91Ó°ÊÓ

As a solution gets more acidic as H+ increases, the pH diminishes. As a solution gets more basic (higher [OH-]), the pH increases. As the pH of a solution diminishes by one pH unit, the concentration of H+ increases by ten times

Given, pH = $10$

We know, pH = -log

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{−\mathrm{pH}}$and$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{−}\right]=1×{10}^{−14}$

$$\begin{gathered} ⇒\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{−10} \\ =1.0×{10}^{−10}\mathrm{M} \end{gathered}$$

$\left[{\mathrm{OH}}^{−}\right]=\frac{1.0×{10}^{−14}}{1.0×{10}^{10}}=1×{10}^{−4}\mathrm{M}$

Given pH = $5$

We know, pH = -log

$\begin{array}{r}\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{−\mathrm{pH}}\text{and}\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{−}\right]=1×{10}^{−14}\end{array}$

$⇒\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{−5}$

$=1.0×{10}^{−5}\mathrm{M}$

$\begin{array}{r}\left[{\mathrm{OH}}^{−}\right]=\frac{1.0×{10}^{−14}}{1.0×{10}^{-5}}=1×{10}^{−9}\mathrm{M}\end{array}$

Given pH = $7$

We know, pH = -log
$\begin{array}{r}\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{−\mathrm{pH}}\text{and}\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{−}\right]=1×{10}^{−14}\end{array}$

$⇒\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{−7}$$=1.0×{10}^{−7}\mathrm{M}$

$\begin{array}{r}\left[{\mathrm{OH}}^{−}\right]=\frac{1.0×{10}^{−14}}{1.0×{10}^{-7}}=1×{10}^{−7}\mathrm{M}\end{array}$

Given pH = $1.82$

We know, pH = -log

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{−\mathrm{pH}}\text{and}\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{−}\right]=1×{10}^{−14}$

$⇒\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{−1.82}$

$=1.51×{10}^{−2}\mathrm{M}$

$\begin{array}{r}\left[{\mathrm{OH}}^{−}\right]=\frac{1.0×{10}^{−14}}{1.51×{10}^{-2}}=6.6×{10}^{−13}\mathrm{M}\end{array}$