This problem involves acid-base equilibrium in precipitation-dissolution equilibrium.  

To solve this problem, the following preliminary calculations are required:  

First, calculate the $\mathrm{pH}$ of the solution obtained by dissolving $2.50\ \mathrm{g}\ \mathrm{KH}_{3}(\mathrm{C}_{2}\mathrm{O}_{4})_{2}{\cdot}2\mathrm{H}_{2}\mathrm{O}$ in $100\ \mathrm{mL}$ of water.  

Second, given that the $\mathrm{pH}$ of saturated $\mathrm{Ca(OH)}_{2}$ is $12.00$, calculate the $K_{\mathrm{sp}}$ of $\mathrm{Ca(OH)}_{2}$.  

Based on these results, calculate the concentrations of each ion ($[\mathrm{H^{+}}]$, $[\mathrm{{C_{2}O_{4}}^{2-}}]$, $[\mathrm{Ca^{2+}}]$) in the solution obtained by mixing the solution described in the first step with saturated calcium hydroxide in a $1:1$ ratio.  

Select the answer closest to the calculated values.  

Relevant data: $\mathrm{p}K_{\mathrm{a1}}(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}) = 1.22$, $\mathrm{p}K_{\mathrm{a2}}(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}) = 4.19$, $\mathrm{p}K_{\mathrm{sp}}(\mathrm{CaC_{2}O_{4}}) = 8.70$.
本题涉及沉淀溶解平衡中的酸碱平衡。

为了解决本题，需要预先进行如下计算：

第一，请计算将 $2 . 5 0\ \mathrm { g } \ \mathrm{K H } _ { 3 } ( \mathrm { C } _ { 2 } \mathrm { O } _ { 4 } ) _ { 2 } { \cdot } 2 \mathrm { H } _ { 2 } \mathrm { O }$ 溶于 $1 0 0 \ \mathrm { m L }$ 水后溶液的 $\mathrm { p H }$ 值。
  
第二，已知饱和 $\mathrm { C a ( O H ) } _ { 2 }$ 的 $\mathrm { p H }$ 值为$12.00%$，请计算 $\mathrm { C a ( O H ) } _ { 2 }$ 的 $K _ { \mathrm { s p } }$ 。

在此基础上，计算将第一条中所述溶液与饱和氢氧化钙以$1:1$混合后溶液得到的各离子浓度（$[\mathrm{H^{+}}]$, $[\mathrm {{C_{2}O_{4}}^{2 -}}]$, $[\mathrm{Ca^{2 +}}]$）。

