## The Quantitative Calculation of pH in Monoacid/Monobasic Solution in Teaching of Inorganic Chemistry

Wang Ting, Zhang Zongpei, He Zhanhang,, Li Kai,

 基金资助: 郑州大学化学学科拔尖计划2.0项目2020郑州大学本科教学改革研究与实践项目2020郑州大学大学生创新创业训练计划项目

Abstract

The quantitative calculation of pH of monoacid/monobasic solution is a basic problem for inorganic chemistry teaching. The qualitative description and quantitative calculation of pH are connected by approximation conditions. It is very important for the students to understand these approximation conditions, which will help them solve problems about ionic equilibrium in aqueous solution. However, how to use these approximation conditions correctly in solution of low concentration and weak acids is often difficult. Even in some textbooks, the description of concepts is not accurate and not consistent with the associated courses. Thus, to solve the problems mentioned above, the derivations of the approximation conditions are introduced and examples are analyzed in detail in this paper. The problems of quantitative calculation of pH of monoacid/monobasic solution which should be paid attention to in the course teaching of inorganic chemistry are discussed.

Keywords： Inorganic chemistry ; Course teaching ; pH ; Quantitative calculation

Wang Ting. The Quantitative Calculation of pH in Monoacid/Monobasic Solution in Teaching of Inorganic Chemistry. University Chemistry[J], 2020, 35(8): 129-134 doi:10.3866/PKU.DXHX202003085

## 1 一元酸碱溶液pH定量计算的近似条件的推导

${\rm H^{+} + A^{−}} \rightleftharpoons {\rm HA} \\ {K_{\rm{a}}} = \frac{{[{{\rm{H}}^ + }][{{\rm{A}}^ - }]}}{{[{\rm{HA}}]}}$

${\rm H_{2}O} \rightleftharpoons {\rm H^{+} + OH^{−}} \\ {K_{\rm{w}}} = [{{\rm{H}}^ + }][{\rm{O}}{{\rm{H}}^ - }]$

$[{\operatorname{H} ^ + }] = [{{\rm{A}}^ - }] + [{\rm{O}}{{\rm{H}}^ - }]$

$[{{\rm{H}}^ + }] = \frac{{{K_{\rm{a}}}[{\rm{HA}}]}}{{[{{\rm{H}}^ + }]}} + \frac{{{K_{\rm{w}}}}}{{[{{\rm{H}}^ + }]}}$

$[{{\rm{H}}^ + }] = \sqrt {{K_{\rm{a}}}[{\rm{HA}}] + {K_{\rm{w}}}}$

$c = [{\rm{HA}}] + [{{\rm{A}}^ - }] = [{\rm{HA}}] + \frac{{{K_{\rm{a}}}[{\rm{HA}}]}}{{[{{\rm{H}}^ + }]}}$

$[{\rm{HA}}] = \frac{{c[{{\rm{H}}^ + }]}}{{[{{\rm{H}}^ + }] + {K_{\rm{a}}}}}$

$[{{\rm{H}}^ + }] = \sqrt {\frac{{{K_{\rm{a}}}c[{{\rm{H}}^ + }]}}{{[{{\rm{H}}^ + }] + {K_{\rm{a}}}}} + {K_{\rm{w}}}}$

${[{{\rm{H}}^ + }]^3} + {K_{\rm{a}}}{[{{\rm{H}}^ + }]^2} - ({K_{\rm{a}}}c + {K_{\rm{w}}})[{{\rm{H}}^ + }] - {K_{\rm{a}}}{K_{\rm{w}}} = 0$

${E_1} = \frac{{\sqrt {{K_{\rm{a}}}[{\rm{HA}}]} - \sqrt {{K_{\rm{a}}}[{\rm{HA}}] + {K_{\rm{w}}}} }}{{\sqrt {{K_{\rm{a}}}[{\rm{HA}}] + {K_{\rm{w}}}} }}$

 相对误差< 1% 2% 2.5% 4% 5% Ka[HA] > 49.3Kw 24.3Kw 19.3Kw 11.8Kw 9.26Kw

$[{{\rm{H}}^ + }] = \sqrt {{K_{\rm{a}}}[{\rm{HA}}]}$

$[{\rm{HA}}] = c - [{{\rm{H}}^ + }] + [{\rm{O}}{{\rm{H}}^ - }]$

$[{\rm{HA}}] = c - [{{\rm{H}}^ + }]$

$[{{\rm{H}}^ + }] = \sqrt {{K_{\rm{a}}}(c - [{{\rm{H}}^ + }])}$

${[{{\rm{H}}^ + }]^2} + {K_{\rm{a}}}[{{\rm{H}}^ + }] - {K_{\rm{a}}}c = 0$

$[{{\rm{H}}^ + }] = \frac{{ - {K_{\rm{a}}} + \sqrt {K_{\rm{a}}^2 + 4{K_{\rm{a}}}c} }}{2}$

$[{{\rm{H}}^ + }] = \sqrt {{K_{\rm{a}}}c}$

${E_2} = \frac{{\sqrt {{K_{\rm{a}}}c} - \frac{{ - {K_{\rm{a}}} + \sqrt {{K_{\rm{a}}}^2 + 4{K_{\rm{a}}}c} }}{2}}}{{\frac{{ - {K_{\rm{a}}} + \sqrt {{K_{\rm{a}}}^2 + 4{K_{\rm{a}}}c} }}{2}}}$

