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Statement-1 : 0.20 M solution of NaCN is...

Statement-1 : 0.20 M solution of NaCN is more basic than 0.20 M solution of NaF.
Statement-2 :0.20 M solution of NaCN is more basic than 0.20 M solution of `CH_3COONa`

A

Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1

B

Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1

C

Statement-1 is True, Statement-2 is False.

D

Statement-1 is False, Statement-2 is True.

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The correct Answer is:
To solve the question, we need to analyze both statements regarding the basicity of the solutions of sodium cyanide (NaCN), sodium fluoride (NaF), and sodium acetate (CH₃COONa). ### Step 1: Analyze Statement 1 - **Statement 1** claims that a 0.20 M solution of NaCN is more basic than a 0.20 M solution of NaF. - **Reasoning**: - Both NaCN and NaF are salts that dissociate in water to give Na⁺ ions and their respective anions (CN⁻ and F⁻). - The basicity of a solution is determined by the strength of the conjugate base. - CN⁻ is the conjugate base of HCN, while F⁻ is the conjugate base of HF. - The acid dissociation constant (Kₐ) of HCN is greater than that of HF (since HF is a weak acid and HCN is even weaker), which means that CN⁻ is a stronger base than F⁻. - **Conclusion**: Therefore, 0.20 M NaCN is indeed more basic than 0.20 M NaF. **Statement 1 is true.** ### Step 2: Analyze Statement 2 - **Statement 2** claims that a 0.20 M solution of NaCN is more basic than a 0.20 M solution of CH₃COONa. - **Reasoning**: - Again, both NaCN and CH₃COONa dissociate in water to give Na⁺ ions and their respective anions (CN⁻ and CH₃COO⁻). - The basicity comparison now involves CN⁻ and CH₃COO⁻. - CN⁻ is a stronger base than CH₃COO⁻ because CH₃COO⁻ is the conjugate base of acetic acid (CH₃COOH), which is a weak acid, but HCN is even weaker than acetic acid. - Thus, CN⁻ can accept protons more readily than CH₃COO⁻ can, indicating that NaCN is more basic than CH₃COONa. - **Conclusion**: Therefore, 0.20 M NaCN is more basic than 0.20 M CH₃COONa. **Statement 2 is true.** ### Step 3: Determine the Relationship Between the Statements - Both statements are true, but they refer to different comparisons. - Statement 2 does not provide an explanation for Statement 1, as they are independent comparisons of different salts. ### Final Answer - The correct option is **B**: Statement 1 is true, Statement 2 is true, but Statement 2 is not a correct explanation for Statement 1.

To solve the question, we need to analyze both statements regarding the basicity of the solutions of sodium cyanide (NaCN), sodium fluoride (NaF), and sodium acetate (CH₃COONa). ### Step 1: Analyze Statement 1 - **Statement 1** claims that a 0.20 M solution of NaCN is more basic than a 0.20 M solution of NaF. - **Reasoning**: - Both NaCN and NaF are salts that dissociate in water to give Na⁺ ions and their respective anions (CN⁻ and F⁻). - The basicity of a solution is determined by the strength of the conjugate base. - CN⁻ is the conjugate base of HCN, while F⁻ is the conjugate base of HF. ...
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