Consider the diprotic acid${\mathbf{H}}_{\mathbf{2}}\mathbf{}\mathbf{A}$with${\mathbf{K}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$and${\mathbf{K}}_2\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}$.Find the pH and concentrations of${\mathbf{H}}_{\mathbf{2}}\mathbf{A}\mathbf{,}{\mathbf{HA}}^{\mathbf{-}}$, and${\mathbf{A}}^{\mathbf{2}\mathbf{-}}$in

(a)role="math" localid="1654926233413" $\mathbf{0}\mathbf{.}\mathbf{100}{\mathbf{MH}}_{\mathbf{2}}\mathbf{}\mathbf{A}\mathbf{;}$

(b) role="math" localid="1654926403235" $\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{MNaHA}$ ;

(c) $\mathbf{0}\mathbf{.}\mathbf{100}{\mathbf{MNa}}_{\mathbf{2}}\mathbf{}\mathbf{A}$ .

• The negative logarithm of hydrogen ion concentration is used to calculate the pH of a solution.
• Concentration is a notion used frequently in chemistry and related sciences. It's a metric for determining how much of one chemical is combined with another.

(a)

The content and pH of it must be discovered.

$\left[{\mathrm{H}}^{+}\right]=\sqrt{\frac{{\mathrm{K}}_1{\mathrm{K}}_2\mathrm{F}+{\mathrm{K}}_1{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_1+\mathrm{F}}}$

The pH of a solution can be determined using the formula:

$\mathrm{pH}=-\log {\left[{\mathrm{H}}^{+}\right]}_{\mathrm{y}}\mathrm{H}$

The negative logarithm of Hydrogen ion activity can be calculated using the equation above.

Supplied, the given acid's dissociation

localid="1654928956651" $$\begin{gathered} \frac{{\mathrm{x}}^2}{0.100-\mathrm{x}}={\mathrm{K}}_1 \\ \mathrm{x}=3.11\times {10}^{-3} \\ \left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{HA}}^{-}\right] \\ \mathrm{pH}=2.51 \end{gathered}$$

$$\begin{gathered} \left[{\mathrm{HA}}^{-}\right]=0.100-\mathrm{x} \\ =0.0969\mathrm{M} \\ \left[{\mathrm{A}}^{2-}\right]=\frac{{\mathrm{K}}_2\left[{\mathrm{HA}}^{-}\right]}{\left[{\mathrm{H}}^{+}\right]} \\ =1.00\times {10}^{-3}\mathrm{M} \end{gathered}$$

The pH and the concentration of ${\mathrm{H}}_2\mathrm{A}$is 2.51 and$1.00\times {10}^{-8}\mathrm{M}$

(b)

The ion concentration and pH in the water ${\mathrm{N}}_{\mathrm{a}}{\mathrm{HSO}}_3$It must be discovered.

$\left[{\mathrm{H}}^{+}\right]=\sqrt{\frac{{\mathrm{K}}_1{\mathrm{K}}_2\mathrm{F}+{\mathrm{K}}_1{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_1+\mathrm{F}}}$

The pH of a solution can be determined using the formula:

$\mathrm{pH}=-\log {\left[{\mathrm{H}}^{+}\right]}_{\mathrm{y}}\mathrm{H}$

The negative logarithm of Hydrogen ion activity can be calculated using the equation above.

The pH of a solution can be determined using the formula:

$\left[{\mathrm{H}}^{+}\right]=\sqrt{\frac{{\mathrm{K}}_1{\mathrm{K}}_2\mathrm{F}+{\mathrm{K}}_1{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_1+\mathrm{F}}}$

The pH and the concentration of is 6.00 and$1.00\times {10}^{-3}\mathrm{M}$

(c)

The ion concentration and pH in the water ${\mathrm{Na}}_2$It must be discovered.

$\left[{\mathrm{H}}^{+}\right]=\sqrt{\frac{{\mathrm{K}}_1{\mathrm{K}}_2\mathrm{F}+{\mathrm{K}}_1{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_1+\mathrm{F}}}$

The pH of a solution can be determined using the formula:

$\mathrm{pH}=-\log {\left[{\mathrm{H}}^{+}\right]}_{\mathrm{y}}\mathrm{H}$

The negative logarithm of Hydrogen ion activity can be calculated using the equation above.

The pH of a solution can be determined using the formula:

localid="1655098735494" $$\begin{gathered} \frac{{\mathrm{x}}^2}{0.100-\mathrm{x}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_2}=\mathrm{x}=\left[{\mathrm{OH}}^{-}\right]=\left[{\mathrm{HA}}^{-}\right]=3.16\times {10}^{-4}\mathrm{M} \\ \mathrm{pH}=10.50 \\ \left[{\mathrm{A}}^{2-}\right]=0.100-\mathrm{x}=9.97\times {10}^{-2}\mathrm{M} \\ \left[{\mathrm{H}}_2\mathrm{A}\right]\gg \frac{[\mathrm{H}+1\left[{\mathrm{HA}}^{-}\right]}{{\mathrm{K}}_1}=1.00\times {10}^{-10}\mathrm{M} \end{gathered}$$

The solution's concentration is$1.00\times {10}^{-10}\mathrm{M}$

The pH and the concentration of${\mathrm{Na}}_2\mathrm{A}$is 10.50 and$1.00\times {10}^{-10}\mathrm{M}$