 相对误差< 1% 2% 2.5% 4% 5% c/Ka > 2130 637 417 162 105

$[{{\rm{H}}^ + }] = \sqrt {{K_{\rm{a}}}c + {K_{\rm{w}}}}$

 Kac c/Ka 计算方法 < 20Kw < 400 ${[{{\rm{H}}^ + }]^3} + {K_{\rm{a}}}{[{{\rm{H}}^ + }]^2} - ({K_{\rm{a}}}c + {K_{\rm{w}}})[{{\rm{H}}^ + }] - {K_{\rm{a}}}{K_{\rm{w}}} = 0$ < 20Kw > 400 $[{{\rm{H}}^ + }] = \sqrt {{K_{\rm{a}}}c + {K_{\rm{w}}}}$ > 20Kw < 400 $[{{\rm{H}}^ + }] = \frac{{ - {K_{\rm{a}}} + \sqrt {{K_{\rm{a}}}^2 + 4{K_{\rm{a}}}c} }}{2}$ > 20Kw > 400 $[{{\rm{H}}^ + }] = \sqrt {{K_{\rm{a}}}c}$

### 2.1 一元强酸pH计算实例

$1.0 \times {10^{ - 3}}{\rm{mol}} \cdot {\rm{d}}{{\rm{m}}^{ - {\rm{3}}}} \times \frac{{1.0 \times {{10}^{ - 3}}{\rm{d}}{{\rm{m}}^3}}}{{10.0{\rm{ d}}{{\rm{m}}^3} + 1.0 \times {{10}^{ - 3}}{\rm{ d}}{{\rm{m}}^3}}} = 1.0 \times {10^{ - 7}}{\rm{mol}} \cdot {\rm{d}}{{\rm{m}}^{ - {\rm{3}}}}$

\begin{align} \;\;\;\;\;\; {{\text{H}}_{\text{2}}}\text{O} = {{\text{H}}^{+}} + \text{O}{{\text{H}}^{-}} \\ \begin{array}{*{35}{l}} { {{c}_{{起始}}}(\text{mol}\cdot \text{d}{{\text{m}}^{-\text{3}}}) } & { 1.0\times {{10}^{-7}} }& {0} \\ {{{c}_{{平衡}}}(\text{mol}\cdot \text{d}{{\text{m}}^{-\text{3}}}) } & {1.0\times {{10}^{-7}}\text{ + }x } & { x} \\ \end{array} \end{align}

${K_{\rm{w}}} = [{{\rm{H}}^ + }][{\rm{O}}{{\rm{H}}^ - }] = (1.0 \times {10^{ - 7}} + x)x = 1.0 \times {10^{ - 14}}$

$x = 0.6 \times {10^{ - 7}}$

$[{{\rm{H}}^ + }] = \left( {1.0 \times {{10}^{ - 7}} + 0.6 \times {{10}^{ - 7}}} \right){\rm{mol}} \cdot {\rm{d}}{{\rm{m}}^{ - {\rm{3}}}} = 1.6 \times {10^{ - 7}}{\rm{mol}} \cdot {\rm{d}}{{\rm{m}}^{ - {\rm{3}}}}$

$1.0 \times {10^{ - 3}}{\rm{ mol}} \cdot {\rm{d}}{{\rm{m}}^{ - {\rm{3}}}} \times \frac{{1.0 \times {{10}^{ - 3}}{\rm{d}}{{\rm{m}}^3}}}{{10.0{\rm{ d}}{{\rm{m}}^3} + 1.0 \times {{10}^{ - 3}}{\rm{ d}}{{\rm{m}}^3}}} = 1.0 \times {10^{ - 7}}{\rm{mol}} \cdot {\rm{d}}{{\rm{m}}^{ - {\rm{3}}}}$

$[{{\rm{H}}^ + }] = \frac{{ - {K_{\rm{a}}} + \sqrt {{K_{\rm{a}}}^2 + 4{K_{\rm{a}}}c} }}{2} = 1.0 \times {10^{ - 7}}{\rm{mol}} \cdot {\rm{d}}{{\rm{m}}^{ - {\rm{3}}}}$

### 2.2 一元弱酸pH计算实例

$[{{\rm{H}}^ + }] = \sqrt {{K_{\rm{a}}}c + {K_{\rm{w}}}} {\rm{ = }}2.7 \times {10^{ - 7}}{\rm{mol}} \cdot {\rm{d}}{{\rm{m}}^{ - {\rm{3}}}}$

$[{{\rm{H}}^ + }] = \frac{{ - {K_{\rm{a}}} + \sqrt {{K_{\rm{a}}}^2 + 4{K_{\rm{a}}}c} }}{2} = 2.5 \times {10^{ - 7}}{\rm{mol}} \cdot {\rm{d}}{{\rm{m}}^{ - {\rm{3}}}}，误差7.8\%。$

## 3 教学过程中的一些建议

1) 《无机化学》教材中应适当加入一元酸碱pH计算的近似条件的推导过程，帮助学生清晰理解这些近似条件的来源。这部分内容也可作为阅读材料让学生自学，加深其对溶液中离子平衡计算问题的认识。

2)在课堂讲解过程中，如果课时允许，应加入适当的推导演算环节，避免学生对公式死记硬背。同时，着重强调这些近似计算条件的适用前提。

3)在课程中补充介绍一些应用计算机软件计算酸碱pH的方法，如CurTiPot等程序或软件的使用等。

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