选择与计算值最接近的答案。

有关数据： $\mathrm { p } K _ { \mathrm { a 1 } } ( \mathrm { H } _ { 2 } \mathrm { C } _ { 2 } \mathrm { O } _ { 4 } ) = 1 . 2 2$ ， $\mathrm { p } K _ { \mathrm { a 2 } } ( \mathrm { H } _ { 2 } \mathrm { C } _ { 2 } \mathrm { O } _ { 4 } ) = 4 . 1 9$ ， $\mathrm { p } K _ { \mathrm { s p } } (\mathrm {CaC_{2}O_{4} } ) = 8 . 7 0$。
$[\mathrm{H^{+}}]= 0.52\ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm {{C_{2}O_{4}}^{2 -}}]=6.2\times 10^{-7} \ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm{Ca^{2 +}}]=2.5\times 10^{-3} \ \mathrm{mol \cdot L^{ - 1}}$
$[\mathrm{H^{+}}]= 0.186\ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm {{C_{2}O_{4}}^{2 -}}]=3.23\times 10^{-5} \ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm{Ca^{2 +}}]=6.16\times 10^{-5} \ \mathrm{mol \cdot L^{ - 1}}$
$[\mathrm{H^{+}}]= 0.0225\ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm {{C_{2}O_{4}}^{2 -}}]=2.04\times 10^{-4} \ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm{Ca^{2 +}}]=9.73\times 10^{-6} \ \mathrm{mol \cdot L^{ - 1}}$
$[\mathrm{H^{+}}]= 0.0216\ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm {{C_{2}O_{4}}^{2 -}}]=2.14\times 10^{-4} \ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm{Ca^{2 +}}]=9.48\times 10^{-6} \ \mathrm{mol \cdot L^{ - 1}}$
$[\mathrm{H^{+}}]= 0.0933\ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm {{C_{2}O_{4}}^{2 -}}]=2.53\times 10^{-5} \ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm{Ca^{2 +}}]=7.88\times 10^{-5} \ \mathrm{mol \cdot L^{ - 1}}$
$[\mathrm{H^{+}}]= 0.0162\ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm {{C_{2}O_{4}}^{2 -}}]=1.33\times 10^{-5} \ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm{Ca^{2 +}}]=0.005 \ \mathrm{mol \cdot L^{ - 1}}$
None of the other options are correct
$[\mathrm{H^{+}}]= 0.186\ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm {{C_{2}O_{4}}^{2 -}}]=3.23\times 10^{-5} \ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm{Ca^{2 +}}]=6.16\times 10^{-5} \ \mathrm{mol \cdot L^{ - 1}}$
$[\mathrm{H^{+}}]= 0.186\ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm {{C_{2}O_{4}}^{2 -}}]=3.23\times 10^{-5} \ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm{Ca^{2 +}}]=6.16\times 10^{-5} \ \mathrm{mol \cdot L^{ - 1}}$
$[\mathrm{H^{+}}]= 0.0225\ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm {{C_{2}O_{4}}^{2 -}}]=2.04\times 10^{-4} \ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm{Ca^{2 +}}]=9.73\times 10^{-6} \ \mathrm{mol \cdot L^{ - 1}}$
$[\mathrm{H^{+}}]= 0.0216\ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm {{C_{2}O_{4}}^{2 -}}]=2.14\times 10^{-4} \ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm{Ca^{2 +}}]=9.48\times 10^{-6} \ \mathrm{mol \cdot L^{ - 1}}$
$[\mathrm{H^{+}}]= 0.0933\ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm {{C_{2}O_{4}}^{2 -}}]=2.53\times 10^{-5} \ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm{Ca^{2 +}}]=7.88\times 10^{-5} \ \mathrm{mol \cdot L^{ - 1}}$
$[\mathrm{H^{+}}]= 0.0162\ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm {{C_{2}O_{4}}^{2 -}}]=1.33\times 10^{-5} \ \mathrm{mol \cdot L^{ - 1}}$

$[\mathrm{Ca^{2 +}}]=0.005 \ \mathrm{mol \cdot L^{ - 1}}$
其他选项均不正确
First Question  

$n ( \mathrm { K H _ { 3 } } ( \mathrm { C _ { 2 } O _ { 4 } } ) _ { 2 } \cdot 2 \mathrm { H _ { 2 } O } ) = m / M = 9 . 8 3 \times 1 0 ^ { - 3 }\ \mathrm { m o l } $  

Concentration $c = n / V = 9 . 8 3 \times 1 0 ^ { - 2 } \ \mathrm{mol \cdot L^{ - 1}}$  

<Checkpoint>[Concentration $c = n / V = 9 . 8 3 \times 1 0 ^ { - 2 } \ \mathrm{mol \cdot L^{ - 1}}$ ](1)</Checkpoint>  

The charge balance equation is:  

$[\mathrm{K^{+}}]+[\mathrm{H^{+}}]=[\mathrm{OH^{-}}]+[\mathrm{{HC_{2}O_{4}}^{-}}]+2[\mathrm{{C_{2}O_{4}}^{2-}}]$  

<Checkpoint>[The charge balance equation is:  $[\mathrm{K^{+}}]+[\mathrm{H^{+}}]=[\mathrm{OH^{-}}]+[\mathrm{{HC_{2}O_{4}}^{-}}]+2[\mathrm{{C_{2}O_{4}}^{2-}}]$](2)</Checkpoint>  

Let $[ \mathrm { H } ^ { + } ] = x\ \mathrm { m o l } / \mathrm { L }$  

Based on the distribution fraction, we have:  

$[\mathrm{{HC_{2}O_{4}}^{-}}]= c_{0}(\mathrm{H_{2}C_{2}O_{4}}) \times(10^{-1.22}x)/(x^{2}+10^{-1.22}x+10^{-1.22}\times 10^{-4.19})$  

