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Chapter 14: Q14.3-53 E (page 831)

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 \(NO_2^ - (aq) + {H_2}O(l) \rightleftharpoons HN{O_2}(aq) + O{H^ - }(aq)\)

Short Answer

Expert verified

The Le Chatelier principle is used to solve this problem.

\(\begin{aligned}{}{}&{c\left( {{\rm{NO}}_2^ - } \right)}&{c\left( {{\rm{O}}{{\rm{H}}^ - }} \right)}&{c\left( {{\rm{HN}}{{\rm{O}}_2}} \right)}\\{{\rm{ a) }}}&{\rm{D}}&{\rm{D}}&{\rm{I}}\\{{\rm{ b) }}}&{\rm{I}}&{\rm{D}}&{\rm{I}}\\{{\rm{ c) }}}&{\rm{I}}&{\rm{I}}&{{\rm{I}})}\\{{\rm{ d) }}}& - & - & - \\{{\rm{ c) }}}&{\rm{I}}&{\rm{I}}&{\rm{I}}\end{aligned}\)

  • \(D\)indicates that the concentration decreases by adding the substance in the task.
  • I indicates that the concentration increases by adding the substance in the task.

Step by step solution

01

Le Chatelier Principle

  • Let us recall the Le Chatelier principle to solve this problem.
  • The Le Chatelier principle says that increasing the number of reactants or decreasing the number of products will "push" the equilibrium towards the products.
  • Similarly, increasing the number of products or decreasing the number of reactants will "push" the equilibrium towards the reactants.
02

Explanation

a) As \({{\rm{H}}^ + }\)ions from the \({\rm{HCl}}\)react with \({\rm{O}}{{\rm{H}}^ - }\)ions to form water, the concentration of \(O{H^ - }\)ions decreases.

  • As the concentration of a product decreases, the equilibrium gets pushed towards the products.
  • Hence, the concentration of \({\rm{HN}}{{\rm{O}}_2}\)increases and the concentration of \({\rm{NO}}_2^ - \)decreases.

b) By adding \({\rm{HN}}{{\rm{O}}_2},\) the concentration of \({\rm{HN}}{{\rm{O}}_2}\)increases.

  • As the product concentration increases, the equilibrium is shifted towards the reactants.
  • Therefore, the concentration of \({\rm{NO}}_2^ - \)increases and the concentration of \({\rm{O}}{{\rm{H}}^ - }\)decreases.

c) The addition of \({\rm{NaOH}}\)increases the concentration of \({\rm{O}}{{\rm{H}}^ - }\)ions as \({\rm{NaOH}}\)fully ionizes in water as it is a strong base.

\({\rm{NaOH}}(aq) \to {\rm{N}}{{\rm{a}}^ + }(aq) + O{{\rm{H}}^ - }(aq)\)

  • As the product concentration increases, the equilibrium is shifted towards the reactants Therefore, the concentration of \({\rm{NO}}_2^ - \)increases whereas the concentration of \({\rm{HN}}{{\rm{O}}_2}\)decreases.

d) When \({\rm{NaCl}}\) is added to the solution, it does not have any impact on the concentration of either of the chemical species in question.

  • \({\rm{NaCl}}\)is a salt made from a strong base and a strong acid, hence its solution is neutral and its addition does not impact the concentrations of either \({{\rm{H}}^ + }\)or \({\rm{O}}{{\rm{H}}^ - }\)ions. Therefore, all the concentrations remain the same.

e) Addition of \({\rm{KNO}}\)would absolutely have no impact on the concentration of any of the chemical species in question.

  • Let us assume \({\rm{KNO}}\)as \({\rm{KN}}{{\rm{O}}_2}\)
  • \({\rm{KN}}{{\rm{O}}_2}\)ionizes in water and \({{\rm{K}}^ + }\)and \({\rm{NO}}_2^ - \)ions form.
  • The addition of \({\rm{KN}}{{\rm{O}}_2}\) will increase the concentration of \({\rm{NO}}_2^ - .\)
  • As the reactant concentration increases , the equilibrium is shifted towards the products.
  • Therefore, the concentrations of \({\rm{O}}{{\rm{H}}^ - }\) and \({\rm{HN}}{{\rm{O}}_2}\) increases as well.

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Most popular questions from this chapter

What are the\(pH\;and\;pOH\) of a solution of \(2.0\)\(M\)\(HCl\), which ionizes completely?

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) \(H{S^ - }\)

Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:

\({\rm{\;(a)\;NO}}_2^ - + {{\rm{H}}_2}{\rm{O}} \to {\rm{HN}}{{\rm{O}}_2} + {\rm{O}}{{\rm{H}}^ - }\).

\({\rm{\;(b)\;HBr}} + {{\rm{H}}_2}{\rm{O}} \to {{\rm{H}}_3}{{\rm{O}}^ + } + {\rm{B}}{{\rm{r}}^ - }\)

\({\rm{\;(c)\;H}}{{\rm{S}}^ - } + {{\rm{H}}_2}{\rm{O}} \to {{\rm{H}}_2}{\rm{S}} + {\rm{O}}{{\rm{H}}^ - }\)

\({\rm{\;(d)\;}}{{\rm{H}}_2}{\rm{PO}}_4^ - + {\rm{O}}{{\rm{H}}^ - } \to {\rm{HP}}{{\rm{O}}_4}^{2 - } + {{\rm{H}}_2}{\rm{O}}\)

\({\rm{\;(e)\;}}{{\rm{H}}_2}{\rm{PO}}_4^ - + {\rm{HCl}} \to {{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4} + {\rm{C}}{{\rm{l}}^ - }\)

\({\rm{\;(f)\;}}{\left( {{\rm{Fe}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_5}({\rm{OH}})} \right)^{2 + }} + {\left( {{\rm{Al}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_6}} \right)^{3 + }} \to {\left( {{\rm{Fe}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_6}} \right)^{3 + }} + {\left( {{\rm{Al}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_5}({\rm{OH}})} \right)^{2 + }}\)

\({\rm{\;(g)\;C}}{{\rm{H}}_3}{\rm{OH}} + {{\rm{H}}^ - } \to {\rm{C}}{{\rm{H}}_3}{{\rm{O}}^ - } + {{\rm{H}}_2}\)

Propionic acid, \({C_2}{H_5}C{O_2}H\left( {{K_a} = 1.34 \times 1{0^{ - 5}}} \right)\), is used in the manufacture of calcium propionate, a food preservative. What is the hydronium ion concentration in a \(0.698 M\) solution of \({C_2}{H_5}C{O_2}H ? \)

Explain why the ionization constant, Ka, for HI is larger than the ionization constant for HF
See all solutions

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