Question

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^-`

Answer 1

`HA(aq) + H_2O(l) rightleftharpoons H_3O^+ + A^-`

Explanation:

And `K_a=([H_3O^+][A^-])/([HA])`

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_3PO_4(aq) + 2H_2O(l) rightleftharpoonsHPO_4^(-) +2H_3O^+`

`K_a=([HPO_4^(2-)][H_3O^+]^2)/([H_3PO_4])`

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

`HClO_2(aq) + H_2O(l) rightleftharpoonsClO_2^(-) +H_3O^+`

`K_a=([HClO_2^(-)][H_3O^+])/([HClO_2])`

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

`HOAc(aq) + H_2O(l) rightleftharpoonsAcO^(-) +H_3O^+`

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