$[\mathrm{{C_{2}O_{4}}^{2-}}]=c_{0}(\mathrm{H_{2}C_{2}O_{4}}) \times(10^{-1.22}\times10^{-4.19})/(x^{2}+10^{-1.22}x+10^{-1.22}\times 10^{-4.19})$  

<Checkpoint>[Based on the distribution fraction, we have:  $[\mathrm{{HC_{2}O_{4}}^{-}}]= c_{0}(\mathrm{H_{2}C_{2}O_{4}}) \times(K _ { \mathrm { a 1 } }[\mathrm{H^{+}}])/([\mathrm{H^{+}}]^{2}+K _ { \mathrm { a 1 } }[\mathrm{H^{+}}]+K _ { \mathrm { a 1 } }\times K _ { \mathrm { a 2 } })$  ,$[\mathrm{{C_{2}O_{4}}^{2-}}]=c_{0}(\mathrm{H_{2}C_{2}O_{4}}) \times(K _ { \mathrm { a 1 } }\times K _ { \mathrm { a 2 } })/([\mathrm{H^{+}}]^{2}+K _ { \mathrm { a 1 } }[\mathrm{H^{+}}]+K _ { \mathrm { a 1 } }\times K _ { \mathrm { a 2 } })$](3)</Checkpoint>  

Here, $c_{0}(\mathrm{H_{2}C_{2}O_{4}})=2c$  

Substituting into the above equations, we solve for $x = 0 . 0 3 1 4\ \mathrm{mol \cdot L^{ - 1}}$  

Thus, $\mathrm { p H } = 1 . 5 0$  

Second Question  

Calculate $K_{\mathrm{sp}}$ using the $\mathrm{pH}$ of saturated calcium hydroxide:  

$K_{\mathrm{sp}}(\mathrm{Ca(OH)_{2}})=[\mathrm{OH^{-}}]^{2}\times[\mathrm{Ca^{2+}}]=1/2[\mathrm{OH^{-}}]^{3}=5\times 10^{-7}$  

<Checkpoint>[$K_{\mathrm{sp}}(\mathrm{Ca(OH)_{2}})=5\times 10^{-7}$](1)</Checkpoint>  

After mixing the two solutions described in the problem:  

$[\mathrm{K^{+}}]=1/2  c=4.91\times 10^{-2} \ \mathrm{mol \cdot L^{ - 1}}$  

The concentration of saturated calcium hydroxide is $0.005 \mathrm{mol \cdot L^{ - 1}}$, while the oxalate concentration in the first solution is $c  =3\times 9 . 8 3 \times 1 0 ^ { - 2 } \ \mathrm{mol \cdot L^{ - 1}}=0.295\mathrm{mol \cdot L^{ - 1}}$. Therefore, calcium hydroxide is insufficient after mixing.  

<Checkpoint>[Calcium hydroxide is insufficient after mixing](1)</Checkpoint>  

Comparing the solubility product constant with the solution acidity, it can be concluded that, under insufficient calcium hydroxide conditions, calcium hydroxide completely reacts and converts to calcium oxalate, i.e., $c_{0}(\mathrm{H_{2}C_{2}O_{4}})=9.33\times10^{-2}\ \mathrm{mol \cdot L^{ - 1}}$  

<Checkpoint>[Calcium hydroxide completely reacts and converts to calcium oxalate](1)</Checkpoint>  

Since the total oxalate concentration is much higher than the total calcium concentration, the calcium ion concentration can be neglected based on the precipitation-dissolution equilibrium.  

