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If 1 mole of \(\mathrm{HCl}\) and 1 mole of \(\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) are mixed in enough water to make one liter of solution, what will be the concentrations of the species in the final equilibrium? $$ \mathrm{K}_{\text {diss }}=1.8 \times 10^{-5} \text { for } \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} $$

Short Answer

Expert verified
The final equilibrium concentrations for the given mixture are approximately: - [CH鈧僀OOH] 鈮 1.34 脳 10鈦宦 M - [C鈧侶鈧僌鈧傗伝] 鈮 0.9987 M - [H鈦篯 鈮 0.9987 M

Step by step solution

01

Determine the reactions involving HCl and NaC鈧侶鈧僌鈧

When \(\mathrm{HCl}\), a strong acid, is dissolved in water, it dissociates completely into H鈦 and Cl鈦 ions: \[ \mathrm{HCl} \rightarrow \mathrm{H}^{+} + \mathrm{Cl}^- \] Upon dissolution of \(\mathrm{NaC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\), Na鈦 and C鈧侶鈧僌鈧傗伝 ions form. The C鈧侶鈧僌鈧傗伝 ion can accept a proton (H鈦) and becomes a weak acid (acetic acid, CH鈧僀OOH) as follows: \[ \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^- + \mathrm{H}^{+} \rightleftharpoons \mathrm{CH}_3\mathrm{COOH} \]
02

Write down the equilibrium constant expression

The equilibrium constant expression is given by: \[ K_{\text{diss}} = \frac{[\mathrm{CH}_{3}\mathrm{COOH}]}{[\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^-][\mathrm{H}^{+}]} \] With the given value of K\(_\text{diss}\): \[ 1.8 \times 10^{-5} = \frac{[\mathrm{CH}_{3}\mathrm{COOH}]}{[\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^-][\mathrm{H}^{+}]} \]
03

Determine the initial concentrations and set up a table

Initially, 1 mole each of \(\mathrm{HCl}\) and \(\mathrm{NaC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) are mixed in enough water to make one liter of solution. Therefore, initial concentrations are as follows: [CH鈧僀OOH] = 0 M [C鈧侶鈧僌鈧傗伝] = 1 M [H鈦篯 = 1 M The reaction between C鈧侶鈧僌鈧傗伝 and H鈦 consumes H鈦 while forming CH鈧僀OOH. Setting up a change table, we have: CH3COOH C鈧侶鈧僌鈧傗伝 H鈦 Initial 0 1 1 Change +x -x -x Equilibrium x 1-x 1-x
04

Substitute the concentrations into the equilibrium constant expression and solve for x

We now substitute the equilibrium concentrations into the equilibrium constant expression: \[ 1.8 \times 10^{-5} = \frac{x}{(1-x)(1-x)} \] Solving for x, we obtain: \[ x \approx 1.34 \times 10^{-3} \]
05

Determine the final concentrations

Now that we know the value of x, we can determine the equilibrium concentrations: [CH鈧僀OOH] = x 鈮 1.34 脳 10鈦宦 M [C鈧侶鈧僌鈧傗伝] = 1 - x 鈮 0.9987 M [H鈦篯 = 1 - x 鈮 0.9987 M The final equilibrium concentrations are approximately: - [CH鈧僀OOH] 鈮 1.34 脳 10鈦宦 M - [C鈧侶鈧僌鈧傗伝] 鈮 0.9987 M - [H鈦篯 鈮 0.9987 M

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Key Concepts

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

Acid-Base Reactions
Acid-base reactions are fundamental processes in chemistry where an acid donates a proton (H+) to a base. A strong acid, such as hydrochloric acid (HCl), will dissociate completely in water, releasing H+ and its corresponding anion, in this case, Cl-. In contrast, a weak acid, like acetic acid formed when C2H3O2- (acetate ion) accepts an H+ ion, only partially dissociates in water. This partial dissociation is described by an equilibrium wherein the acetate ion acts as a base, and water, the solvent, can provide the H+ ion.

Understanding acid-base chemistry is essential for predicting the behavior of acids and bases in solution and for calculating the pH of a solution.
Equilibrium Constant Expression
The equilibrium constant expression quantitatively relates the concentrations of reactants and products at equilibrium in a reversible chemical reaction. The expression is defined as the product of the concentrations of the products raised to the power of their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to the power of their stoichiometric coefficients. For dissociation reactions, it is represented as Kdiss. The value of Kdiss helps us understand to what extent a reaction will proceed before reaching equilibrium.

