# How do you write the balanced acid equation and the dissociation expression Ka for following compounds in water? a) H_3PO_4 b) HClO_2 c) CH_3COOH d) HCO_3^- e) HSO_4^-

Sep 15, 2017

$H A \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + {A}^{-}$

#### Explanation:

And ${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[{A}^{-}\right]}{\left[H A\right]}$

If ${K}_{a}$ is LARGE, then we gots a strong acid...if ${K}_{a}$ is small, then we got a weak acid, and the equilibrium LIES to the LEFT as written.

For $a .$ we gots a diprotic acid.....

${H}_{3} P {O}_{4} \left(a q\right) + 2 {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s H P {O}_{4}^{-} + 2 {H}_{3} {O}^{+}$

${K}_{a} = \frac{\left[H P {O}_{4}^{2 -}\right] {\left[{H}_{3} {O}^{+}\right]}^{2}}{\left[{H}_{3} P {O}_{4}\right]}$

For $b .$ we gots a monoprotic weak acid.....

$H C l {O}_{2} \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s C l {O}_{2}^{-} + {H}_{3} {O}^{+}$

${K}_{a} = \frac{\left[H C l {O}_{2}^{-}\right] \left[{H}_{3} {O}^{+}\right]}{\left[H C l {O}_{2}\right]}$

For $c .$ we gots a monoprotic weak acid.....

$H O A c \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s A c {O}^{-} + {H}_{3} {O}^{+}$

K_a=??. Have bash at the remaining expressions yourself.