Ionic Equilibrium

1

2

Zero

7

< 7

7 but < 8

8 but < 9

10

6.0 to 6.4

7.36 to 7.42

6.4 to 6.95

7.8 to 8.0

2.4%

4.2%

0.2%

$0.4 \%$

$\mathrm{ClO}_{4}^{2-}$

$\mathrm{ClO}_{4}$

HCl

$\mathrm{ClO}_{4}^{-}$

$HPO_{4}^{-}$

$HPO_{4}^{2-}$

$H_{3}PO_{4}$

$PO_{4}^{3-}$

$pH=pK_{a}-\log\dfrac{\left[ CH_{3}COONa \right]}{\left[ CH_{3}COOH \right]}$

$pH=pK_{a}+\log\dfrac{\left[ CH_{3}COONa \right]}{\left[ CH_{3}COOH \right]}$

$pOH=pK_{a}-\log\dfrac{\left[ CH_{3}COONa \right]}{\left[ CH_{3}COOH \right]}$

$pOH=pK_{a}-\log\dfrac{\left[ CH_{3}COOH \right]}{\left[ CH_{3}COONa \right]}$

For basic solution, $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]$

In basic solution, the excess of $\mathrm{OH}^{-}$ions are present.

In the basic solution, there is excess of $\mathrm{H}_{3} \mathrm{O}^{+}$ions.

The pOH of a basic solution can be calculated by $\mathrm{pOH}=-\log \left[\mathrm{H}^{+}\right]$

11

10

12

3

$\mathrm{NH}_{3}$

NaOH

KOH

HI

$\mathrm{HPO}_{3}^{2-}$ and $\mathrm{SO}_{4}^{2-}$

$\mathrm{HPO}_{3}^{2-}$ and $\mathrm{HSO}_{4}^{-}$

$\mathrm{H}_{2} \mathrm{PO}_{3}^{-}$and $\mathrm{HSO}_{4}^{-}$

$\mathrm{PO}_{3}^{3-}$ and $\mathrm{SO}_{4}^{2-}$

$\mathrm{H}_{3} \mathrm{O}^{+}$and $\mathrm{H}_{2} \mathrm{O}$

$\mathrm{H}_{2} \mathrm{O}$ and $\mathrm{HPO}_{4}^{2-}$

$\mathrm{PO}_{4}^{3-}$ and $\mathrm{H}_{3} \mathrm{O}^{+}$

$\mathrm{H}_{3} \mathrm{O}^{+}$and $\mathrm{HPO}_{4}^{2-}$

$\quad \mathrm{H}_{2} \mathrm{SO}_{3}$ and $\mathrm{HSO}_{3}^{-}$

$\mathrm{HSO}_{3}^{-}$and $\mathrm{H}_{3} \mathrm{O}^{+}$

$\mathrm{H}_{2} \mathrm{SO}_{3}$ and $\mathrm{H}_{2} \mathrm{O}$

$\mathrm{H}_{3} \mathrm{O}^{+}$and $\mathrm{H}_{2} \mathrm{SO}_{3}$

$\mathrm{NH}_{4} \mathrm{OH}$ and $\mathrm{NH}_{2} \mathrm{OH}$

$\mathrm{NH}_{3}$ and $\mathrm{NH}_{2}^{-}$

$\mathrm{NH}_{3}$ and $\mathrm{NH}_{4}{ }^{+}$

$\mathrm{NH}_{4}+$ and $\mathrm{NH}_{3}$

$\mathrm{CH}_{3} \mathrm{COOH}$

$\mathrm{H}_{2} C_{2} \mathrm{O}_{4}$

HCl

$\mathrm{NH}_{3}$