.468)x + 2 \cdot 1.86x^2 = 0.200~(mol\cdot L^{-1})$

Solving for $x = 0.107$, $\rm [F^-] = 0.107~mol\cdot L^{-1}$, $\rm [HF] = 0.050~mol\cdot L^{-1}$, $\rm [HF_2^-] = 0.021~mol\cdot L^{-1}$

<Checkpoint>[$\rm [F^-] = 0.107~mol\cdot L^{-1}$, $\rm [HF_2^-] = 0.021~mol\cdot L^{-1}$](2)</Checkpoint>

Ignoring $\rm [OH^-]$, $\rm [K^+] \approx [F^-] + [HF_2^-] - [H^+] = 0.128~mol\cdot L^{-1}$, $\rm m(KOH) = 0.72~g$

<Checkpoint>[$\rm m(KOH) = 0.72~g$](1)</Checkpoint>

(2)
 
If pH remains constant, the relationship between $\rm F$, $\rm HF$, and $\rm HF_2$ derived from the previous question remains unchanged.

Let $\rm [F] = y~mol\cdot L^{-1}$, $\rm [Li^+] = (1.84 \times 10^{-3}/y)~mol\cdot L^{-1}$

Material balance: $\rm [F] + [HF] + 2[HF_2] = c(HF) + [Li^+]$

<Checkpoint>[Material balance: $\rm [F] + [HF] + 2[HF_2] = c(HF) + [Li^+]$ ](1)</Checkpoint>

$(1 + 0.468)y + 2 \cdot 1.86y^2 = 0.200 + 1.84 \times 10^{-3}/y$  
Solving for $y = 0.114$, $\rm [Li^+] = 0.0161~mol\cdot L^{-1}$, $\rm s = 0.42~g\cdot L^{-1}$

<Checkpoint>[$\rm s = 0.42~g\cdot L^{-1}$](2)</Checkpoint>

(3)

At this point, $\rm [Na^+] = 0.0161~mol\cdot L^{-1}$, and the analytical concentration of fluorine is $\rm 0.200 + 0.0161 = 0.216~(mol\cdot L^{-1})$

At this point, the solution is equivalent to $\rm 0.144~mol\cdot L^{-1}~MF(KF + NaF) + 0.072~mol\cdot L^{-1}~HF$

Let $\rm [HF] = a~mol\cdot L^{-1}$, $\rm [F^-] = b~mol\cdot L^{-1}$, $\rm [HF_2^-] = 3.98ab~mol\cdot L^{-1}$, $\rm [H^+] = 10^{-3.17}(a/b)~mol\cdot L^{-1}$

Material balance: $\rm [HF] + [F^-] + 2[HF_2^-] = a + b + 2 \cdot 3.98ab = 0.216~(mol\cdot L^{-1})$   (*)

<Checkpoint>[Material balance: $\rm [HF] + [F^-] + 2[HF_2^-] = a + b + 2 \cdot 3.98ab = 0.216~(mol\cdot L^{-1})$](1)</Checkpoint>

Proton balance: $\rm [HF] + [HF_2^-] + [H^+] = a + 3.98ab + 10^{-3.17}(a/b) = 0.072~(mol\cdot L^{-1})$   (**)

<Checkpoint>[Proton balance: $\rm [HF] + [HF_2^-] + [H^+] = a + 3.98ab + 10^{-3.17}(a/b) = 0.072~(mol\cdot L^{-1})$](1)</Checkpoint>

From (*), $\rm a = 0.072 / (1 + 3.98b + 10^{-3.17}/b)$

Substituting into (**): $\rm 0.072(1 + 7.96b) / (1 + 3.98b + 10^{-3.17}/b) + b = 0.216$

Solving for $\rm b = 0.121$, $\rm a = 0.0484$
$\rm pH = 3.17 - \lg(a/b) = 3.57$

<Checkpoint>[$\rm pH = 3.57$](2)</Checkpoint>

$u = 0.72, v = 0.42, w = 3.57$, therefore $vw/u = 2.08$, choose C
（1）

设最终$\rm [F^-] = x~mol\cdot L^{-1}$

$\rm [HF] = (10^{-3.50}/10^{-3.17})x = 0.468x~mol\cdot L^{-1}$
$\rm [HF_2^-] = 3.98 \cdot 0.468x \cdot x = 1.86x^2~mol\cdot L^{-1}$