In the given exercise, the equilibrium constant expression was to calculate the extent of dissociation of acetic acid in solution. This understanding aids in making critical interpretations about the nature of weak acids and their equilibrium behaviors in water鈥攖hat is, how readily they give up or take on protons.
Stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions. It is a quantitative assessment that uses the balanced chemical equation to relate the quantities of substances involved in a reaction. In our context, stoichiometry enables us to understand the relationship between the amounts of acetate ion and acetic acid at equilibrium.

When setting up a stoichiometric calculation, the coefficients of the balanced chemical equation provide the ratio in which reactants convert into products. By applying the stoichiometric principles to equilibrium situations, we can construct a 'change table' that reflects the initial concentrations, changes in those concentrations, and the final concentrations when equilibrium is established.
Dissociation of Weak Acids
The dissociation of weak acids, such as acetic acid (CH3COOH), in an aqueous solution is an equilibrium process where the weak acid partially ionizes into its constituent ions. Dissociation occurs to a much lesser extent than with strong acids.

For instance, the sodium acetate (NaC2H3O2) mentioned in the exercise dissociates in water to give Na+ and C2H3O2- ions. The C2H3O2- ion can then react reversibly with H+ to form CH3COOH, creating a dynamic equilibrium where the rates of the forward and reverse reactions are equal. This delicate balance is influenced by the acid's dissociation constant, which provides insight into the acid's strength and behavior in solution.
Concentration Calculations
Concentration calculations involve determining the molarity (M) of species in a solution, which is expressed as moles of solute per liter of solution (mol/L). When dealing with chemical equilibria, concentration calculations are critical for quantifying the species present at equilibrium.

The method involves determining the initial concentrations, establishing the expression for the equilibrium constant, and using algebra to solve for unknowns. Upon obtaining the equilibrium concentrations (as the variable 'x' in the provided exercise), you then apply these values to find the concentrations of all species at equilibrium, thereby completing the solution to the problem. This systematic approach ensures an accurate representation of a system's chemistry at equilibrium.

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Most popular questions from this chapter

If you place \(\mathrm{HClO}_{4}, \mathrm{HNO}_{3}\) or HCl in water, you find that they are strong acids. However, they show distinct differences in acidities when dissolved in acetic acid. Such an occurrence is referred to as the leveling effect of the solvent, water: a) Explain the basis for this leveling effect by comparing acid reactions in the water solvent system to the acetic acid solvent system, b) Discuss the leveling effect in terms of basicities instead of acidities.

Sulfuryl chloride decomposes according to the equation, $$ \mathrm{SO}_{2} \mathrm{CI}_{2}(\mathrm{~g}) \rightleftarrows \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{CI}_{2}(\mathrm{~g}) $$ Given that at \(300^{\circ} \mathrm{K}\) and 1 atm pressure, the degree of dissociation is 121, find the equilibrium constant, \(K_{p}\). What will be the degree of dissociation when pressure \(=10 \mathrm{~atm} ?\)

Find the hydronium ion concentration of \(1 \mathrm{M} \mathrm{HOAC}\) (acetic acid) solution. Assume \(\mathrm{k}_{\mathrm{a}}=1.75 \times 10^{-5}\) for acetic acid.

Given the equilibrium: \(\mathrm{CICH}_{2} \mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftarrows \mathrm{H}_{3} \mathrm{O}^{+}+\) \(\mathrm{ClCH}_{2} \mathrm{COO}^{-}\) exists at \(25^{\circ} \mathrm{C} . \mathrm{K}_{\mathrm{a}}=1.35 \times 10^{-3}\). (a) Determine the \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration for a \(0.1 \mathrm{M}\) solution of monochloroacetic acid \(\left(\mathrm{CICH}_{2} \mathrm{COOH}\right)\) in water \(\left(\mathrm{H}_{2} \mathrm{O}\right)\). (b) Can one make the assumption that the dissociated acid is negligible with respect to the undissociated acid? (c) Calculate to what degree this solution is more acidic than 1.0 M acetic acid \(\left(\mathrm{K}_{\mathrm{a}}=1.8 \times 10^{-5}\right.\) at \(\left.25^{\circ} \mathrm{C}\right)\).

The dissociation sequence of the polyprotic acid \(\mathrm{H}_{3} \mathrm{PO}_{4}\) shows three Bronsted-Lowry acids. Rank them in order of decreasing strengths.

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