We will abbreviate malonic acid, ${\mathbf{CH}}_{\mathbf{2}}\mathbf{(}{\mathbf{CO}}_{\mathbf{2}}\mathbf{H}{\mathbf{)}}_{\mathbf{2}}$ , as${\mathbf{H}}_{\mathbf{2}}\mathbf{M}$ . Find the pH and concentrations of${\mathbf{H}}_{\mathbf{2}}\mathbf{M}\mathbf{,}{\mathbf{HM}}^{\mathbf{-}}$ , and ${\mathbf{M}}^{\mathbf{2}\mathbf{-}}$in

(a) $\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{M}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{M}$;

(b) $\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{MNaHM}$;

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

• The pH of a solution can be determined utilizing the negative logarithm of the hydrogen ion concentration.
• 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 ${\mathrm{H}}_2\mathrm{M}$It is necessary to find a solution.

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

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

$\mathrm{pH}=-\log [{\mathrm{H}}^{+}{]}_{\mathrm{\gamma }}\mathrm{H}$

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

The pH and concentration of a solution can be estimated using the following formula:

role="math" localid="1663560070934" $$\begin{gathered} \frac{{\mathrm{x}}^2}{0.100-\mathrm{x}}={\mathrm{K}}_1\mathrm{x} \\ =1.12\times {10}^{-2} \\ \mathrm{pH}=-\mathrm{logx} \\ =1.95 \\ \left[{\mathrm{H}}_2\mathrm{M}\right]=0.1000-\mathrm{x} \\ =0.089\mathrm{M} \\ \left[{\mathrm{HM}}^{-}\right]=\mathrm{x} \\ =1.12\times {10}^{-2}\mathrm{M} \\ \left[{\mathrm{M}}^{2-}\right]=\frac{{\mathrm{K}}_2\left[\mathrm{HM}\right]}{\left[{\mathrm{H}}^{+}\right]} \\ =2.01\times {10}^{-6}\mathrm{M} \end{gathered}$$

The pH of given solution is 1.95 and the concentration is $2.01\times {10}^{-6}\mathrm{M}$.

(b)

The pH of the NaHM solution, as well as the ion concentration, must be fine-tuned.

$\left[H^{+}\right]=\sqrt{\frac{K_1{K}_{2}F+K_1{K}_w}{K_1+F}}$

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

$\mathrm{pH}=-\log [{\mathrm{H}}^{+}{]}_{\mathrm{\gamma }}{\mathrm{H}}^{-}$

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

The pH and concentration of a solution can be estimated using the following formula:

$\left[{\mathrm{H}}^{+}\right]»\sqrt{\frac{{\mathrm{K}}_1{\mathrm{K}}_2(0.100)+{\mathrm{K}}_1{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_1+(0.100)}}$

$$\begin{gathered} =5.30\times {10}^{-5}\mathrm{M}=\mathrm{pH}=4.28 \\ \left[{\mathrm{HM}}^{-}\right]»0.100\mathrm{M} \end{gathered}$$

$$\begin{gathered} \left[{\mathrm{H}}_2\mathrm{M}\right]\frac{\left[\mathrm{H}+\parallel {\mathrm{HM}}^{-}\right]}{{\mathrm{K}}_1}=3.7\times {10}^{-3}\mathrm{M}\left[{\mathrm{M}}^{2-}\right] \\ =\frac{{\mathrm{K}}_2\left[{\mathrm{HM}}^{-}\right]}{\left[{\mathrm{H}}^{+}\right]} \\ =3.8\times {10}^{-3}\mathrm{M} \end{gathered}$$

The pH of given solution is 4.28 and the concentration is $3.8\times {10}^{-3}\mathrm{M}$.

(c)

The ion concentration and pH in the water the answer must be fine-tuned..

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

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

$\mathrm{pH}=-\log [{\mathrm{H}}^{+}{]}_{\mathrm{\gamma }}{\mathrm{H}}^{-}$

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

The pH and concentration of a solution can be estimated using the following formula:

${\mathrm{M}}^{2-}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{HM}}^{-}+{\mathrm{OH}}^{-}$

$$\begin{gathered} \frac{{\mathrm{x}}^2}{0.100-\mathrm{x}}={\mathrm{K}}_{\mathrm{bl}} \\ \mathrm{x}=2.23\times {10}^{-5} \\ \mathrm{pH}=-\log \frac{\mathrm{Kw}}{\mathrm{x}} \\ =9.35 \\ \frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}2}}=4.98\times {10}^{-9} \\ \left[{\mathrm{M}}^{2-}\right]=0.100-\mathrm{x} \\ =0.100\mathrm{M} \\ \left[{\mathrm{HM}}^{-}\right]=\mathrm{x} \\ =2.23\times {10}^{-5}\mathrm{M} \\ \left[{\mathrm{H}}_2\mathrm{A}\right]\gg \frac{\left|\mathrm{H\_}+\right||{\mathrm{HA}}^{-1}}{{\mathrm{K}}_1} \\ =7.04\times {10}^{-12}\mathrm{M} \end{gathered}$$

The pH of ${\mathrm{Na}}_2\mathrm{M}$solution is 9.35 and the concentration is $7.04\times {10}^{-12}\mathrm{M}$