Question #4e4c2

1 Answer
May 4, 2017

Answer:

#HNO_2 + H_2O rightleftharpoonsNO_2^(-) + H_3O^+#

Explanation:

And thus, #K_a=([NO_2^-][H_3O^+])/([HNO_2])#

And for the bases, ..........

#N(CH_3)_3 + H_2O rightleftharpoonsHstackrel(+)N(CH_3)_3 +HO^-#

#K_b=([Hstackrel(+)N(CH_3)_3][HO^-])/([ddotN(CH_3)_3])#

You still have a bit of work to do, but in each case, the acid DONATES a proton TO the solvent. And the base ACCEPTS a proton FROM the solvent. The equilibrium may be measured and quantified by using the #K_a# and #K_b# values; these must be measured.

Note that with polyprotic acids, we simply accept that the acid can donate MORE than one proton. #H_3PO_4# can act as a diacid in aqueous solution:

#H_3PO_4+H_2Orightleftharpoons H_2PO_4^(-) +H_3O^+#

And..........

#H_2PO_4^(-) + H_2O rightleftharpoonsHPO_4^(2-) + H_3O^+#.