Question

Question #4e4c2

Answer 1

`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^+`.