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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.

Short Answer

Expert verified

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

Step by step solution

01

Determine concentration

  • 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—its 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:
    \(\begin{aligned}c\left( {NO_2^ - } \right) &= 0.10M \cdot 1.5 \cdot {10^{ - 5}}\\c\left( {N_2^ - } \right) &= 1.5 \cdot {10^{ - 6}}M.\end{aligned}\)
02

Determine dissociation constant

  • 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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Most popular questions from this chapter

What will be the\(pH\)of a buffer solution prepared from\(0.20\;mol N{H_3}, 0.40\;mol N{H_4}N{O_3}\), and just enough water to give\(1.00\;L\)of solution?

Draw a curve similar to that shown in Figure 14.23 for a series of solutions of\(N{H_3}\) . Plot [OH] on the vertical axis and the total concentration of \(N{H_3}\) (both ionized and nonionized \(N{H_3}\) molecules) on the horizontal axis. Let the total concentration of \(N{H_3}\) vary from \(1 \times 1{0^{ - 10}}M\) to \(1 \times 1{0^{ - 2}}M.\)

Nitric acid reacts with insoluble copper (II) oxide to form soluble copper (II) nitrate,Cu (NO3)2, a compound that has been used to prevent the growth of algae in swimming pools. Write the balanced chemical equation for the reaction of an aqueous solution of HNO3 with CuO.

Calculate the percent ionization of a 0.10-M solution of acetic acid with a pH of 2.89.

Question: Show by suitable net ionic equations that each of the following species can act as a Bronsted-Lowry acid: (a) \(HN{O_3}\) (b) \(PH_4^ + \) (c) \({H_2}S\) (d) \(C{H_3}C{H_2}COOH\) (e) \({H_2}PO_4^ - \) (f) \(H{S^ - }\)
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