Q14.3-58 E Determine \({K_b}\)for the nitri... [FREE SOLUTION]

Chapter 14: Q14.3-58 E (page 832)

Determine \({K_b}\)for the nitrite ion, \(NO_2^ - \). In a \(0.10 - M\)solution this base is \(0.0015\% \)ionized.

Expert verified

The value of \({K_b}\) for the nitrite ion is \(6666.67\).

The nitrite ion reacts with water in the following manner:
\({\rm{NO}}_2^ - (aq) + {{\rm{H}}_2}{\rm{O}}(l){\rm{ }} \to {\rm{HN}}{{\rm{O}}_2}(aq) + {\rm{O}}{{\rm{H}}^ - }(aq).\)

From the equation, the \({K_b}\)value for the nitrite ion is \({K_b} = \frac{{c\left( {H{\rm{N}}{{\rm{O}}_2}} \right) \cdot c\left( {{\rm{O}}{{\rm{H}}^ - }} \right)}}{{c\left( {{\rm{NO}}_2^ - } \right)}}.\)
If this base is \(0.0015\% \)ionized in the solution, it means that the ratio of its ionized form concentration versus its total鈥攊ts ionized plus its unionized form is \(0.0015\% .\)
\(\frac{{c\left( {{\rm{NO}}_2^ - } \right)}}{{c\left( {{\rm{NO}}_2^ - } \right) + c\left( {{\rm{H}}{{\rm{O}}_2}} \right)}} = 0.0015\% .\)
As the total concentration is \(0.10{\rm{ }}M,\)we can write the last equation as \(\frac{{c\left( {NO_2^ - } \right)}}{{0.10M}} = 0.0015\% .\)
Calculating \(c\left( {{\rm{NO}}_2^ - } \right)\), we get:
\(\begin{aligned}c\left( {NO_2^ - } \right) &= 0.10M \cdot 1.5 \cdot {10^{ - 5}}\\c\left( {N_2^ - } \right) &= 1.5 \cdot {10^{ - 6}}M.\end{aligned}\)

Once we know the nitrite ion concentration, we can calculate the equilibrium concentration of its conjugate acid, \({\rm{HN}}{{\rm{O}}_2}\):
\(\begin{aligned}c\left( {{\rm{HO}}{{\rm{O}}_2}} \right) &= 0.10M - 1.5 \cdot {10^{ - 6}}M\\c\left( {{\rm{HN}}{{\rm{O}}_2}} \right) &= 0.10M.\end{aligned}\)

Having a look at the ionization reaction of the nitrite ion, let us assume that the concentrations of \({\rm{HN}}{{\rm{O}}_2}\)and \(O{H^ - }\) are equal and \(0.10{\rm{M}}.\)
Now, as we have all the date, we can calculate the \({K_b}\).
\(\begin{aligned}{K_b} &= \frac{{c\left( {{\rm{HN}}{{\rm{O}}_2}} \right) \cdot c\left( {{\rm{O}}{{\rm{H}}^ - }} \right)}}{{c\left( {{\rm{NO}}_2^ - } \right)}}\\Kb &= \frac{{0.10{\rm{M}} \cdot 0.1{\rm{M}}}}{{1.5 \cdot {{10}^{ - 6}}{\rm{M}}}}\\{K_b} &= 6666.67.\end{aligned}\)
Therefore, the \({K_b}\) for the nitrite ion is \(6666.67\).

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