酸碱中和滴定

$\ce{NaOH}$滴定$\ce{CH_3COOH}$

image-20220705073822131

image-20220705073728070

当等浓度的碱滴定酸时

1. 滴定开始

滴定开始,观察开头pH,确认溶液酸碱性

计算$K_a$值

已知$c(\ce{CH3COOH})$,由图可知pH值(约为3),可知$c(\ce{H+})$和$c(\ce{CH3COO-})$
$$
K_a = \frac{c(\ce{H+})\cdot c(\ce{CH3COO-})}{c(\ce{CH3COOH})} = \frac{10^{-3} \cdot 10^{-3}}{10^{-1}} = 10^{-5}
$$

2. 滴定一半(图中①)

当滴定一半时,若不发生电离或水解,则有

可视作$\ce{CH3COOH * CH3COONa}$两溶液等体积混合

$c(\ce{CH3COOH})=c(\ce{CH3COO-})$

由于醋酸电离>水解(醋酸的性质)

则有c($\ce{H^+}$)>c($\ce{OH^-}$)

粒子浓度大小

$$
c(\ce{CH3COO-}) > c(\ce{Na+}) > c(\ce{CH3COOH}) > c(\ce{H+}) > c(\ce{OH-})
$$

电荷守恒

$$
c(\ce{Na+}) + c(\ce{H+}) = c(\ce{CH3COO-}) + c(\ce{OH-})
$$

物料守恒

$$
2c(\ce{Na+}) = c(\ce{CH3COOH}) + c(\ce{CH3COO-})
$$

质子守恒

$$
2c(\ce{H+}) + c(\ce{CH3COOH}) = 2c(\ce{OH-}) + c(\ce{CH3COO-})
$$

不相信?上图:

image-20220705102223173

3. $c(\ce{CH3COOH})$与$c(\ce{CH3COO-})$相等时

浓度与上式基本不变,则有:

粒子浓度大小

$$
c(\ce{CH3COO-}) = c(\ce{CH3COOH}) > c(\ce{Na+}) > c(\ce{H+}) > c(\ce{OH-})
$$

4. pH = 7时(图中②)

当pH=7时,有$c(\ce{H+}) = c(\ce{OH-})$

电荷守恒

$$
c(\ce{Na+}) + c(\ce{H+}) = c(\ce{OH-}) + c(\ce{CH3COO-})
$$

由电荷守恒,可知粒子浓度大小:

粒子浓度大小

$$
c(\ce{Na+}) = c(\ce{CH3COO-}) > c(\ce{CH3COOH}) > c(\ce{H+}) = c(\ce{OH-})
$$

由于不知溶液具体比值,无法精确写出物料守恒与质子守恒

5. 溶液完全反应时(图中③)

当溶液完全反应时,若不发生电离或水解,则可将溶液视作$\ce{CH3COONa}$溶液

可认为$c(\ce{Na+}) = c(\ce{CH3COO-})$

由于$\ce{CH3COO-}$会水解,则有:

粒子浓度大小

$$
c(\ce{Na+}) > c(\ce{CH3COO-}) > c(\ce{OH-})> c(\ce{CH3COOH}) > c(\ce{H+})
$$

电荷守恒

$$
c(\ce{Na+}) + c(\ce{H+}) = c(\ce{OH-}) + c(\ce{CH3COO-})
$$

物料守恒

$$
c(\ce{Na+}) = c(\ce{CH3COOH}) + c(\ce{CH3COO-})
$$

质子守恒

$$
c(\ce{H+}) + c(\ce{CH3COOH}) = c(\ce{OH-})
$$

image-20220705154009554

6. 滴定过量一倍时(图中滴定40mL)

此溶液可视作$\ce{CH3COONa} \cdot \ce{NaOH}$的混合溶液(1:1)

此溶液中,$\ce{CH3COO-}$水解生成$\ce{OH-}$

粒子浓度大小

$$
c(\ce{Na+}) > c(\ce{OH-}) > c(\ce{CH3COO-}) > c(\ce{CH3COOH}) > c(\ce{H+})
$$

电荷守恒

$$
c(\ce{Na+}) + c(\ce{H+}) = c(\ce{OH-}) + c(\ce{CH3COO-})
$$

物料守恒

$$
c(\ce{Na+}) = 2c(\ce{CH3COOH}) + 2c(\ce{CH3COO-})
$$

质子守恒

$$
c(\ce{H+}) + c(\ce{CH3COO-}) + 2c(\ce{CH3COOH}) = c(\ce{OH-})
$$

$\ce{HCl}$滴定$\ce{NH3*H2O}$

现有20ml 0.1mol/L的$\ce{NH3*H2O}$,用0.1mol/L的$\ce{HCl}$滴定:

