91影视

The representation of a chemical reaction in the type of substance is known as a chemical equation. The equation in which the quantity of particles of the relative multitude of atoms is equivalent on the two sides of the equation is known as a reasonable chemical equation.

Given pH =$3.4$

We know,

$pH=鈭�\log 鈦�\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$=$\log 鈦�\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=鈭�3.4$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=3.98脳{10}^{鈭�4}$

Now using,

role="math" localid="1653146271755" $$\begin{gathered} \left[{\mathrm{OH}}^{鈭�}\right]=\frac{1脳{10}^{鈭�14}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}=\frac{1脳{10}^{-14}}{3.98脳{10}^{-4}} \\ \left[{\mathrm{OH}}^{鈭�}\right]=2.5脳{10}^{-11} \end{gathered}$$

Given pH = $6$

We know,

$pH=鈭�\log 鈦�\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\log 鈦�\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=鈭�6$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=6.02脳{10}^{-7}$

Now using,

$$\begin{gathered} \left[{\mathrm{OH}}^{鈭�}\right]=\frac{1脳{10}^{鈭�14}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}=\frac{1脳{10}^{鈭�14}}{6.02脳{10}^{-7}} \\ \left[{\mathrm{OH}}^{鈭�}\right]=1.66脳{10}^{-8} \end{gathered}$$

Given pH = $8$

We know,

$pH=鈭�\log 鈦�\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\log 鈦�\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=鈭�8$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-8}$

Using,

$$\begin{gathered} \left[{\mathrm{OH}}^{鈭�}\right]=\frac{{10}^{鈭�14}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}=\frac{{10}^{鈭�14}}{{10}^{-8}} \\ \left[{\mathrm{OH}}^{鈭�}\right]={10}^{-6} \end{gathered}$$

Given pH = $11$

We know,

$\text{pH}鈭�\log 鈦�\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\log 鈦�\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=-11$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-11}$

Using,

role="math" localid="1653147202977" $\left[{\mathrm{OH}}^{鈭�}\right]=\frac{{10}^{鈭�14}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}=\frac{{10}^{鈭�14}}{{10}^{-11}}={10}^{鈭�3}$

Given pH =$9.2$

We know,

$\text{pH=}鈭�\log 鈦�\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\log 鈦�\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=-9.2$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=6.3脳{10}^{-10}$

Using,

$$\begin{gathered} \begin{array}{r}\left[{\mathrm{OH}}^{鈭�}\right]=\frac{1脳{10}^{-14}}{{\mathrm{H}}_3{\mathrm{O}}^{+}}\end{array} \\ \left[{\mathrm{OH}}^{鈭�}\right]=\frac{{10}^{鈭�14}}{6.3脳{10}^{-10}}=2.51脳{10}^{鈭�11} \end{gathered}$$