91Ó°ÊÓ

• 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:

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

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

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

(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:

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

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:

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

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

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

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

(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:

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

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:

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

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

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