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To increase the ionization of formic acid, \(\mathrm{HCOOH}(\mathrm{aq})\) which of the following should be added to the solution? (a) \(\mathrm{NaCl} ;\) (b) \(\mathrm{NaHCOO} ;(\mathrm{c}) \mathrm{H}_{2} \mathrm{SO}_{4} ;\) (d) \(\mathrm{NaHCO}_{3}\)

Short Answer

Expert verified
To increase the ionization of formic acid, sulphuric acid, H2SO4, should be added to the solution.

Step by step solution

01

Analyze NaCl

Sodium chloride, NaCl, is a salt which when added to the solution will increase the ionic strength of the solution. However, it will not contribute to the ionization of HCOOH, because NaCl won't interact with the acid or the base formed during the ionization.
02

Analyze NaHCOO

Sodium formate, NaHCOO, is the salt result from the ionization of formic acid. If we add NaHCOO to the solution, it will provide more formate ions, HCOO-. This will push the equilibrium towards the left, reducing the ionization of formic acid due to Le Chatelier's principle.
03

Analyze H2SO4

Sulphuric acid, H2SO4, it is a strong acid. If this is added to the solution, it can react with the formate ion, HCOO-, produced in the ionization of formic acid and convert it back to formic acid. Thus, addition of H2SO4 increases the ionization of formic acid, because the ion produced by ionization is consumed, removing it from the equilibrium and pushing the reaction towards right making more acid ionize.
04

Analyze NaHCO3

Sodium bicarbonate, NaHCO3, is a weak base. If added to the solution, it reacts with the acid but as a weak base it can't shift the equilibrium as much as a strong acid like H2SO4.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Formic Acid
Formic acid, represented by the chemical formula \(HCOOH\), is the simplest carboxylic acid. It is commonly found in various natural sources, such as the venom of ants. In chemical solutions, formic acid behaves as a weak acid. This means it doesn't completely dissociate in water; rather, it only partially ionizes. The ionization of formic acid in water results in hydrogenions \( (H^+) \) and formate ions \( (HCOO^-) \). When formic acid ionizes, the equilibrium can be affected by changes in the solution,such as the concentration of certain ions. Understanding what can increase the ionization of formic acid can help in manipulating equilibrium reactions involving weak acids.
  • Weak Acid: Partially ionizes in water.
  • Formate Ion: \( HCOO^- \), the result of ionized formic acid.
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemistry thatexplains how a system at equilibrium reacts to disturbances and stresses. According to this principle, if the conditions of a chemical equilibrium are changed (such as pressure, temperature, or concentrations), the equilibrium will shift to counteract the change and create a new equilibrium state. This principle is crucial in understanding reactions involving ionization, like with formic acid.
If in the exercise, we add sodium formate \((NaHCOO)\), the equilibrium will shift according to Le Chatelier's Principle. The addition of more formate ions means the system will attempt to reduce the disturbance by shifting the equilibrium towards forming more non-ionized formic acid, thus reducing ionization.
  • Counteraction: Shifts in reaction to changes.
  • Equilibrium Shift: System adjusts to restore balance.
Equilibrium in Chemistry
In chemistry, equilibrium refers to the state in which the forward and reverse reactions occur at the same rate, resulting in the concentrations of reactants and products remaining constant over time. Equilibrium doesn’t imply equal concentrations but indicates a balance in the reaction dynamics. When discussing the ionization of formic acid and any influences on it, equilibrium concepts are vital.
Adding a strong acid like sulfuric acid \((H_2SO_4)\) to a solution containing formic acid influences the equilibrium. The strong acid consumes products (like formate ions), thereby shifting the equilibrium to favor more ionization of formic acid to restore balance. This results in more hydrogen ions being released into the solution, increasing the ionization.
  • State of Balance: Forward and reverse reactions occur at the same rate.
  • Influences: Changes can shift equilibrium and affect concentrations.

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Most popular questions from this chapter

Using appropriate equilibrium constants but without doing detailed calculations, determine whether a solution can be simultaneously: (a) \(0.10 \mathrm{M} \mathrm{NH}_{3}\) and \(0.10 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl},\) with \(\mathrm{pH}=6.07\) (b) \(0.10 \mathrm{M} \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) and \(0.058 \mathrm{M} \mathrm{HI}\) (c) \(0.10 \mathrm{M} \mathrm{KNO}_{2}\) and \(0.25 \mathrm{M} \mathrm{KNO}_{3}\) (d) \(0.050 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) and \(0.65 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) (e) \(0.018 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) and \(0.018 \mathrm{M} \mathrm{NaC}_{6} \mathrm{H}_{5} \mathrm{COO}\) with \(\mathrm{pH}=4.20\) (f) \(0.68 \mathrm{M} \mathrm{KCl}, 0.42 \mathrm{M} \mathrm{KNO}_{3}, 1.2 \mathrm{M} \mathrm{NaCl},\) and \(0.55 \mathrm{M}\) \(\mathrm{NaCH}_{3} \mathrm{COO},\) with \(\mathrm{pH}=6.4\)

Explain whether the equivalence point of each of the following titrations should be below, above, or at pH 7: (a) \(\mathrm{NaHCO}_{3}(\text { aq) titrated with } \mathrm{NaOH}(\mathrm{aq}) ; \text { (b) } \mathrm{HCl}(\mathrm{aq})\) titrated with \(\mathrm{NH}_{3}(\mathrm{aq}) ;\) (c) KOH(aq) titrated with HI(aq).

Calculate the \(\mathrm{pH}\) of a \(0.5 \mathrm{M}\) solution of \(\mathrm{Ca}(\mathrm{HSe})_{2}\), given that \(\mathrm{H}_{2}\) Se has \(K_{\mathrm{a}_{1}}=1.3 \times 10^{-4}\) and \(K_{\mathrm{a}_{2}}=1 \times 10^{-11}\)

Calculate the pH at the points in the titration of \(25.00 \mathrm{mL}\) of \(0.160 \mathrm{M} \mathrm{HCl}\) when (a) \(10.00 \mathrm{mL}\) and \((\mathrm{b}) 15.00 \mathrm{mL}\) of 0.242 M KOH have been added.

In the titration of \(25.00 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) calculate the number of milliliters of \(0.200 \mathrm{M} \mathrm{NaOH}\) that must be added to reach a pH of (a) \(3.85,\) (b) 5.25 (c) 11.10.

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