A 0.1 M solution of a salt of a weak acid (HA) and a weak base (BOH) has a pH of 8.2. Given $K_a(HA) = 1 \times 10^{-5}$ and $K_b(BOH) = 2 \times 10^{-6}$, what is the value of $K_h$ for the salt?

## CORE INFORMATION\n✓ **Answer**: B\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Hydrolysis of salts of weak acids and weak bases\n\t**Related Topics**: pH, $K_a$, $K_b$, $K_w$, $K_h$\n\t**Prerequisites**: Understanding of acid-base equilibria, hydrolysis, and the relationship between $K_a$, $K_b$, $K_w$, and $K_h$.\n## SOLUTION PATHWAY\n**Given Information**:\n\t$C = 0.1$ M (Concentration of the salt)\n\tpH = 8.2\n\t$K_a(HA) = 1 \times 10^{-5}$\n\t$K_b(BOH) = 2 \times 10^{-6}$\n**Method & Steps**:\n\t1. Calculate $pOH$:\n\t\tReasoning: pH + pOH = 14\n\t\tFormula/Rule: $pOH = 14 - pH$\n\t2. Calculate $[OH^-]$:\n\t\tWorking: $pOH = 14 - 8.2 = 5.8$\n\t\t$[OH^-] = 10^{-pOH} = 10^{-5.8} = 1.58 \times 10^{-6}$ M\n\t\tKey Point: The hydrolysis of the salt produces $OH^-$ ions, making the solution slightly basic.\n\t3. Calculate $K_h$:\n\t\tFormula/Rule: $K_h = \frac{K_w}{K_a \times K_b} = \frac{[OH^-]^2}{C - [OH^-]}$ (Since $[OH^-]$ is small compared to C, we can approximate $C - [OH^-] \approx C$)\n\t\tWorking: $K_h = \frac{(1.58 \times 10^{-6})^2}{0.1} \approx 2.5 \times 10^{-11}$\n\t\tResult: Using the relationship $K_h = \frac{K_w}{K_aK_b}$:\n\t\t$K_h = \frac{1 \times 10^{-14}}{(1 \times 10^{-5})(2 \times 10^{-6})} = 5 \times 10^{-4}$\n\t\tVerification: The calculated $K_h$ values using the two methods are different. This is due to the approximation made in the first method. The second method, using $K_h = \frac{K_w}{K_aK_b}$, is more accurate and does not involve approximation.\n## OPTION ANALYSIS\n[A] 2 x 10⁻⁵ :x: Incorrect. This value is not obtained from the correct calculation.\n[B] 5 x 10⁻⁴ ✓: Correct. This value is obtained using the formula $K_h = \frac{K_w}{K_aK_b}$.\n[C] 1 x 10⁻³ :x: Incorrect. This value is not obtained from the correct calculation.\n[D] 2 x 10⁻⁴ :x: Incorrect. This value is not obtained from the correct calculation.\n## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1. Using the wrong formula for $K_h$: Make sure to use the correct formula $K_h = \frac{K_w}{K_aK_b}$ for salts of weak acids and weak bases.\n\t\t2. Incorrect approximation: While the approximation $C - [OH^-] \approx C$ is often valid, be mindful of its limitations, especially when $[OH^-]$ is not significantly smaller than $C$.\n\t**Quick Tips**:\n\t\t1. Remember the relationship between $K_a$, $K_b$, $K_w$, and $K_h$.\n\t\t2. Understand the concept of hydrolysis and its effect on pH.

The correct Option is "5 x 10⁻⁴"

Which of the following salts will have the highest degree of hydrolysis in a 0.1 M solution?\n(A) $NaCN$ ($K_a(HCN) = 6.2 \times 10^{-10}$) \n(B) $NH_4Cl$ ($K_b(NH_3) = 1.8 \times 10^{-5}$)\n(C) $CH_3COONa$ ($K_a(CH_3COOH) = 1.8 \times 10^{-5}$) \n(D) $KF$ ($K_a(HF) = 7.2 \times 10^{-4}$)

## CORE INFORMATION\n✓ **Answer**: A\n## CONCEPT FRAMEWORK\n **Primary Concept**: Hydrolysis of salts\n **Related Topics**: Acid-base equilibrium, $K_a$, $K_b$, $K_w$\n **Prerequisites**: Understanding of weak acids and bases, equilibrium constants.\n## SOLUTION PATHWAY\n**Given Information**:\n 0.1 M solutions of different salts\n $K_a$ values for the conjugate acids of the anions\n $K_b$ value for the conjugate base of the cation\n**Method & Steps**:\n 1. Identify the nature of each salt and its hydrolysis reaction.\n Reasoning: Salts of weak acids and strong bases undergo hydrolysis, producing basic solutions. Salts of strong acids and weak bases undergo hydrolysis, producing acidic solutions.\n 2. Calculate the hydrolysis constant ($K_h$) for each salt.\n Working: For salts of weak acids and strong bases, $K_h = \frac{K_w}{K_a}$. For salts of strong acids and weak bases, $K_h = \frac{K_w}{K_b}$. $K_w = 1.0 \times 10^{-14}$ at 25°C.\n $K_h(NaCN) = \frac{10^{-14}}{6.2 \times 10^{-10}} = 1.6 \times 10^{-5}$\n $K_h(NH_4Cl) = \frac{10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$\n $K_h(CH_3COONa) = \frac{10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$\n $K_h(KF) = \frac{10^{-14}}{7.2 \times 10^{-4}} = 1.4 \times 10^{-11}$\n 3. The salt with the largest $K_h$ will have the highest degree of hydrolysis.\n Result: $NaCN$ has the largest $K_h$ ($1.6 \times 10^{-5}$).\n Verification: The higher the $K_h$, the greater the extent of hydrolysis.\n ## OPTION ANALYSIS\n[A] ✓: $NaCN$ has the highest $K_h$ value, indicating the highest degree of hydrolysis.\n[B] :x: $NH_4Cl$ has a lower $K_h$ value compared to $NaCN$.\n[C] :x: $CH_3COONa$ has the same $K_h$ value as $NH_4Cl$, which is lower than that of $NaCN$.\n[D] :x: $KF$ has the lowest $K_h$ value among the given options.\n ## LEARNING AIDS\n **Common Mistakes**:\n 1.Confusing $K_a$ and $K_b$: Remember to use the correct equilibrium constant ($K_a$ for the conjugate acid or $K_b$ for the conjugate base) when calculating $K_h$.\n 2.Ignoring the concentration: The concentration of the salt solution affects the degree of hydrolysis, but in this case, all solutions are 0.1M so it doesn't change the relative comparison.\n **Quick Tips**:\n 1.The weaker the conjugate acid (or base), the higher the degree of hydrolysis.\n 2.The larger the $K_h$ value, the greater the extent of hydrolysis.

