# Question #4e4c2

May 4, 2017

$H N {O}_{2} + {H}_{2} O r i g h t \le f t h a r p \infty n s N {O}_{2}^{-} + {H}_{3} {O}^{+}$

#### Explanation:

And thus, ${K}_{a} = \frac{\left[N {O}_{2}^{-}\right] \left[{H}_{3} {O}^{+}\right]}{\left[H N {O}_{2}\right]}$

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

$N {\left(C {H}_{3}\right)}_{3} + {H}_{2} O r i g h t \le f t h a r p \infty n s H \stackrel{+}{N} {\left(C {H}_{3}\right)}_{3} + H {O}^{-}$

${K}_{b} = \frac{\left[H \stackrel{+}{N} {\left(C {H}_{3}\right)}_{3}\right] \left[H {O}^{-}\right]}{\left[\ddot{N} {\left(C {H}_{3}\right)}_{3}\right]}$

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. ${H}_{3} P {O}_{4}$ can act as a diacid in aqueous solution:

${H}_{3} P {O}_{4} + {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{2} P {O}_{4}^{-} + {H}_{3} {O}^{+}$

And..........

${H}_{2} P {O}_{4}^{-} + {H}_{2} O r i g h t \le f t h a r p \infty n s H P {O}_{4}^{2 -} + {H}_{3} {O}^{+}$.