R-C

1. 滴定开始

计算$K_b$

已知氨水的pH值为11,可知$c(\ce{OH-})$和$c(\ce{NH4+})$
$$
K_b = \frac{c(\ce{OH-}) \cdot c(\ce{NH4+})}{c(\ce{NH3*H2O})} = \frac{10^{-3} \cdot 10^{-3}}{10^{-1}} = 10^{-5}
$$

2. 滴定一半(图中①)

该溶液可视作$\ce{NH3*H2O}$与$\ce{NH4Cl}$等体积混合物

由于此时pH>7,可知电离>水解

则有:

离子浓度大小

$$
c(\ce{NH4+}) > c(\ce{Cl-}) > c(\ce{NH3*H2O}) > c(\ce{OH-}) > c(\ce{H+})
$$

电荷守恒

$$
c(\ce{NH4+}) + c(\ce{H+}) = c(\ce{Cl-}) + c(\ce{OH-})
$$

物料守恒

$$
2c(\ce{Cl-}) = c(\ce{NH4+}) + c(\ce{NH3*H2O})
$$

质子守恒

$$
2c(\ce{H+}) + c(\ce{NH4+}) = 2c(\ce{OH-}) + c(\ce{NH4*H2O})
$$

经验之图:

image-20220708185559192

3. 当$c(\ce{NH4+})$与$c(\ce{NH3*H2O})$相等时

此时,可认为是滴定一半时少滴几滴,则有:

粒子浓度大小

$$
c(\ce{NH4+}) = c(\ce{NH3*H2O}) > c(\ce{Cl-}) > c(\ce{OH-}) > c(\ce{H+})
$$

4. 当pH=7时(图中②)

当pH=7时,有$c(\ce{H+}) = c(\ce{OH-})$

电荷守恒

$$
c(\ce{NH4+}) + c(\ce{H+}) = c(\ce{Cl-}) + c(\ce{OH-})
$$

由电荷守恒,得$c(\ce{NH4+}) = c(\ce{Cl-})$

粒子浓度大小

$$
c(\ce{NH4+}) = c(\ce{Cl-}) > c(\ce{NH3*H2O}) > c(\ce{H+}) = c(\ce{OH-})
$$

由于不知溶液具体比值,无法精确写出物料守恒与质子守恒

5. 溶液完全反应时(图中③)

若不发生电离或水解,此溶液可视作$\ce{NH4Cl}$纯溶液

其中$\ce{NH4+}$会水解,则有:

粒子浓度大小

$$
c(\ce{Cl-}) > c(\ce{NH4+}) > c(\ce{H+})> c(\ce{NH3*H2O}) > c(\ce{OH-})
$$

电荷守恒

$$
c(\ce{NH4+}) + c(\ce{H+}) = c(\ce{Cl-}) + c(\ce{OH-})
$$

物料守恒

$$
c(\ce{Cl-}) = c(\ce{NH4+}) + c(\ce{NH3*H2O})
$$

质子守恒

$$
c(\ce{H+}) = c(\ce{OH-}) + c(\ce{NH3*H2O})
$$

image-20220708191351609

6. 滴定过量一倍时(滴定40mL)

此溶液可视作$\ce{NH4Cl}$和$\ce{HCl}$的等体积混合溶液(1:1)

$\ce{NH4+}$水解生成$\ce{NH3*H2O}$和$\ce{\ce{H+}}$

则$c(\ce{H+})> c(\ce{NH3*H2O})$

电荷守恒

$$
c(\ce{NH4+}) + c(\ce{H+}) = c(\ce{Cl-}) + c(\ce{OH-})
$$

物料守恒

$$
c(\ce{Cl-}) = 2c(\ce{NH4+}) + 2c(\ce{NH3*H2O})
$$

质子守恒

$$
c(\ce{H+}) = c(\ce{NH4+}) + 2c(\ce{NH3*H2O}) + c(\ce{OH-})
$$

最后修改日期: 2022年7月8日

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郭 涛
4 月 前

那个……有没有一种可能……我们这是个编程网站……