Question

Phosphonocarboxylic acid

effectively inhibits the replication of the herpes virus. Structurally, it is a

combination of phosphoric acid and acetic acid. It can donate three protons.

The equilibrium constant values are

${\mathbf{Ka}}_{{}_1}\mathbf{1}.\mathbf{0}\times {\mathbf{10}}^{22},{\mathbf{Ka}}_2\mathbf{7}.\mathbf{8}\times {\mathbf{10}}^{26}\mathbf{and}{\mathbf{Ka}}_3\mathbf{2}.\mathbf{0}\times {\mathbf{10}}^{29}$.

Enough phosphonocarboxylic acid is added to blood (pH 7.40) to make its

total concentration 1.0 × 1025 M. The pH of the blood does not

change. Determine the concentrations of all four forms of the acid in this

mixture

Short answer

$\begin{array}{l}{\mathrm{H}}_3{\mathrm{O}}^{+}=3{.98.10}^{-8}\mathrm{M}\\ {\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_4=3{.98.10}^{-8}\mathrm{M}\\ {\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_5=9{.96.10}^{-6}\mathrm{M}\\ {\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_3=7{.8.10}^{-6}\mathrm{M}\\ {\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_2=3{.92.10}^{-7}\mathrm{M}\end{array}$

Step-by-step solution

The three reactions of k

The first reaction is:

${\mathrm{CPO}}_5{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{CPO}}_5{\mathrm{H}}_4+{\mathrm{H}}_3{\mathrm{O}}^{+}$

The ${\mathrm{K}}_{\mathrm{a}1}$is ${\mathrm{K}}_{\mathrm{a}1}=1{.10}^{-2}$

The second reaction is:

${\mathrm{CPO}}_5{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{CPO}}_5{\mathrm{H}}_4+{\mathrm{H}}_3{\mathrm{O}}^{+}$

The ${\mathrm{K}}_{\mathrm{a}2}$is ${\mathrm{K}}_{\left({\mathrm{n}}_2\right)}=7.8·{10}^{(-6)}$

The third reaction is

${\mathrm{CPO}}_5{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{CPO}}_5{\mathrm{H}}_4+{\mathrm{H}}_3{\mathrm{O}}^{+}$

The ${\mathrm{K}}_{\mathrm{a}3}$ is ${\mathrm{K}}_{\mathrm{a}3}=2{.10}^{-9}$

The pH of Hydronium

Because $\mathrm{pH}$is $7.4$, the concentration of ${\mathrm{H}}_3{\mathrm{O}}^{+}$is:

$\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{]=10}}^{\text{(-pH)}}{\text{=10}}^{\text{(-7.4)}}\\ {\text{[H}}_{\text{3}}{\text{O]=3.98g10}}^{\text{(-8)}}\text{M}\end{array}$

The concentration of Ka1

The ${\mathrm{K}}_{\mathrm{a}1}$is

${\text{K}}_{\text{a1}}\text{=}\frac{{\text{[C}}_{\text{2}}{\text{PO}}_{\text{5}}{\text{H}}_{\text{5}}{\text{]H}}_{\text{3}}\text{O}}{{\text{[C}}_{\text{2}}{\text{PO}}_{\text{5}}{\text{H}}_{\text{5}}\text{]}}$

The concentration of${\mathrm{H}}_3\mathrm{O}$ is

${\mathrm{H}}_3\mathrm{O}=3{.98.10}^{-8}\mathrm{M}$

The concentration of$\left[{\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_4\right]$ is

$\left[{\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_4\right]=3{.98.10}^{-8}\mathrm{M}$

The concentration of$\left[{\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_5\right]$ is

$\begin{array}{l}\left[{\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_5\right]=1{.10}^{(-3)}-\mathrm{x}\\ \left[{\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_5\right]=1{.10}^{(-5)}-3{.98.10}^{(-8)}\\ \left[{\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_5\right]=9{.96.10}^{(-6)}\mathrm{M})\end{array}$

The $\mathrm{pH}$and the concentration of solution is determined by tirst dissociation of acid.

The concentration of Ka2

The${\mathrm{K}}_{\mathrm{a}2}$ is

$\begin{array}{l}{\mathrm{K}}_{\mathrm{a}1}=\frac{\left[{\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_3\right]{\mathrm{H}}_3\mathrm{O}}{\left[{\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_5\right]}\\ \left[{\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_3\right]=\frac{{\mathrm{K}}_{\mathrm{a}2}\left[{\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_3\right]}{{\mathrm{H}}_3\mathrm{O}}\\ {\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_3=\frac{7{.8.10}^{-6}.3{.98.10}^{-5}}{3{.98.10}^{-8}}\\ {\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_3=7{.8.10}^{-6}\mathrm{M}\end{array}$

The concentration of Ka3

The ${\mathrm{K}}_{\mathrm{a}3}$ is

$\begin{array}{l}{\mathrm{K}}_{\mathrm{a}1}=\frac{\left[\right[{\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_2]{\mathrm{H}}_3\mathrm{O}}{\left[{\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_5\right]}\\ \left[{\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_2\right]=\frac{{\mathrm{K}}_{\mathrm{a}1}\left[{\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_3\right]}{{\mathrm{H}}_3\mathrm{O}}\\ \left[{\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_2\right]=\frac{2{.10}^{-9}.7{.8.10}^{-6}}{3{.98.10}^{-8}}\\ \left[{\mathrm{C}}_2{\mathrm{PO}}_5{\mathrm{H}}_2\right]=3{.92.10}^{-7}\mathrm{M}\end{array}$