Formic acid, \(\mathrm{HCOOH},\) is a weak electrolyte. What solute particles are present in an aqueous solution of this compound? Write the chemical equation for the ionization of \(\mathrm{HCOOH}\).

Short Answer

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In an aqueous solution of formic acid, \(\mathrm{HCOOH}\), the solute particles present are \(\mathrm{H^{+}}\), \(\mathrm{HCOO^{-}}\), and \(\mathrm{HCOOH}\). The chemical equation for the ionization of formic acid is: \[\mathrm{HCOOH_{(aq)}} \rightleftharpoons \mathrm{H^{+}_{(aq)}} + \mathrm{HCOO^{-}_{(aq)}}\]

Step by step solution

01

Identify the solute particles in an aqueous solution of formic acid

In an aqueous solution of formic acid, some of the \(\mathrm{HCOOH}\) molecules will be ionized, and some will stay as undissociated molecules. The solute particles present in the solution are the ionized ions and the undissociated molecules. In this case, the solute particles are \(\mathrm{H^{+}}\), \(\mathrm{HCOO^{-}}\), and \(\mathrm{HCOOH}\).
02

Write the chemical equation for the ionization of formic acid

For the ionization of formic acid, \(\mathrm{HCOOH}\), in water, we can write the following equilibrium chemical equation: \[\mathrm{HCOOH_{(aq)}} \rightleftharpoons \mathrm{H^{+}_{(aq)}} + \mathrm{HCOO^{-}_{(aq)}}\] In this equation, the double arrow indicates that the ionization of formic acid is reversible and in equilibrium. The left-hand side shows the formic acid molecules, and the right-hand side shows the ionized hydrogen ions and formate ions in the aqueous solution.

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Key Concepts

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

Weak Electrolyte
Understanding the behavior of formic acid begins with recognizing it as a weak electrolyte. This classification means that formic acid does not completely dissociate into its ions when dissolved in water. Unlike strong electrolytes, which break apart fully, a weak electrolyte exists in a dynamic balance between its undissociated molecules and its constituent ions once in solution.

A weak electrolyte like formic acid partially ionizes, resulting in a relatively low concentration of ions in solution, which influences the electrical conductivity of the solution. The weak electrolyte's characteristics stem from the strength of the acid or base; in the case of formic acid, it is a weak acid, and thus, behaves as a weak electrolyte.
Chemical Equation
The ionization of formic acid in water is best represented by a chemical equation, which conveys the substances involved and the process they undergo. Writing chemical equations is fundamental to understanding chemical reactions, as it provides a clear and succinct description of the reactants, products, and their states of matter.

For formic acid, the chemical equation is balanced by ensuring that the number of each type of atom on the reactant side equals that on the product side. Additionally, the chemical equation for a weak acid incorporates a double arrow, signifying the establishment of a dynamic equilibrium between the reactants and products.
Equilibrium
Equilibrium is a state in which the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products over time. When we say that formic acid ionization is an equilibrium process, we mean that its molecules are constantly both dissociating into ions and recombining to form undissociated molecules.

The concept of equilibrium is crucial because it dictates the concentrations of all species in the solution when the system is at rest. The presence of both the undissociated formic acid and its ions in solution is due to this reversible reaction reaching a point where the rates of the forward and reverse reactions are balanced.
Solute Particles in Solution
When formic acid is dissolved in water, the solution contains a mixture of solute particles. These include undissociated formic acid molecules, hydrogen ions, and formate ions. The relative amounts of these particles are governed by the degree to which the acid ionizes.

The understanding of solute particles in solution is essential for aspects like calculating the pH, understanding the solution's conductivity, and predicting the direction of certain chemical reactions. In solutions of weak electrolytes, such as formic acid, the presence of both ions and undissociated molecules further emphasizes the dynamic aspect of chemical equilibrium.

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Most popular questions from this chapter

Hard water contains \(\mathrm{Ca}^{2+}, \mathrm{Mg}^{2+},\) and \(\mathrm{Fe}^{2+},\) which interfere with the action of soap and leave an insoluble coating on the insides of containers and pipes when heated. Water softeners replace these ions with \(\mathrm{Na}^{+}\). (a) If \(1500 \mathrm{~L}\) of hard water contains \(0.020 \mathrm{M} \mathrm{Ca}^{2+}\) and \(0.0040 \mathrm{M} \mathrm{Mg}^{2+}\), how many moles of \(\mathrm{Na}^{+}\) are needed to replace these ions? (b) If the sodium is added to the water softener in the form of \(\mathrm{NaCl}\), how many grams of sodium chloride are needed?

Using the activity series (Table 4.5 ), write balanced chemical equations for the following reactions. If no reaction occurs, simply write NR. (a) Nickel metal is added to a solution of copper(II) nitrate; (b) a solution of zinc nitrate is added to a solution of magnesium sulfate; (c) hydrochloric acid is added to gold metal; (d) chromium metal is immersed in an aqueous solution of cobalt(II) chloride; (e) hydrogen gas is bubbled through a solution of silver nitrate.

Glycerol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3},\) is a substance used extensively in the manufacture of cosmetics, foodstuffs, antifreeze, and plastics. Glycerol is a water-soluble liquid with a density of \(1.2656 \mathrm{~g} / \mathrm{mL}\) at \(15^{\circ} \mathrm{C}\). Calculate the molarity of a solution of glycerol made by dissolving \(50.000 \mathrm{~mL}\) glycerol at \(15^{\circ} \mathrm{C}\) in enough water to make \(250.00 \mathrm{~mL}\) of solution.

The commercial production of nitric acid involves the following chemical reactions: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ (a) Which of these reactions are redox reactions? (b) In each redox reaction identify the element undergoing oxidation and the element undergoing reduction.

(a) How many milliliters of \(0.120 \mathrm{M} \mathrm{HCl}\) are needed to completely neutralize \(50.0 \mathrm{~mL}\) of \(0.101 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution? (b) How many milliliters of \(0.125 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) are needed to neutralize \(0.200 \mathrm{~g}\) of \(\mathrm{NaOH}\) ? (c) If \(55.8 \mathrm{~mL}\) of \(\mathrm{BaCl}_{2}\) solution is needed to precipitate all the sulfate ion in a \(752-\mathrm{mg}\) sample of \(\mathrm{Na}_{2} \mathrm{SO}_{4},\) what is the molarity of the solution? (d) If \(42.7 \mathrm{~mL}\) of \(0.208 \mathrm{M}\) HCl solution is needed to neutralize a solution of \(\mathrm{Ca}(\mathrm{OH})_{2},\) how many grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) must be in the solution?

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