物料守恒：$\rm [F^-] + [HF] + 2[HF_2^-] = (1 + 0.468)x + 2 \cdot 1.86x^2 = 0.200~(mol\cdot L^{-1})$

解得$x = 0.107$，$\rm [F^-] = 0.107~mol\cdot L^{-1}$，$\rm [HF] = 0.050~mol\cdot L^{-1}$，$\rm [HF_2^-] = 0.021~mol\cdot L^{-1}$

<Checkpoint>[$\rm [F^-] = 0.107~mol\cdot L^{-1}$，$\rm [HF_2^-] = 0.021~mol\cdot L^{-1}$](2)</Checkpoint>

忽略$\rm [OH^-]$，$\rm [K^+] \approx [F^-] + [HF_2^-] - [H^+] = 0.128~mol\cdot L^{-1}$，$\rm m(KOH) = 0.72~g$

<Checkpoint>[$\rm m(KOH) = 0.72~g$](1)</Checkpoint>

（2）
 
pH不变，则上问得出的$\rm F$、$\rm HF$、$\rm HF_2$关系不变

设$\rm [F] = y~mol\cdot L^{-1}$，$\rm [Li^+] = (1.84 \times 10^{-3}/y)~mol\cdot L^{-1}$

物料守恒：$\rm [F] + [HF] + 2[HF_2] = c(HF) + [Li^+]$

<Checkpoint>[物料守恒：$\rm [F] + [HF] + 2[HF_2] = c(HF) + [Li^+]$ ](1)</Checkpoint>

$(1 + 0.468)y + 2 \cdot 1.86y^2 = 0.200 + 1.84 \times 10^{-3}/y$  
解得$y = 0.114$，$\rm [Li^+] = 0.0161~mol\cdot L^{-1}$，$\rm s = 0.42~g\cdot L^{-1}$

<Checkpoint>[$\rm s = 0.42~g\cdot L^{-1}$](2)</Checkpoint>

（3）

此时$\rm [Na^+] = 0.0161~mol\cdot L^{-1}$，氟的分析浓度为$\rm 0.200 + 0.0161 = 0.216~(mol\cdot L^{-1})$

此时溶液相当于$\rm 0.144~mol\cdot L^{-1}~MF(KF + NaF) + 0.072~mol\cdot L^{-1}~HF$

设$\rm [HF] = a~mol\cdot L^{-1}$，$\rm [F^-] = b~mol\cdot L^{-1}$，$\rm [HF_2^-] = 3.98ab~mol\cdot L^{-1}$，$\rm [H^+] = 10^{-3.17}(a/b)~mol\cdot L^{-1}$

物料守恒：$\rm [HF] + [F^-] + 2[HF_2^-] = a + b + 2 \cdot 3.98ab = 0.216~(mol\cdot L^{-1})$   (*)

<Checkpoint>[物料守恒：$\rm [HF] + [F^-] + 2[HF_2^-] = a + b + 2 \cdot 3.98ab = 0.216~(mol\cdot L^{-1})$](1)</Checkpoint>

质子守恒：$\rm [HF] + [HF_2^-] + [H^+] = a + 3.98ab + 10^{-3.17}(a/b) = 0.072~(mol\cdot L^{-1})$   (**)

<Checkpoint>[质子守恒：$\rm [HF] + [HF_2^-] + [H^+] = a + 3.98ab + 10^{-3.17}(a/b) = 0.072~(mol\cdot L^{-1})$](1)</Checkpoint>

由(*)，$\rm a = 0.072 / (1 + 3.98b + 10^{-3.17}/b)$

代入(**)：$\rm 0.072(1 + 7.96b) / (1 + 3.98b + 10^{-3.17}/b) + b = 0.216$

解得$\rm b = 0.121$，$\rm a = 0.0484$
$\rm pH = 3.17 - \lg(a/b) = 3.57$

<Checkpoint>[$\rm pH = 3.57$](2)</Checkpoint>

$u = 0.72, v = 0.42, w = 3.57$，故$vw/u = 2.08$，选C