<Checkpoint>[Calcium ion concentration can be neglected](1)</Checkpoint>  

Thus, the charge balance equation is the same as in the first question:  

$[\mathrm{K^{+}}]+[\mathrm{H^{+}}]=[\mathrm{OH^{-}}]+[\mathrm{{HC_{2}O_{4}}^{-}}]+2[\mathrm{{C_{2}O_{4}}^{2-}}]$  

Following the same logic as the first question and using the distribution fraction, we calculate:  

$[\mathrm{H^{+}}]= 0.0216\ \mathrm{mol \cdot L^{ - 1}}$  

<Checkpoint>[Following the same logic as the first question and using the distribution fraction, we calculate:  $[\mathrm{H^{+}}]= 0.0216\ \mathrm{mol \cdot L^{ - 1}}$](1)</Checkpoint>  

This naturally yields:  

$[\mathrm {{C_{2}O_{4}}^{2 -}}]=2.14\times 10^{-4} \ \mathrm{mol \cdot L^{ - 1}}$  

Based on the precipitation-dissolution equilibrium, it follows that:  

$[\mathrm{Ca^{2 +}}]=9.48\times 10^{-6} \ \mathrm{mol \cdot L^{ - 1}}$  

<Checkpoint>[$[\mathrm{Ca^{2 +}}]=9.48\times 10^{-6} \ \mathrm{mol \cdot L^{ - 1}}$](1)</Checkpoint>  

<Checkpoint>[$[\mathrm {{C_{2}O_{4}}^{2 -}}]=2.14\times 10^{-4} \ \mathrm{mol \cdot L^{ - 1}}$](1)</Checkpoint>  

<Checkpoint>[$[\mathrm{H^{+}}]= 0.0216\ \mathrm{mol \cdot L^{ - 1}}$](1)</Checkpoint>  

Thus, option D is correct.  

<Checkpoint>[Option D is correct](1)</Checkpoint>
第一问

$n ( \mathrm { K H _ { 3 } } ( \mathrm { C _ { 2 } O _ { 4 } } ) _ { 2 } \cdot 2 \mathrm { H _ { 2 } O } ) = m / M = 9 . 8 3 \times 1 0 ^ { - 3 }\ \mathrm { m o l } $ 

浓度$c = n / V = 9 . 8 3 \times 1 0 ^ { - 2 } \ \mathrm{mol \cdot L^{ - 1}}$ 

<Checkpoint>[浓度$c = n / V = 9 . 8 3 \times 1 0 ^ { - 2 } \ \mathrm{mol \cdot L^{ - 1}}$ ](1)</Checkpoint>

列电荷守恒式得：

$[\mathrm{K^{+}}]+[\mathrm{H^{+}}]=[\mathrm{OH^{-}}]+[\mathrm{{HC_{2}O_{4}}^{-}}]+2[\mathrm{{C_{2}O_{4}}^{2-}}]$

<Checkpoint>[列电荷守恒式得：  $[\mathrm{K^{+}}]+[\mathrm{H^{+}}]=[\mathrm{OH^{-}}]+[\mathrm{{HC_{2}O_{4}}^{-}}]+2[\mathrm{{C_{2}O_{4}}^{2-}}]$](2)</Checkpoint>

设 $[ \mathrm { H } ^ { + } ] = x\ \mathrm { m o l } / \mathrm { L }$ 

根据分布分数，有

$[\mathrm{{HC_{2}O_{4}}^{-}}]= c_{0}(\mathrm{H_{2}C_{2}O_{4}}) \times(10^{-1.22}x)/(x^{2}+10^{-1.22}x+10^{-1.22}\times 10^{-4.19})$

$[\mathrm{{C_{2}O_{4}}^{2-}}]=c_{0}(\mathrm{H_{2}C_{2}O_{4}}) \times(10^{-1.22}\times10^{-4.19})/(x^{2}+10^{-1.22}x+10^{-1.22}\times 10^{-4.19})$

<Checkpoint>[根据分布分数，有  $[\mathrm{{HC_{2}O_{4}}^{-}}]= c_{0}(\mathrm{H_{2}C_{2}O_{4}}) \times(K _ { \mathrm { a 1 } }[\mathrm{H^{+}}])/([\mathrm{H^{+}}]^{2}+K _ { \mathrm { a 1 } }[\mathrm{H^{+}}]+K _ { \mathrm { a 1 } }\times K _ { \mathrm { a 2 } })$  ,$[\mathrm{{C_{2}O_{4}}^{2-}}]=c_{0}(\mathrm{H_{2}C_{2}O_{4}}) \times(K _ { \mathrm { a 1 } }\times K _ { \mathrm { a 2 } })/([\mathrm{H^{+}}]^{2}+K _ { \mathrm { a 1 } }[\mathrm{H^{+}}]+K _ { \mathrm { a 1 } }\times K _ { \mathrm { a 2 } })$](3)</Checkpoint>

