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

As titration reactions are rapid and yield is almost 100%, so \({K_c} > 1.\)

Step by step solution

01

Definition of Titration reactions

The term titration reaction refers to a reaction in which one substance is entirely neutralized by another substance. This titration could be an acid-base titration, a redox titration, or something else entirely.

02

Find the reaction yield

Let's consider an acid base titration reaction as

\(\begin{aligned}{}{\rm{HCl}}(aq) + {\rm{NaOH}}(aq) \to {\rm{NaCl}}(aq) + {{\rm{H}}_2}{\rm{O}}(aq)\\{\rm{HCl}}(aq) + {\rm{NaOH}}(aq) \to {\rm{NaCl}}(aq) + {{\rm{H}}_2}{\rm{O}}(aq)\end{aligned}\)

This reaction has a 100% yield, which means that all of the starting materials have been transformed to products and there are no leftovers. So,\({K_c} > 1\).

Thus,\({K_c} > 1\)for all the titration reactions which have 100% yield.

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Most popular questions from this chapter

Question: The hydrolysis of the sugar sucrose to the sugars glucose and fructose follows a first-order rate equation for the disappearance of sucrose.

C12 H22 O11(aq) + H2O(l)⟶C6 H12 O6 (aq) + C6 H12 O6 (aq)

Rate = k[C12H22O11]

In neutral solution, k = 2.1 × 10−11/s at 27 °C. (As indicated by the rate constant, this is a very slow reaction. In the human body, the rate of this reaction is sped up by a type of catalyst called an enzyme.) (Note: That is not a mistake in the equation—the products of the reaction, glucose and fructose, have the same molecular formulas, C6H12O6, but differ in the arrangement of the atoms in their molecules). The equilibrium constant for the reaction is 1.36 × 105 at 27 °C. What are the concentrations of glucose, fructose, and sucrose after a 0.150 M aqueous solution of sucrose has reached equilibrium? Remember that the activity of a solvent (the effective concentration) is 1.

Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas.

(a) Write the expression for the equilibrium constant \(\left( {{K_c}} \right.)\)for the reversible reaction

\(F{e_2}{O_3}(s) + 3{H_2}(g) \rightleftharpoons 2Fe(s) + 3{H_2}O(g)\) \(\Delta H = 98.7kJ\)

(b) What will happen to the concentration of each reactant and product at equilibrium if more \(Fe\)is added?

(c) What will happen to the concentration of each reactant and product at equilibrium if \({H_2}O\) is removed?

(d) What will happen to the concentration of each reactant and product at equilibrium if \({H_2}\) is added?

(e) What will happen to the concentration of each reactant and product at equilibrium if the pressure on the system is increased by reducing the volume of the reaction vessel?

(f) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?

How can the pressure of water vapor are increased in the following equilibrium?

\({H_2}O(l) \rightleftharpoons {H_2}O(g)\) \(\Delta H = 41kJ\)

Question: A reaction is represented by this equation: \({K_c} = 1 \times 1{0^3}\)

(a) Write the mathematical expression for the equilibrium constant.

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