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(a). IF both (A) and (R) are correct and...

(a). IF both (A) and (R) are correct and (R) is the correct explanation of (A).
(b). If both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c). If (A) is correct, but (R) is incorrect.
(d). If (A) is incorrect, but (R) is correct.
(e) if both (A) and (R) are incorrect.
Assertion (A): If x " mL of " 0.2 M `H_(2)SO_(3)` solution requires 10 " mL of " 0.24 M KOH solution then x " mL of " 0.1 M `H_(2)SO_(3)` would require 20 " mL of " 0.01 M acidified `K_(2)Cr_(2)O_(7)`
solution.
Reason (R): `H_(2)SO_(3)` is dibasic acid.

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To solve the problem, we need to analyze the assertion (A) and the reason (R) provided in the question. ### Step 1: Understand Assertion (A) Assertion (A) states that if x mL of 0.2 M \( H_2SO_3 \) solution requires 10 mL of 0.24 M KOH solution, then x mL of 0.1 M \( H_2SO_3 \) would require 20 mL of 0.01 M acidified \( K_2Cr_2O_7 \) solution. ### Step 2: Analyze the Reaction with KOH 1. **Calculate the equivalence of \( H_2SO_3 \) with KOH**: - The reaction between \( H_2SO_3 \) (a dibasic acid) and KOH (a strong base) can be expressed as: \[ \text{Equivalence of } H_2SO_3 = n \times \text{Molarity} \times \text{Volume} \] - Here, \( n \) (the number of ionizable protons) for \( H_2SO_3 \) is 2 because it is dibasic. - For KOH, \( n \) is 1. - Therefore, we can write: \[ 2 \times 0.2 \times x = 1 \times 0.24 \times 10 \] - Simplifying this gives: \[ 0.4x = 2.4 \implies x = \frac{2.4}{0.4} = 6 \text{ mL} \] ### Step 3: Analyze the Reaction with \( K_2Cr_2O_7 \) 2. **Calculate the equivalence of \( H_2SO_3 \) with \( K_2Cr_2O_7 \)**: - In acidic medium, \( K_2Cr_2O_7 \) acts as an oxidizing agent. The reaction involves the oxidation of \( SO_3^{2-} \) to \( SO_4^{2-} \). - The n-factor for \( K_2Cr_2O_7 \) can be calculated based on the change in oxidation state of chromium from +6 to +3, which gives a change of 3 per chromium atom. Since there are 2 chromium atoms, the total change is 6. - Therefore, the n-factor for \( K_2Cr_2O_7 \) is 6. - The equivalence of \( K_2Cr_2O_7 \) can be expressed as: \[ \text{Equivalence of } K_2Cr_2O_7 = n \times \text{Molarity} \times \text{Volume} = 6 \times 0.01 \times 20 = 1.2 \] ### Step 4: Compare the Equivalences 3. **Equate the equivalences**: - The equivalence of \( H_2SO_3 \) calculated from the first part is \( 0.1 \times 6 \times 2 = 1.2 \). - The equivalence of \( K_2Cr_2O_7 \) is also \( 1.2 \). - Therefore, the assertion is correct. ### Step 5: Analyze Reason (R) Reason (R) states that \( H_2SO_3 \) is a dibasic acid. This is indeed true as \( H_2SO_3 \) can donate two protons. ### Step 6: Determine the Relationship - Both assertion (A) and reason (R) are correct. - However, the reason (R) does not adequately explain the assertion (A) because the reaction with \( K_2Cr_2O_7 \) is an oxidation-reduction reaction rather than an acid-base reaction. ### Conclusion The correct answer is (b): If both (A) and (R) are correct but (R) is not the correct explanation of (A). ---
To solve the problem, we need to analyze the assertion (A) and the reason (R) provided in the question. ### Step 1: Understand Assertion (A) Assertion (A) states that if x mL of 0.2 M \( H_2SO_3 \) solution requires 10 mL of 0.24 M KOH solution, then x mL of 0.1 M \( H_2SO_3 \) would require 20 mL of 0.01 M acidified \( K_2Cr_2O_7 \) solution. ### Step 2: Analyze the Reaction with KOH 1. **Calculate the equivalence of \( H_2SO_3 \) with KOH**: - The reaction between \( H_2SO_3 \) (a dibasic acid) and KOH (a strong base) can be expressed as: ...
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