How do you write the chemical equation and the #K_a# expression for the acid dissociation of the following acids in aqueous solution?

Question

#C_6H_5COOH#
#HCO_3^-#

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Answer 1 · 2016-08-23T15:50:22

Attend:

#C_6H_5CO_2H(aq) + H_2O rightleftharpoons H_3O^+ + C_6H_5CO_2^-#

#K_a=([H_3O^+][C_6H_5CO_2^-])/([C_6H_5CO_2H])#

#HCO_3^(-) +H_2O(l) rightleftharpoons CO_3^(2-) + H_3O^+#

#K_a=???#

All I have done here is conserve mass and charge. The second equilibrium will lies to the left inasmuch as bicarbonate is a very indifferent acid.