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) $HSO_4^-$

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Answer 1 · 2017-09-15T05:29:57

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

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.

How do you write the balanced acid equation and the dissociation expression Ka for following compounds in water? a) H_3PO_4H_3PO_4 b) HClO_2HClO_2 c) CH_3COOHCH_3COOH d) HCO_3^-HCO_3^- e) HSO_4^-HSO_4^-