其中，$c_{0}(\mathrm{H_{2}C_{2}O_{4}})=2c$

代入上式，解得 $x = 0 . 0 3 1 4\ \mathrm{mol \cdot L^{ - 1}}$ 

即 $\mathrm { p H } = 1 . 5 0$

第二问

通过饱和氢氧化钙的$\mathrm{pH}$计算$K_{\mathrm{sp}}$

$K_{\mathrm{sp}}(\mathrm{Ca(OH)_{2}})=[\mathrm{OH^{-}}]^{2}\times[\mathrm{Ca^{2+}}]=1/2[\mathrm{OH^{-}}]^{3}=5\times 10^{-7}$

<Checkpoint>[$K_{\mathrm{sp}}(\mathrm{Ca(OH)_{2}})=5\times 10^{-7}$](1)</Checkpoint>

将题中所述两溶液混合之后，$[\mathrm{K^{+}}]=1/2  c=4.91\times 10^{-2} \ \mathrm{mol \cdot L^{ - 1}}$

饱和氢氧化钙的浓度为$0.005 \mathrm{mol \cdot L^{ - 1}}$，第一条中溶液草酸根浓度$c  =3\times 9 . 8 3 \times 1 0 ^ { - 2 } \ \mathrm{mol \cdot L^{ - 1}}=0.295\mathrm{mol \cdot L^{ - 1}}$ ，从而混合后氢氧化钙不足量。

<Checkpoint>[混合后氢氧化钙不足量](1)</Checkpoint>

对比沉淀溶解平衡常数与溶液酸度，在氢氧化钙不足量的情况下，可以认为氢氧化钙完全反应，转化为草酸钙，亦即$c_{0}(\mathrm{H_{2}C_{2}O_{4}})=9.33\times10^{-2}\ \mathrm{mol \cdot L^{ - 1}}$

<Checkpoint>[可以认为氢氧化钙完全反应，转化为草酸钙](1)</Checkpoint>

由于草酸根总浓度远大于钙的总浓度，根据沉淀溶解平衡，可以忽略钙离子浓度。

<Checkpoint>[可以忽略钙离子浓度](1)</Checkpoint>

因此，电荷守恒式与第一问相同：

$[\mathrm{K^{+}}]+[\mathrm{H^{+}}]=[\mathrm{OH^{-}}]+[\mathrm{{HC_{2}O_{4}}^{-}}]+2[\mathrm{{C_{2}O_{4}}^{2-}}]$

根据分布分数，同第一问理可以计算得

$[\mathrm{H^{+}}]= 0.0216\ \mathrm{mol \cdot L^{ - 1}}$

<Checkpoint>[根据分布分数，同第一问理可以计算得  $[\mathrm{H^{+}}]= 0.0216\ \mathrm{mol \cdot L^{ - 1}}$](1)</Checkpoint>

此时自然得到

$[\mathrm {{C_{2}O_{4}}^{2 -}}]=2.14\times 10^{-4} \ \mathrm{mol \cdot L^{ - 1}}$

根据沉淀溶解平衡，显然有

$[\mathrm{Ca^{2 +}}]=9.48\times 10^{-6} \ \mathrm{mol \cdot L^{ - 1}}$

<Checkpoint>[$[\mathrm{Ca^{2 +}}]=9.48\times 10^{-6} \ \mathrm{mol \cdot L^{ - 1}}$](1)</Checkpoint>

<Checkpoint>[$[\mathrm {{C_{2}O_{4}}^{2 -}}]=2.14\times 10^{-4} \ \mathrm{mol \cdot L^{ - 1}}$](1)</Checkpoint>

<Checkpoint>[$[\mathrm{H^{+}}]= 0.0216\ \mathrm{mol \cdot L^{ - 1}}$](1)</Checkpoint>

从而选项D正确。

<Checkpoint>[选项D正确](1)</Checkpoint>