91影视

Fraction ofwould be almost same as ;

$$\begin{gathered} 伪\left(H_{2}A\right)=\frac{{\left[H^{+}\right]}^2}{{\left[H^{+}\right]}^2+{\left[H^{+}\right]}^2{K}_1+K_1{K}_2} \\ 伪\left(H_{3}A\right)=\frac{\left[H^{+}\right]}{{\left[H^{+}\right]}^3+{\left[H^{+}\right]}^2{K}_1+\left[H^{+}\right]K_1{K}_2+K_1{K}_2{K}_3} \end{gathered}$$

Fraction of $H_2{A}^{-}$would be almost same as $H{A}^{-}$;

role="math" localid="1663560545247" $$\begin{gathered} 伪\left(H{A}^{-}\right)=\frac{K_1\left[H^{+}\right]}{{\left[H^{+}\right]}^2+\left[H^{+}\right]K_1+K_1{K}_2} \\ 伪\left(H_2{A}^{-}\right)=\frac{K_1{\left[H^{+}\right]}^2}{{\left[H^{+}\right]}^3+{\left[H^{+}\right]}^2{K}_1+\left[H^{+}\right]K_1{K}_2+K_1{K}_2{K}_3} \end{gathered}$$

Fraction of$H{A}^{2-}$would be almost same as $H{A}^{-}$;

$$\begin{gathered} 伪\left(H{A}^{-}\right)=\frac{K_1\left[H^{+}\right]}{{\left[H^{+}\right]}^2+\left[H^{+}\right]K_1+K_1{K}_2} \\ 伪\left(H{A}^{2-}\right)=\frac{K_1{K}_2{\left[H^{+}\right]}^2}{{\left[H^{+}\right]}^3+{\left[H^{+}\right]}^2{K}_1+\left[H^{+}\right]K_1{K}_2+K_1{K}_2{K}_3} \end{gathered}$$

Fraction of$A^{3-}$would be almost same as $A^{2-}$;

$$\begin{gathered} 伪\left(A^{2-}\right)=\frac{K_1{K}_2}{{\left[H^{+}\right]}^2+\left[H^{+}\right]K_1+K_1{K}_2} \\ 伪\left(A^{2-}\right)=\frac{K_1{K}_2{K}_3}{{\left[H^{+}\right]}^3+{\left[H^{+}\right]}^2{K}_1+\left[H^{+}\right]K_1{K}_2+K_1{K}_2{K}_3} \end{gathered}$$

Hence, all equations are derived.

We will calculate the values of pK_a for phosphoric acid:

$$\begin{gathered} K_{1,2}={10}^{-p{K}_{a,1,2....}} \\ p{K}_{a1}=2.148鈫�K_1={10}^{-2.148},p{K}_{a2}=7.198鈫�K_2={10}^{-7.198},p{K}_{a3}=12.375鈫�K_3={10}^{-12.375} \end{gathered}$$

And also form$pH=7.00鈫�\left[H^{+}\right]={10}^{-7}$

Fraction of $H_{3}A$would be;

$$\begin{gathered} 伪\left(H_{3}A\right)=\frac{\left[H^{+}\right]}{{\left[H^{+}\right]}^3+{\left[H^{+}\right]}^2{K}_1+\left[H^{+}\right]K_1{K}_2+K_1{K}_2{K}_3} \\ 伪\left(H_{3}A\right)=\frac{\left({10}^{-7}\right)}{{\left({10}^{-7}\right)}^3+\left({\left({10}^{-7}\right)}^2{10}^{-2.148}\right)+\left({10}^{-7}{10}^{-2.148}{10}^{-7.198}\right)+\left({10}^{-2.148}{10}^{-7.198}{10}^{-12.375}\right)} \\ 伪\left(H_{3}A\right)=8.60脳{10}^{-6} \end{gathered}$$

Fraction of $H_{2}A$would be;

role="math" localid="1663561375345" $$\begin{gathered} 伪\left(H_{2}A\right)=\frac{K_1{\left[H^{+}\right]}^2}{{\left[H^{+}\right]}^3+{\left[H^{+}\right]}^2{K}_1+\left[H^{+}\right]K_1{K}_2+K_1{K}_2{K}_3} \\ 伪\left(H_{2}A\right)=\frac{{\left({10}^{-2.148}{10}^{-7}\right)}^2}{{\left({10}^{-7}\right)}^3+\left({\left({10}^{-7}\right)}^2{10}^{-2.148}\right)+\left({10}^{-7}{10}^{-2.148}{10}^{-7.198}\right)+\left({10}^{-2.148}{10}^{-7.198}{10}^{-12.375}\right)} \\ 伪\left(H_{2}A\right)=0.610 \end{gathered}$$

Fraction of would be;

$$\begin{gathered} 伪\left(H{A}^{2-}\right)=\frac{K_1{K}_2{\left[H^{+}\right]}^2}{{\left[H^{+}\right]}^3+{\left[H^{+}\right]}^2{K}_1+\left[H^{+}\right]K_1{K}_2+K_1{K}_2{K}_3} \\ 伪\left(H{A}^{2-}\right)=\frac{\left({10}^{-7.198}{10}^{-2.148}{10}^{-7}\right)}{{\left({10}^{-7}\right)}^3+\left({\left({10}^{-7}\right)}^2{10}^{-2.148}\right)+\left({10}^{-7}{10}^{-2.148}{10}^{-7.198}\right)+\left({10}^{-2.148}{10}^{-7.198}{10}^{-12.375}\right)} \\ 伪\left(H{A}^{2-}\right)=0.390 \end{gathered}$$

Fraction of $A^3$would be;

$$\begin{gathered} 伪\left(A^{3-}\right)=\frac{K_1{K}_2{K}_3}{{\left[H^{+}\right]}^3+{\left[H^{+}\right]}^2{K}_1+\left[H^{+}\right]K_1{K}_2+K_1{K}_2{K}_3} \\ 伪\left(A^{3-}\right)=\frac{{10}^{-2.148}{10}^{-7.198}{10}^{-12.375}}{{\left({10}^{-7}\right)}^3+\left({\left({10}^{-7}\right)}^2{10}^{-2.148}\right)+\left({10}^{-7}{10}^{-2.148}{10}^{-7.198}\right)+\left({10}^{-2.148}{10}^{-7.198}{10}^{-12.375}\right)} \\ 伪\left(A^{3-}\right)=1.6脳{10}^{-6} \end{gathered}$$

The final answer is:

$$\begin{gathered} 伪\left(H_{3}A\right)=8.60脳{10}^{-6} \\ 伪\left(H_{2}A\right)=0.610 \\ 伪\left(H{A}^{2-}\right)=0.390 \\ 伪\left(A^{2-}\right)=1.6脳{10}^{-6} \end{gathered}$$