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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

Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, and HCl. Milk of Magnesia, a suspension of solid Mg(OH)2 in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs

The odour of vinegar is due to the presence of acetic acid, \(C{H_3}C{O_2}H\), a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a \(1 - \)M aqueous solution of this acid.

What is the effect on the concentrations of \(N{O_2}^ - ,HN{O_2},\)and \(O{H^ - }\)when the following are added to a solution of \(KN{O_2}\)in water?

\(\begin{aligned}{l}(a)HCl\\(b)HN{O_2}\\(c)NaOH\\(d)NaCl\\(e)KNO\end{aligned}\)

The equation for the equilibrium is \(NO_2^ - (aq) + {H_2}O(l) \rightleftharpoons HN{O_2}(aq) + O{H^ - }(aq)\)

The ionization constant of lactic acid, \(C{H_3}CH(OH)C{O_2}H\), an acid found in the blood after strenuous exercise, is \(1.36 \times 1{0^{ - 4}}\). If \(20.0\;g\) of lactic acid is used to make a solution with a volume of \(1.00L\), what is the concentration of hydronium ion in the solution?

What is the effect on the concentration of ammonia, hydroxide ion, and ammonium ion when the following are added to a basic buffer solution of equal concentrations of ammonia and ammonium nitrate:

(a)\(KI\)

(b)\(N{H_3}\)

(c)\(HI\)

(d)\(NaOH\)

(e)\(N{H_4}Cl\)
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