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\((\mathbf{A}): 0.33 \mathrm{M}\) solution of \(\mathrm{KCN}\) is more basic than \(0.33 \mathrm{M}\) solution of \(\mathrm{KF}\). (R): \(0.33 \mathrm{M}\) solution of \(\mathrm{KCN}\) is more basic than \(0.33 \mathrm{M}\) solution of \(\mathrm{CH}_{3} \mathrm{COOK}\).

Short Answer

Expert verified
Both (A) and (R) are true, and (R) is a correct explanation of (A).

Step by step solution

01

Understanding the Basics

First, recognize that both KCN and KF are salts that can undergo hydrolysis in aqueous solutions. These compounds come from strong bases (KOH) and weak acids (HCN and HF, respectively). Thus, both KCN and KF solutions will be basic, but to different extents.
02

Comparing Basicity

The basicity of the solutions depends on the strength of the conjugate acid of the anion from the salt. For KCN, the hydrolysis involves the CN鈦 ion, whose conjugate acid is HCN, a very weak acid. For KF, the F鈦 ion's conjugate acid is HF, a stronger acid than HCN. Because HCN is weaker, CN鈦 is a stronger base than F鈦, making the KCN solution more basic than the KF solution.
03

Evaluating the Given Assertion (A)

Analyzing assertion (A), we note that it states a 0.33 M KCN solution is more basic than a 0.33 M KF solution. This is true based on our understanding of conjugate acids and the relative strengths of HCN and HF.
04

Analyzing the Reason (R)

For reason (R), it claims that a 0.33 M KCN solution is more basic than a 0.33 M CH鈧僀OOK solution. This is also true since CH鈧僀OO鈦, the ion in CH鈧僀OOK, comes from acetic acid which is weaker than HCN. Therefore, CN鈦 is a stronger base than CH鈧僀OO鈦.

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Key Concepts

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

Basicity Comparisons
When comparing the basicity of different solutions, it's crucial to understand the role of the conjugate acids of the anions formed during the hydrolysis of salts. For example, let's compare a 0.33 M solution of potassium cyanide (KCN) and a 0.33 M solution of potassium fluoride (KF). Both KCN and KF are derived from strong bases (KOH) and weak acids, namely hydrocyanic acid (HCN) and hydrofluoric acid (HF), respectively. Here's the key: the solution's basicity is largely influenced by the strength of the conjugate acid formed from the anion of the salt. The weaker the conjugate acid, the stronger the base's tendency to attract hydrogen ions, thus making the solution more basic. HCN is a very weak acid compared to HF, thus the cyanide ion (CN鈦) is a stronger base compared to fluoride ion (F鈦). This means in a solution, KCN will result in more OH鈦 ions than KF, leading to a higher pH and therefore a more basic solution.
Conjugate Acid-Base Pairs
Understanding conjugate acid-base pairs is fundamental to grasping the concept of acidity and basicity in aqueous solutions. A conjugate acid-base pair consists of two species that can transform into each other by the gain or loss of a proton. In the context of KCN and KF, the conjugate acid-base pairs involve the conjugate acids HCN and HF. - **HCN and CN鈦**: In the KCN solution, CN鈦 acts as the base, and its conjugate acid is HCN. Because HCN is a very weak acid, CN鈦 is a strong conjugate base. - **HF and F鈦**: In the KF solution, F鈦 is the base, paired with the conjugate acid, HF. Since HF is stronger than HCN, F鈦 is a weaker base compared to CN鈦. Recognizing which species are conjugate acids or bases is essential for predicting the behavior of different salt solutions in water. The weaker the conjugate acid, the stronger its base, leading to a more basic solution.
Aqueous Solutions of Salts
Aqueous solutions of salts commonly involve hydrolysis, where the water solvent interacts with the ions of the dissolved salt, potentially affecting the solution鈥檚 pH. The salt's ability to affect the pH depends on the properties of the constituent ions, particularly focusing on the conjugate acids and bases formed. When salts like KCN dissolve in water, they dissociate into K鈦 and CN鈦. The cyanide ion (CN鈦) undergoes hydrolysis, reacting with water to produce OH鈦 ions, making the solution basic. Similarly, KF dissociates into K鈦 and F鈦 ions, with F鈦 undergoing hydrolysis, though less intensely than CN鈦, producing fewer OH鈦 ions. The process is driven by the strength of the conjugate acids. Since HCN is weaker than HF, the CN鈦 ion is a stronger base, which means the aqueous KCN solution has a higher pH than that of KF. Understanding these interactions helps explain why different salts result in solutions with varying degrees of acidity or basicity.

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Most popular questions from this chapter

Which of the following statement is correct? 1\. the \(\mathrm{pH}\) of \(1.0 \times 10^{-8} \mathrm{M}\) solution of \(\mathrm{HCl}\) is 8 2\. the conjugate base of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)is \(\mathrm{HPO}_{4}^{2-}\) 3\. autoprotolysis constant of water increases with temperature 4\. when a solution of a weak monoprotic acid is titrated against a strong base at half neutralization point, \(\mathrm{pH}=(1 / 2) \mathrm{pKa}\). a. 2,3 b. \(1,2,3\) c. 3,4 d. \(2,3,4\)

Potassium chromate is slowly added to a solution containing \(0.20 \mathrm{M} \mathrm{AgNO}_{3}\) and \(0.20 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) Describe what happens if the Ksp for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) is \(1.1 \times 10^{-12}\) and the \(\mathrm{Ksp}\) of \(\mathrm{BaCrO}_{4}\) is \(1.2 \times 10^{-10}\) a. The \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) precipitates first out of solution and then \(\mathrm{BaCrO}_{4}\) precipitates. b. The \(\mathrm{BaCrO}_{4}\) precipitates first out of solution. c. Both \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) and \(\mathrm{BaCrO}_{4}\) precipitate simultaneously out of solution. d. Neither \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) nor \(\mathrm{BaCrO}_{4}\) precipitates out of solution.

Which of the following is/are correct? a. The equilibrium constant does not depend upon pressure b. When pressure is applied on ice \(\rightleftharpoons\) water equilibrium more water will be formed c. The equilibrium constant increases when a catalyst is introduced d. Changes with temperature

What is the value for \(K_{e q}\) for the following reaction: \(\mathrm{PbCl}_{2}(\mathrm{~s}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})\), if \(\mathrm{PbCl}_{2}(\mathrm{~s})\) \(=1.50 \mathrm{~g},\left[\mathrm{~Pb}^{2+}\right]=1.6 \times 10^{-2} \mathrm{M}\) and \(\left[\mathrm{Cl}^{-}\right]=3.2 \times\) \(10^{-2} \mathrm{M}\) at equilibrium? (The molar mass of \(\mathrm{PbCl}_{2}\) (s) is \(278 \mathrm{~g} /\) mol and its density is \(5.85 \mathrm{~g} / \mathrm{cm}^{3}\) ). a. \(2.6 \times 10^{-6}\) b. \(0.8 \times 10^{-6}\) c. \(1.6 \times 10^{-5}\) d. \(6.4 \times 10^{-4}\)

Match the following: Column I \(\quad\) Column II (p) Increase of pressure A. \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{NH}_{3}(\mathrm{~g})\) favours forward reaction B. \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow\) (q) Constant pressure \(2 \mathrm{NO}(\mathrm{g})\) favours forward reactior C. \(\mathrm{NH}_{2} \mathrm{COONH}_{4}(\mathrm{~s}) \rightleftharpoons\) (r) D. \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s}) \quad(\mathrm{s})\) \(+\mathrm{CO}_{2}(\mathrm{~g})\)

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