A solution of $NH_4CN$ ($K_a(HCN) = 6.2 \times 10^{-10}$, $K_b(NH_3) = 1.8 \times 10^{-5}$) is:\n(A) Acidic \n(B) Basic \n(C) Neutral \n(D) Cannot be determined

## CORE INFORMATION\n✓ **Answer**: B\n## CONCEPT FRAMEWORK\n **Primary Concept**: Hydrolysis of salts\n **Related Topics**: Acid-base equilibrium, $K_a$, $K_b$, $K_w$\n **Prerequisites**: Understanding of weak acids and bases, equilibrium constants\n## SOLUTION PATHWAY\n**Given Information**:\n $K_a(HCN) = 6.2 \times 10^{-10}$\n $K_b(NH_3) = 1.8 \times 10^{-5}$\n Salt: $NH_4CN$\n**Method & Steps**:\n 1. Identify the nature of the ions.\n Reasoning: $NH_4CN$ is a salt formed from a weak acid (HCN) and a weak base ($NH_3$).\n Formula/Rule: The solution's pH depends on the relative strengths of the conjugate acid ($NH_4^+$) and conjugate base ($CN^-$).\n 2. Determine the relative strengths of the conjugate acid and base.\n Working: We need to find the $K_a$ of $NH_4^+$ and the $K_b$ of $CN^-$.\n $K_a(NH_4^+) = \frac{K_w}{K_b(NH_3)} = \frac{10^{-14}}{1.8 \times 10^{-5}} \approx 5.6 \times 10^{-10}$\n $K_b(CN^-) = \frac{K_w}{K_a(HCN)} = \frac{10^{-14}}{6.2 \times 10^{-10}} \approx 1.6 \times 10^{-5}$\n Key Point: Since $K_b(CN^-) > K_a(NH_4^+)$, the solution will be basic.\n 3. Conclude the nature of the solution.\n Result: The solution is basic.\n Verification: The stronger base ($CN^-$) dominates, leading to a higher concentration of $OH^-$ ions than $H^+$ ions.\n ## OPTION ANALYSIS\nA :x: The solution is not acidic because $K_b(CN^-) > K_a(NH_4^+)$.\nB ✓: The solution is basic due to the stronger base ($CN^-$) compared to the conjugate acid ($NH_4^+$).\nC :x: The solution is not neutral; the relative strengths of the conjugate acid and base dictate basicity.\nD :x: The nature of the solution can be determined by comparing the $K_a$ and $K_b$ values.\n ## LEARNING AIDS\n **Common Mistakes**:\n 1.Incorrectly calculating $K_a$ and $K_b$ for the conjugate acid and base: Use the relationships $K_a \times K_b = K_w$ and carefully consider which $K_a$ or $K_b$ to use.\n 2.Ignoring the relative strengths of the conjugate acid and base: Simply stating that the salt is formed from a weak acid and weak base is insufficient; comparison of their strengths is crucial.\n **Quick Tips**:\n 1.If $K_b$ of the conjugate base is greater than $K_a$ of the conjugate acid, the solution will be basic. If $K_a > K_b$, the solution will be acidic.\n 2.Remember that $K_w = K_a \times K_b = 10^{-14}$ at 25°C.

NEET Chemistry Chemical Equilibrium MCQs

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Characteristic Of Equillibrium

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Le-Chatlier'S Principle

A weak base, BOH, has a $K_b$ of $1.0 imes 10^{-5}$ . A 0.1 M solution of BOH is titrated with 0.1 M HCl. At the equivalence point, the pH of the resulting solution will be closest to:

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📚 Subject:Chemistry

📑 Chapter:Chemical Equilibrium

🎯Topic:Acid-Base Theories

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :3 mins

🔄Ncert Concepts :Acid-Base Theories, Titration, Hydrolysis, Equilibrium

A 0.1 M solution of a salt of a weak acid (HA) and a weak base (BOH) has a pH of 8.2. Given $K_a(HA) = 1 imes 10^{-5}$ and $K_b(BOH) = 2 imes 10^{-6}$ , what is the value of $K_h$ for the salt?