Which is the stronger base, \({\left( {C{H_3}} \right)_3}N\)or\({H_2}BO_3^ - \)?

It is a substance capable of removing a proton from a very weak acid. In water, they either entirely dissociate into their ions or they completely dissociate into their ions.

We calculate the \({K_b}\)values. The \({K_a}\;and\;{K_b}\)values for weak acids and bases are checked. The \({K_b}\)values for \(\left( {C{H_3}} \right)N\;is\;6.3 \times 1{0^{ - 5}}\). No \({H_2}BO_3^ - \)is in the table and so we need to calculate \({K_b}\)from \({K_a}\)of its conjugated acid. The \({K_a}\)of its conjugated acid \({H_3}B{O_3}\)is \(5.4 \times 1{0^{ - 10}}\). We calculate\({K_b}\)from \({K_a}\)of its conjugated acid by using the formula

\({K_b} = \frac{{{K_w}}}{{{K_a}}}\)

where \({K_w}\)is the ionic product of water equals \(1{0^{ - 14}}\).

\(\begin{aligned}{{K_b} = \frac{{1{0^{ - 14}}}}{{5.4 \times 1{0^{ - 10}}}}}\\{{K_b} = 1.9 \times 1{0^{ - 5}}}\end{aligned}\)The two values are used for comparing the strength of the bases. \(\left( {C{H_3}} \right)N\) has higher \({K_b}\)values than \({H_2}BO_3^ - \)and so is the stronger base.