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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

Calculate the concentration of each species present in a \(0.010M\) solution of phthalic acid, \({C_6}{H_4}{\left( {C{O_2}H} \right)_2}\).

\(\begin{array}{*{20}{c}}{{C_6}{H_4}{{\left( {C{O_2}H} \right)}_2}(aq) + {H_2}O(l) \rightleftharpoons {H_3}{O^ + }(aq) + {C_6}{H_4}\left( {C{O_2}H} \right){{\left( {C{O_2}} \right)}^ - }(aq)}&{{K_a} = 1.1 \times 1{0^{ - 3}}} \\ {{C_6}{H_4}\left( {C{O_2}H} \right)\left( {C{O_2}} \right)(aq) + {H_2}O(l) \rightleftharpoons {H_3}{O^ + }(aq) + {C_6}{H_4}{{\left( {C{O_2}} \right)}_2}^{2 - }(aq)}&{{K_a} = 3.9 \times 1{0^{ - 6}}} \end{array}\)

Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution.

Salicylic acid, \(HO{C_6}{H_4}C{O_2}H\), and its derivatives have been used as pain relievers for a long time. Salicylic acid occurs in small amounts in the leaves, bark, and roots of some vegetation (most notably historically in the bark of the willow tree). Extracts of these plants have been used as medications for centuries. The acid was first isolated in the laboratory in 1838.

(a) Both functional groups of salicylic acid ionize in water, with \({K_a} = 1.0 \times 1{0^{ - 3}}\)for the \( - C{O_2}H\) group and \(4.2 \times 1{0^{ - 13}}\) for the \( - OH \) group. What is the pH of a saturated solution of the acid (solubility \( = 1.8\;g/L)\).

(b) Aspirin was discovered as a result of efforts to produce a derivative of salicylic acid that would not be irritating to the stomach lining. Aspirin is acetylsalicylic acid, \(C{H_3}C{O_2}{C_6}{H_4}C{O_2}H\). The \(C{O_2}H\)functional group is still present, but its acidity is reduced, \({K_a} = 3.0 \times 1{0^{ - 4}}\). What is the pH of a solution of aspirin with the same concentration as a saturated solution of salicylic acid (See Part a).

(c) Under some conditions, aspirin reacts with water and forms a solution of salicylic acid and acetic acid: \(C{H_3}C{O_2}{C_6}{H_4}C{O_2}H(aq) + {H_2}O(l) \to HO{C_6}{H_4}C{O_2}H(aq) + C{H_3}C{O_2}H(aq)\)

i. Which of the acids, salicylic acid or acetic acid, produces more hydronium ions in such a solution?

ii. What are the concentrations of molecules and ions in a solution produced by the hydrolysis of \(0.50\;g\)of aspirin dissolved in enough water to give \(75ml\) of solution?

What is the ionization constant at 25oC for the weak acid \(C{H_3}NH_3^ + \), the conjugate acid of the weak base \(C{H_3}N{H_2}\)\({K_b} = 4.4 \times 1{0^{ - 4}}\) ,.

White vinegar is a \(5.0 \% \) by mass solution of acetic acid in water. If the density of white vinegar is \(1.007\;g/c{m^3}\), what is the \(pH?\)
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