For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be \(100 \%\) Is \(K_{c} > 1, < 1,\) or \(\approx 1\) for a titration reaction?

Short Answer

Expert verified
\(K_{c} > 1\text{ for a titration reaction, as the reaction must go to completion with a yield essentially of }100 \text{%}\text{.}\)

Step by step solution

01

Understanding Titration

Titration is a technique in which a solution of known concentration (titrant) is used to determine the concentration of an unknown solution. The point at which the reaction is complete is known as the equivalence point, ideally, the amount of titrant added equals the amount of substance present in the unknown solution.
02

Analyzing the Reaction Yield

For titration, the reaction between the titrant and the substance in the unknown solution must yield to completion, which means the products are formed quantitatively with essentially no reverse reaction and thus a yield nearly 100%.
03

Relating Yield to the Equilibrium Constant (\(K_{c}\))

The yield of a reaction is related to the equilibrium constant (\(K_{c}\)). A reaction that goes to completion (yield of 100%) indicates that at equilibrium, the concentration of the products is much greater than the concentration of the reactants. For such a reaction, the equilibrium constant (\(K_{c}\)) would be much greater than 1.

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Key Concepts

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

Understanding the Equivalence Point in Titration
Titration is a laboratory method of quantitative chemical analysis used to determine the concentration of an identified analyte. During this process, a reaction is set up between a known volume of a solution with an unknown concentration, and a known concentration of a standard solution—called the titrant. The moment in a titration when the amount of titrant added is just enough to completely react with the analyte is called the equivalence point. It’s the point of perfect stoichiometric balance, where the moles of titrant equal the moles of analyte.

Understanding this critical point allows you to quantify the reaction. To visibly recognize when the equivalence point is achieved, indicators are often used. These change color at or near the equivalence point. The accuracy of titration depends on the precise identification of the equivalence point, which is fundamental for calculating the concentration of the unknown solution accurately.
The Importance of Reaction Yield in Titration
In titration, the aim is to achieve a reaction yield that is as close to 100% as possible. A high yield ensures that when the equivalence point is reached, nearly all of the analyte has reacted with the titrant. In real-world laboratory conditions, attaining an exact 100% yield is challenging due to potential side reactions and other variables that might influence the process. However, for a titration reaction to be considered effective, the reaction yield should be high enough that it does not materially affect the accuracy of the analysis.

High yields in titration reactions suggest that the reaction goes to completion and that any reverse reaction is negligible. Assessing the reaction yield can often indicate the purity and efficiency of the reactants and help to validate the reliability of the titration results. Deviations from the expected yield might signal issues with the process or reactants that could require further investigation to ensure accurate outcomes.
Relationship Between Equilibrium Constant and Titration
The concept of an equilibrium constant (\(K_c\)), while not directly measured during a titration, relates to the fundamental understanding of the reaction's dynamics. The equilibrium constant is a number that expresses the ratio of the concentration of the products to the reactants at equilibrium. In a titration reaction aiming for a high yield, we expect the reaction to move almost entirely towards the product side, leaving very little to no reactants. This implies that the value of \(K_c\) is significantly greater than 1.

For successful titrations, optimal conditions favor the formation of the products to an extent where reactants are no longer present in significant amounts when equilibrium is established. The larger the \(K_c\), the further the reaction goes to completion. In some cases, depending on the reactants and conditions, titration can even be used to determine the \(K_c\) of a reaction indirectly by analyzing the molar concentrations of reactants and products at the equivalence point.

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Most popular questions from this chapter

Acetic acid is a weak acid that reacts with water according to this equation: \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{H}_{2} \mathrm{O}(a q) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(a q)\) Will any of the following increase the percent of acetic acid that reacts and produces \(\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\) ion? (a) Addition of HCl (b) Addition of NaOH (c) Addition of \(\mathrm{NaCH}_{3} \mathrm{CO}_{2}\)

Calculate the value of the equilibrium constant \(K_{P}\) for the reaction \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\) from these equilibrium pressures: \(\mathrm{NO}, 0.050 \mathrm{atm} ; \mathrm{Cl}_{2}, 0.30\) atm; \(\mathrm{NOCl}, 1.2 \mathrm{atm}\)

How can the pressure of water vapor be increased in the following equilibrium? \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H=41 \mathrm{kJ}\)

Calculate the equilibrium constant at \(25^{\circ} \mathrm{C}\) for each of the following reactions from the value of \(\Delta G^{\circ}\) given. (a) \(\mathrm{I}_{2}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{ICl}(g) \quad \Delta G^{\circ}=-10.88 \mathrm{kJ}\) (b) \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow 2 \mathrm{HI}(g) \quad \Delta G^{\circ}=3.4 \mathrm{kJ}\) (c) \(\mathrm{CS}_{2}(g)+3 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)+\mathrm{S}_{2} \mathrm{Cl}_{2}(g) \quad \Delta G^{\circ}=-39 \mathrm{kJ}\) (d) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) \quad \Delta G^{\circ}=-141.82 \mathrm{kJ}\) (e) \(\mathrm{CS}_{2}(g) \longrightarrow \mathrm{CS}_{2}(l) \quad \Delta G^{\circ}=-1.88 \mathrm{kJ}\)

Consider the equilibrium \(4 \mathrm{NO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)+7 \mathrm{O}_{2}(g)\) (a) What is the expression for the equilibrium constant \(\left(K_{c}\right)\) of the reaction? (b) How must the concentration of \(\mathrm{NH}_{3}\) change to reach equilibrium if the reaction quotient is less than the equilibrium constant? (c) If the reaction were at equilibrium, how would an increase in the volume of the reaction vessel affect the pressure of \(\mathrm{NO}_{2} ?\) (d) If the change in the pressure of \(\mathrm{NO}_{2}\) is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of \(\mathrm{O}_{2}\) change?

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