91影视

The pH of the solution can be calculated as:

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

(a)

The concentration of the given solution is $0.010\mathrm{M}$.

${\mathrm{HNO}}_3鈫�{\mathrm{H}}^{+}+{\mathrm{NO}}_3^{鈥�}$

The hydrogen ion concentration is equal to the actual concentration of the solution.

Therefore, $\left[{\mathrm{H}}^{+}\right]=0.010\mathrm{M}$

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ =-\log \left(0.010\right) \\ =2.00 \end{gathered}$$

The concentration of the given solution is 0.035M .

$\mathrm{KOH}鈫�{\mathrm{K}}^{+}+{\mathrm{OH}}^{鈥�}$

The concentration of the hydroxyl ion is equal to the actual concentration of the solution.

Therefore, $\left[{\mathrm{OH}}^{-}\right]=0.035\mathrm{M}$

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]=\frac{\mathrm{K}}{\left[{\mathrm{OH}}^{-}\right]} \\ =\frac{1.0脳{10}^{-14}}{0.035\mathrm{M}} \\ =2.86脳{10}^{-13}\mathrm{M} \\ \mathrm{pH}=-\log \left(2.86脳{10}^{-13}\right) \\ =12.54 \end{gathered}$$

The concentration of the given solution is 0.030M

$\mathrm{HCI}鈫�{\mathrm{H}}^{+}+{\mathrm{CI}}^{-}$

The hydrogen ion concentration is equal to the actual concentration of the solution.

Therefore, $\left[{\mathrm{H}}^{+}\right]=0.030\mathrm{M}$

localid="1663390304057" $$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ =-\log \left(0.030\right) \\ =1.52 \end{gathered}$$

The concentration of the given solution is 3.0M .

$\mathrm{HCI}鈫�{\mathrm{H}}^{+}+{\mathrm{CI}}^{-}$

The hydrogen ion concentration is equal to the actual concentration of the solution.

Therefore, $\left[{\mathrm{H}}^{+}\right]=3.0\mathrm{M}$

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ =-\log \left(3.0\right) \\ =-0.48 \end{gathered}$$

The concentration of the given solution is 0.010M

The concentration of hydroxyl ion is equal to the actual concentration of the solution.

Therefore, $\left[{\mathrm{OH}}^{-}\right]=0.010\mathrm{M}$

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{OH}}^{-}\right]} \\ =\frac{1.0脳{10}^{-14}}{0.010\mathrm{M}} \\ =1.0脳{10}^{-12}\mathrm{M} \\ \mathrm{pH}=-\log \left(1.0脳{10}^{-12}\right) \\ =12.00 \end{gathered}$$